Deck 20: Principles of Chemical Reactivity: Electron Transfer Reactions

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Question
Which of the following statements concerning voltaic cells is not true?

A) A salt bridge allows cations and anions to move between half-cells.
B) Electrons flow from a cathode to an anode in the external circuit.
C) Oxidation occurs at an anode.
D) A voltaic cell can be used as a source of energy.
E) A voltaic cell consists of two half-cells.
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Question
Which of the following statements is true concerning the voltaic cell shown below? <strong>Which of the following statements is true concerning the voltaic cell shown below?  </strong> A) The Zn anode mass decreases as the cell discharges. B) The Zn cathode mass increases as the cell discharges. C) The Zn cathode mass decreases as the cell discharges. D) The Zn anode mass increases as the cell discharges. E) The mass of the Zn electrode neither increases nor decreases as the cell discharges. <div style=padding-top: 35px>

A) The Zn anode mass decreases as the cell discharges.
B) The Zn cathode mass increases as the cell discharges.
C) The Zn cathode mass decreases as the cell discharges.
D) The Zn anode mass increases as the cell discharges.
E) The mass of the Zn electrode neither increases nor decreases as the cell discharges.
Question
Balance the following half-reaction occurring in an acidic solution. NO3-(aq)→ NO(aq)

A) NO3-(aq)+ 4 H+(aq)+ 3 e− → NO(g)+ 2 H2O(l)
B) NO3-(aq)+ 2 H2O(l)+ 3 e− → NO(g)+ 4 H+(aq)
C) NO3-(aq)+ 4 H+(aq)→ NO(g)+ 2 H2O(l)+ 3 e−
D) NO3-(aq)+ 3 e− → NO(g)+ 4 H+(aq)+ 2 H2O(l)
E) NO3-(aq)+ 4 H+(aq)→ NO(g)+ 2 H2O(l)
Question
Write a balanced chemical equation for the oxidation of solid cadmium by concentrated nitric acid,producing nitrogen dioxide gas and Cd2+(aq)ion.

A) HNO3(aq)+ Cd(s)→ Cd2+(aq)+ NO2(g)+ OH-(aq)
B) 2 HNO3(aq)+ Cd(s)→ Cd2+(aq)+ 2 NO2(g)+ 2 OH-(aq)
C) HNO3(aq)+ Cd(s)+ H+(aq)→ Cd2+(aq)+ NO2(g)+ H2O( <strong>Write a balanced chemical equation for the oxidation of solid cadmium by concentrated nitric acid,producing nitrogen dioxide gas and Cd<sup>2+</sup>(aq)ion.</strong> A) HNO<sub>3</sub>(aq)+ Cd(s)→ Cd<sup>2+</sup>(aq)+ NO<sub>2</sub>(g)+ OH<sup>-</sup>(aq) B) 2 HNO<sub>3</sub>(aq)+ Cd(s)→ Cd<sup>2+</sup>(aq)+ 2 NO<sub>2</sub>(g)+ 2 OH<sup>-</sup>(aq) C) HNO<sub>3</sub>(aq)+ Cd(s)+ H<sup>+</sup>(aq)→ Cd<sup>2+</sup>(aq)+ NO<sub>2</sub>(g)+ H<sub>2</sub>O(   ) D) 4 HNO<sub>3</sub>(aq)+ Cd(s)→ Cd<sup>2+</sup>(aq)+ 2 NO<sub>2</sub>(g)+ 2 H<sub>2</sub>O(   )+ 2 NO<sub>3</sub><sup>-</sup>(aq) E) HNO<sub>3</sub>(aq)+ Cd(s)→ Cd<sup>2+</sup>(aq)+ NO<sub>2</sub>(g) <div style=padding-top: 35px> )
D) 4 HNO3(aq)+ Cd(s)→ Cd2+(aq)+ 2 NO2(g)+ 2 H2O( <strong>Write a balanced chemical equation for the oxidation of solid cadmium by concentrated nitric acid,producing nitrogen dioxide gas and Cd<sup>2+</sup>(aq)ion.</strong> A) HNO<sub>3</sub>(aq)+ Cd(s)→ Cd<sup>2+</sup>(aq)+ NO<sub>2</sub>(g)+ OH<sup>-</sup>(aq) B) 2 HNO<sub>3</sub>(aq)+ Cd(s)→ Cd<sup>2+</sup>(aq)+ 2 NO<sub>2</sub>(g)+ 2 OH<sup>-</sup>(aq) C) HNO<sub>3</sub>(aq)+ Cd(s)+ H<sup>+</sup>(aq)→ Cd<sup>2+</sup>(aq)+ NO<sub>2</sub>(g)+ H<sub>2</sub>O(   ) D) 4 HNO<sub>3</sub>(aq)+ Cd(s)→ Cd<sup>2+</sup>(aq)+ 2 NO<sub>2</sub>(g)+ 2 H<sub>2</sub>O(   )+ 2 NO<sub>3</sub><sup>-</sup>(aq) E) HNO<sub>3</sub>(aq)+ Cd(s)→ Cd<sup>2+</sup>(aq)+ NO<sub>2</sub>(g) <div style=padding-top: 35px> )+ 2 NO3-(aq)
E) HNO3(aq)+ Cd(s)→ Cd2+(aq)+ NO2(g)
Question
When the following oxidation-reduction reaction in acidic solution is balanced,what is the lowest whole-number coefficient for Na+(aq)ion? Na(s)+ Ca2+(aq)→ Na+(aq)+ Ca(s)

A) 5
B) 4
C) 1
D) 3
E) 2
Question
Write the balanced oxidation half-reaction for the following overall reaction: 2 H+(aq)+ Ca(s)→ Ca2+(aq)+ H2(g)

A) Ca(s) → 2e- + Ca2+ (aq)
B) 2 H+(aq)→ H2(g)+ 2 e-
C) H2(g)→ 2 H+(aq)+ 2 e-
D) Ca(s)+ 2 e- → Ca2+(aq)
E) 2Ca(s)→ Ca2+(aq)+ 2 e-
Question
Balance the following oxidation-reduction reaction occurring in an acidic solution. MnO4-(aq)+ Cr2+(aq)→ Mn2+(aq)+ Cr3+(aq)

A) MnO4-(aq)+ 8 H+(aq)+ 5 Cr2+(aq)→ Mn2+(aq)+ 4 H2O(l)+ 5 Cr3+(aq)
B) MnO4-(aq)+ 8 H+(aq)+ Cr2+(aq)→ Mn2+(aq)+ 4 H2O(l)+ Cr3+(aq)
C) MnO4-(aq)+ 4 H2(g)+ 5 Cr2+(aq)→ Mn2+(aq)+ 4 H2O(l)+ 5 Cr3+(aq)
D) MnO4-(aq)+ 8 H+(aq)+ 2 Cr2+(aq)→ Mn2+(aq)+ 4 H2O(l)+ 2 Cr3+(aq)
E) MnO4-(aq)+ Cr2+(aq)→ Mn2+(aq)+ 2 O2(g)+ Cr3+(aq)
Question
Balance the following half-reaction occurring in a basic solution. MnO2(s)→ Mn(OH)2(s)

A) MnO2(s)+ 2 H2O(l)+ 2e−→ Mn(OH)2(s)+ 2 OH−(aq)
B) MnO2(s)+ 2 H2O(l)+ 4e−→ Mn(OH)2(s)+ (OH)2−(aq)
C) MnO2(s)+ H22+(aq)+ 2e−→ Mn(OH)2(s)
D) MnO2(s)+ H2(g)→ Mn(OH)2(s)+ 2 e−
E) MnO2(s)+ 2 H2O(l)→ Mn(OH)2(s)+ 2 OH−(aq)
Question
Write a balanced net ionic equation for the reaction below in an acidic solution. Cr2O72-(aq)+ Ni(s)→ Cr3+(aq)+ Ni2+(aq)

A) Cr2O72-(aq)+ 3 Ni(s)+ 14 H+(aq)→ 2 Cr3+(aq)+ 3 Ni2+(aq)+ 7 H2O( <strong>Write a balanced net ionic equation for the reaction below in an acidic solution. Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ Ni(s)→ Cr<sup>3+</sup>(aq)+ Ni<sup>2+</sup>(aq)</strong> A) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ 3 Ni(s)+ 14 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ 3 Ni<sup>2+</sup>(aq)+ 7 H<sub>2</sub>O(   ) B) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ Ni(s)+ 14 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ Ni<sup>2+</sup>(aq)+ 7 H<sub>2</sub>O(   ) C) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ 3 Ni(s)→ 2 Cr<sup>3+</sup>(aq)+ 3 Ni<sup>2+</sup>(aq)+ O<sup>2-</sup>(aq) D) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ Ni(s)+ 14 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ Ni<sup>2+</sup>(aq)+ 7 H<sub>2</sub>O(   ) E) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ 3 Ni(s)+ 7 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ 3 Ni<sup>2+</sup>(aq)+ 7 OH<sup>-</sup>(aq) <div style=padding-top: 35px> )
B) Cr2O72-(aq)+ Ni(s)+ 14 H+(aq)→ 2 Cr3+(aq)+ Ni2+(aq)+ 7 H2O( <strong>Write a balanced net ionic equation for the reaction below in an acidic solution. Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ Ni(s)→ Cr<sup>3+</sup>(aq)+ Ni<sup>2+</sup>(aq)</strong> A) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ 3 Ni(s)+ 14 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ 3 Ni<sup>2+</sup>(aq)+ 7 H<sub>2</sub>O(   ) B) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ Ni(s)+ 14 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ Ni<sup>2+</sup>(aq)+ 7 H<sub>2</sub>O(   ) C) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ 3 Ni(s)→ 2 Cr<sup>3+</sup>(aq)+ 3 Ni<sup>2+</sup>(aq)+ O<sup>2-</sup>(aq) D) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ Ni(s)+ 14 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ Ni<sup>2+</sup>(aq)+ 7 H<sub>2</sub>O(   ) E) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ 3 Ni(s)+ 7 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ 3 Ni<sup>2+</sup>(aq)+ 7 OH<sup>-</sup>(aq) <div style=padding-top: 35px> )
C) Cr2O72-(aq)+ 3 Ni(s)→ 2 Cr3+(aq)+ 3 Ni2+(aq)+ O2-(aq)
D) Cr2O72-(aq)+ Ni(s)+ 14 H+(aq)→ 2 Cr3+(aq)+ Ni2+(aq)+ 7 H2O( <strong>Write a balanced net ionic equation for the reaction below in an acidic solution. Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ Ni(s)→ Cr<sup>3+</sup>(aq)+ Ni<sup>2+</sup>(aq)</strong> A) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ 3 Ni(s)+ 14 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ 3 Ni<sup>2+</sup>(aq)+ 7 H<sub>2</sub>O(   ) B) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ Ni(s)+ 14 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ Ni<sup>2+</sup>(aq)+ 7 H<sub>2</sub>O(   ) C) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ 3 Ni(s)→ 2 Cr<sup>3+</sup>(aq)+ 3 Ni<sup>2+</sup>(aq)+ O<sup>2-</sup>(aq) D) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ Ni(s)+ 14 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ Ni<sup>2+</sup>(aq)+ 7 H<sub>2</sub>O(   ) E) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ 3 Ni(s)+ 7 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ 3 Ni<sup>2+</sup>(aq)+ 7 OH<sup>-</sup>(aq) <div style=padding-top: 35px> )
E) Cr2O72-(aq)+ 3 Ni(s)+ 7 H+(aq)→ 2 Cr3+(aq)+ 3 Ni2+(aq)+ 7 OH-(aq)
Question
Write a balanced half-reaction for the reduction of hydrogen peroxide to water in an acidic solution.

A) 2 H2O2( <strong>Write a balanced half-reaction for the reduction of hydrogen peroxide to water in an acidic solution.</strong> A) 2 H<sub>2</sub>O<sub>2</sub>(   )→ 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) B) 2 H<sub>2</sub>O<sub>2</sub>(   )+ 2e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) C) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) D) H<sub>2</sub>O<sub>2</sub>(   )+ 4 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ H<sub>2</sub>(g) E) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g) <div style=padding-top: 35px> )→ 2 H2O( <strong>Write a balanced half-reaction for the reduction of hydrogen peroxide to water in an acidic solution.</strong> A) 2 H<sub>2</sub>O<sub>2</sub>(   )→ 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) B) 2 H<sub>2</sub>O<sub>2</sub>(   )+ 2e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) C) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) D) H<sub>2</sub>O<sub>2</sub>(   )+ 4 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ H<sub>2</sub>(g) E) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g) <div style=padding-top: 35px> )+ O2(g)
B) 2 H2O2( <strong>Write a balanced half-reaction for the reduction of hydrogen peroxide to water in an acidic solution.</strong> A) 2 H<sub>2</sub>O<sub>2</sub>(   )→ 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) B) 2 H<sub>2</sub>O<sub>2</sub>(   )+ 2e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) C) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) D) H<sub>2</sub>O<sub>2</sub>(   )+ 4 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ H<sub>2</sub>(g) E) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g) <div style=padding-top: 35px> )+ 2e- → 2 H2O( <strong>Write a balanced half-reaction for the reduction of hydrogen peroxide to water in an acidic solution.</strong> A) 2 H<sub>2</sub>O<sub>2</sub>(   )→ 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) B) 2 H<sub>2</sub>O<sub>2</sub>(   )+ 2e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) C) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) D) H<sub>2</sub>O<sub>2</sub>(   )+ 4 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ H<sub>2</sub>(g) E) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g) <div style=padding-top: 35px> )+ O2(g)
C) H2O2( <strong>Write a balanced half-reaction for the reduction of hydrogen peroxide to water in an acidic solution.</strong> A) 2 H<sub>2</sub>O<sub>2</sub>(   )→ 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) B) 2 H<sub>2</sub>O<sub>2</sub>(   )+ 2e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) C) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) D) H<sub>2</sub>O<sub>2</sub>(   )+ 4 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ H<sub>2</sub>(g) E) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g) <div style=padding-top: 35px> )+ 2 H+(aq)+ 2 e- → 2 H2O( <strong>Write a balanced half-reaction for the reduction of hydrogen peroxide to water in an acidic solution.</strong> A) 2 H<sub>2</sub>O<sub>2</sub>(   )→ 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) B) 2 H<sub>2</sub>O<sub>2</sub>(   )+ 2e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) C) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) D) H<sub>2</sub>O<sub>2</sub>(   )+ 4 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ H<sub>2</sub>(g) E) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g) <div style=padding-top: 35px> )
D) H2O2( <strong>Write a balanced half-reaction for the reduction of hydrogen peroxide to water in an acidic solution.</strong> A) 2 H<sub>2</sub>O<sub>2</sub>(   )→ 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) B) 2 H<sub>2</sub>O<sub>2</sub>(   )+ 2e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) C) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) D) H<sub>2</sub>O<sub>2</sub>(   )+ 4 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ H<sub>2</sub>(g) E) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g) <div style=padding-top: 35px> )+ 4 H+(aq)+ 2 e- → 2 H2O( <strong>Write a balanced half-reaction for the reduction of hydrogen peroxide to water in an acidic solution.</strong> A) 2 H<sub>2</sub>O<sub>2</sub>(   )→ 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) B) 2 H<sub>2</sub>O<sub>2</sub>(   )+ 2e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) C) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) D) H<sub>2</sub>O<sub>2</sub>(   )+ 4 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ H<sub>2</sub>(g) E) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g) <div style=padding-top: 35px> )+ H2(g)
E) H2O2( <strong>Write a balanced half-reaction for the reduction of hydrogen peroxide to water in an acidic solution.</strong> A) 2 H<sub>2</sub>O<sub>2</sub>(   )→ 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) B) 2 H<sub>2</sub>O<sub>2</sub>(   )+ 2e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) C) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) D) H<sub>2</sub>O<sub>2</sub>(   )+ 4 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ H<sub>2</sub>(g) E) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g) <div style=padding-top: 35px> )+ 2 H+(aq)+ 4 e- → 2 H2(g)+ O2(g)
Question
Which of the following reactions will require the use of an inert electrode when used in a voltaic cell?

A) Co(s)+ 2 Ag+(aq)→ 2 Ag(s)+ Co2+(aq)
B) 3 Ni(s)+ 2 Au3+(aq)→ 3 Ni2+ + 2 Au(s)
C) Cu(s)+ 2 Ag+(aq)→ Cu2+(aq)+ 2 Ag(s)
D) Zn(s)+ 2 MnO2(s)+ 2 NH4+(aq)→ Zn2+(aq)+ Mn2O3(s)+ 2 NH3(aq)+ H2O(l)
E) 3 Zn2+(aq)+ 2 Al(s)→ 3 Zn(s)+ 2 Al3+(aq)
Question
Write a balanced half-reaction for the reduction of CrO42-(aq)to Cr(OH)3(s)in a basic solution.

A) CrO42-(aq)+ 3 OH-(aq)+ 3 e- → Cr(OH)3(s)+ 2 O2(g)
B) CrO42-(aq)+ 3 H+(aq)+ 3 e- → Cr(OH)3(s)
C) CrO42-(aq)+ 3 H+(aq)→ Cr(OH)3(s)+ 2 e-
D) CrO42-(aq)+ 4 H2O( <strong>Write a balanced half-reaction for the reduction of CrO<sub>4</sub><sup>2-</sup>(aq)to Cr(OH)<sub>3</sub>(s)in a basic solution.</strong> A) CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 OH<sup>-</sup>(aq)+ 3 e<sup>-</sup> → Cr(OH)<sub>3</sub>(s)+ 2 O<sub>2</sub>(g) B) CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 H<sup>+</sup>(aq)+ 3 e<sup>-</sup> → Cr(OH)<sub>3</sub>(s) C) CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 H<sup>+</sup>(aq)→ Cr(OH)<sub>3</sub>(s)+ 2 e<sup>-</sup> D) CrO<sub>4</sub><sup>2-</sup>(aq)+ 4 H<sub>2</sub>O(   )+ 3 e<sup>-</sup> → Cr(OH)<sub>3</sub>(s)+ 5 OH<sup>-</sup>(aq) E) CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 OH<sup>-</sup>(aq)→ Cr(OH)<sub>3</sub>(s)+ 2 O<sub>2</sub>(g) <div style=padding-top: 35px> )+ 3 e- → Cr(OH)3(s)+ 5 OH-(aq)
E) CrO42-(aq)+ 3 OH-(aq)→ Cr(OH)3(s)+ 2 O2(g)
Question
Which of the following statements is true for the following reaction,assuming the given reaction proceeds in the forward direction? 3 Sn4+(aq)+ 2 Cr(s)→ 3 Sn2+(aq)+ 2 Cr3+(aq)

A) Sn4+(aq)is the reducing agent and Cr(s)is the oxidizing agent.
B) Cr(s)is the reducing agent and Sn2+(aq)is the oxidizing agent.
C) Sn4+(aq)is the reducing agent and Sn2+(aq)is the oxidizing agent.
D) Cr(s)is the reducing agent and Cr3+(aq)is the oxidizing agent.
E) Cr(s)is the reducing agent and Sn4+(aq)is the oxidizing agent.
Question
Which of the following statements is true for the following reaction,assuming the given reaction proceeds in the forward direction? Fe3+(aq)+ Co(s)→ Fe2+(aq)+ Co2+(aq)

A) Fe3+(aq)is oxidized and Co(s)is reduced.
B) Fe3+(aq)is oxidized and Co2+(aq)is reduced.
C) Co(s)is oxidized and Fe3+(aq)is reduced.
D) Co(s)is oxidized and Co2+(aq)is reduced.
E) Fe2+(aq)is oxidized and Co(s)is reduced.
Question
How many electrons are transferred in the given reaction? Ni + 2 HCl → Ni Cl2 + H2

A) 0
B) 1
C) 2
D) 3
E) 4
Question
When the given oxidation-reduction reaction in an acidic solution is balanced,what is the lowest whole-number coefficient for H+,and on which side of the balanced equation should it appear? MnO4-(aq)+ Br-(aq)→ Mn2+(aq)+ Br2(l)

A) 1; reactant side
B) 2; product side
C) 8; reactant side
D) 16; reactant side
E) 4; product side
Question
Write the balanced reduction half-reaction for the following overall reaction: 2 Fe(s)+ 3 Cl2(aq)→ 2 Fe3+(aq)+ 6 Cl-(aq)

A) Cl2(aq)+ 2 e- → 2 Cl-(aq)
B) Fe(s)+ 3 e- → Fe3+(aq)
C) Fe(s)+ Cl2(aq)→ FeCl3(aq)
D) Cl2(aq)→ 2 Cl-(aq)+ 2 e-
E) 3 Cl2(aq)+ 2 e- → 2 Cl-(aq)
Question
Write a balanced chemical equation for the following reaction in a basic solution. ClO-(aq)+ Cr(OH)3(s)→ Cl-(aq)+ CrO42-(aq)

A) 3 ClO-(aq)+ 2 Cr(OH)3(s)+ 4 OH-(aq)→ 3 Cl-(aq)+ 2 CrO42-(aq)+ 5 H2O( <strong>Write a balanced chemical equation for the following reaction in a basic solution. ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)</strong> A) 3 ClO<sup>-</sup>(aq)+ 2 Cr(OH)<sub>3</sub>(s)+ 4 OH<sup>-</sup>(aq)→ 3 Cl<sup>-</sup>(aq)+ 2 CrO<sub>4</sub><sup>2-</sup>(aq)+ 5 H<sub>2</sub>O(   ) B) ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)+ 3 OH<sup>-</sup>(aq)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 H<sub>2</sub>O(   ) C) 2 ClO<sup>-</sup>(aq)+ 3 Cr(OH)<sub>3</sub>(s)+ 3 OH<sup>-</sup>(aq)→ 2 Cl<sup>-</sup>(aq)+ 3 CrO<sub>4</sub><sup>2-</sup>(aq)+ 6 H<sub>2</sub>O(   ) D) 4 ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)+ 4 OH<sup>-</sup>(aq)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 6 H<sub>2</sub>O(   ) E) ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 H<sup>+</sup>(aq) <div style=padding-top: 35px> )
B) ClO-(aq)+ Cr(OH)3(s)+ 3 OH-(aq)→ Cl-(aq)+ CrO42-(aq)+ 3 H2O( <strong>Write a balanced chemical equation for the following reaction in a basic solution. ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)</strong> A) 3 ClO<sup>-</sup>(aq)+ 2 Cr(OH)<sub>3</sub>(s)+ 4 OH<sup>-</sup>(aq)→ 3 Cl<sup>-</sup>(aq)+ 2 CrO<sub>4</sub><sup>2-</sup>(aq)+ 5 H<sub>2</sub>O(   ) B) ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)+ 3 OH<sup>-</sup>(aq)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 H<sub>2</sub>O(   ) C) 2 ClO<sup>-</sup>(aq)+ 3 Cr(OH)<sub>3</sub>(s)+ 3 OH<sup>-</sup>(aq)→ 2 Cl<sup>-</sup>(aq)+ 3 CrO<sub>4</sub><sup>2-</sup>(aq)+ 6 H<sub>2</sub>O(   ) D) 4 ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)+ 4 OH<sup>-</sup>(aq)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 6 H<sub>2</sub>O(   ) E) ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 H<sup>+</sup>(aq) <div style=padding-top: 35px> )
C) 2 ClO-(aq)+ 3 Cr(OH)3(s)+ 3 OH-(aq)→ 2 Cl-(aq)+ 3 CrO42-(aq)+ 6 H2O( <strong>Write a balanced chemical equation for the following reaction in a basic solution. ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)</strong> A) 3 ClO<sup>-</sup>(aq)+ 2 Cr(OH)<sub>3</sub>(s)+ 4 OH<sup>-</sup>(aq)→ 3 Cl<sup>-</sup>(aq)+ 2 CrO<sub>4</sub><sup>2-</sup>(aq)+ 5 H<sub>2</sub>O(   ) B) ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)+ 3 OH<sup>-</sup>(aq)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 H<sub>2</sub>O(   ) C) 2 ClO<sup>-</sup>(aq)+ 3 Cr(OH)<sub>3</sub>(s)+ 3 OH<sup>-</sup>(aq)→ 2 Cl<sup>-</sup>(aq)+ 3 CrO<sub>4</sub><sup>2-</sup>(aq)+ 6 H<sub>2</sub>O(   ) D) 4 ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)+ 4 OH<sup>-</sup>(aq)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 6 H<sub>2</sub>O(   ) E) ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 H<sup>+</sup>(aq) <div style=padding-top: 35px> )
D) 4 ClO-(aq)+ Cr(OH)3(s)+ 4 OH-(aq)→ Cl-(aq)+ CrO42-(aq)+ 6 H2O( <strong>Write a balanced chemical equation for the following reaction in a basic solution. ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)</strong> A) 3 ClO<sup>-</sup>(aq)+ 2 Cr(OH)<sub>3</sub>(s)+ 4 OH<sup>-</sup>(aq)→ 3 Cl<sup>-</sup>(aq)+ 2 CrO<sub>4</sub><sup>2-</sup>(aq)+ 5 H<sub>2</sub>O(   ) B) ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)+ 3 OH<sup>-</sup>(aq)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 H<sub>2</sub>O(   ) C) 2 ClO<sup>-</sup>(aq)+ 3 Cr(OH)<sub>3</sub>(s)+ 3 OH<sup>-</sup>(aq)→ 2 Cl<sup>-</sup>(aq)+ 3 CrO<sub>4</sub><sup>2-</sup>(aq)+ 6 H<sub>2</sub>O(   ) D) 4 ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)+ 4 OH<sup>-</sup>(aq)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 6 H<sub>2</sub>O(   ) E) ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 H<sup>+</sup>(aq) <div style=padding-top: 35px> )
E) ClO-(aq)+ Cr(OH)3(s)→ Cl-(aq)+ CrO42-(aq)+ 3 H+(aq)
Question
In the context of the diagram given below,which of the following statements is true concerning half-cell II? <strong>In the context of the diagram given below,which of the following statements is true concerning half-cell II?  </strong> A) [Cu<sup>2+</sup>] decreases with time,and [K+] increases with time. B) [Cu<sup>2+</sup>] increases with time,and [K+] increases with time. C) [Cu<sup>2+</sup>] decreases with time,and [K+] decreases with time. D) [Cu<sup>2+</sup>] decreases with time,and [SO<sub>4</sub>2-] decreases with time. E) [Cu<sup>2+</sup>] increases with time,and [SO<sub>4</sub>2-] increases with time. <div style=padding-top: 35px>

A) [Cu2+] decreases with time,and [K+] increases with time.
B) [Cu2+] increases with time,and [K+] increases with time.
C) [Cu2+] decreases with time,and [K+] decreases with time.
D) [Cu2+] decreases with time,and [SO42-] decreases with time.
E) [Cu2+] increases with time,and [SO42-] increases with time.
Question
Which of the following is the cell notation for a voltaic cell based on the following reaction? Cu2+(aq)+ Fe(s)→ Cu(s)+ Fe2+(aq)

A) Cu(s)| Cu2+(aq)|| Fe2+(aq)| Fe(s)
B) Fe(s)|| Fe2+(aq),Cu2+(aq)| Cu(s)
C) Cu(s)|| Cu2+(aq),Fe2+(aq)|| Fe(s)
D) Cu(s)| Fe2+(aq)|| Cu2+(aq)| Fe(s)
E) Fe(s)| Fe2+(aq)|| Cu2+(aq)| Cu(s)
Question
The electrochemical reaction that powers a lead-acid storage battery is as follows: Pb(s)+ PbO2(s)+ 4 H+(aq)+ 2 SO42-(aq)→ 2 PbSO4(s)+ 2 H2O( <strong>The electrochemical reaction that powers a lead-acid storage battery is as follows: Pb(s)+ PbO<sub>2</sub>(s)+ 4 H<sup>+</sup>(aq)+ 2 SO<sub>4</sub><sup>2-</sup>(aq)→ 2 PbSO<sub>4</sub>(s)+ 2 H<sub>2</sub>O(   ) A single cell of this battery consists of a Pb electrode and a PbO<sub>2</sub> electrode,each submerged in sulfuric acid.Which of the following reactions occurs at the cathode during discharge?</strong> A) Pb(s)is reduced to PbSO<sub>4</sub>(s). B) PbO<sub>2</sub>(s)is reduced to PbSO<sub>4</sub>(s). C) PbO<sub>2</sub>(s)is oxidized to PbSO<sub>4</sub>(s). D) Pb(s)is oxidized to PbSO<sub>4</sub>(s). E) H<sup>+</sup> is oxidized to H<sub>2</sub>O(   ). <div style=padding-top: 35px> )
A single cell of this battery consists of a Pb electrode and a PbO2 electrode,each submerged in sulfuric acid.Which of the following reactions occurs at the cathode during discharge?

A) Pb(s)is reduced to PbSO4(s).
B) PbO2(s)is reduced to PbSO4(s).
C) PbO2(s)is oxidized to PbSO4(s).
D) Pb(s)is oxidized to PbSO4(s).
E) H+ is oxidized to H2O( <strong>The electrochemical reaction that powers a lead-acid storage battery is as follows: Pb(s)+ PbO<sub>2</sub>(s)+ 4 H<sup>+</sup>(aq)+ 2 SO<sub>4</sub><sup>2-</sup>(aq)→ 2 PbSO<sub>4</sub>(s)+ 2 H<sub>2</sub>O(   ) A single cell of this battery consists of a Pb electrode and a PbO<sub>2</sub> electrode,each submerged in sulfuric acid.Which of the following reactions occurs at the cathode during discharge?</strong> A) Pb(s)is reduced to PbSO<sub>4</sub>(s). B) PbO<sub>2</sub>(s)is reduced to PbSO<sub>4</sub>(s). C) PbO<sub>2</sub>(s)is oxidized to PbSO<sub>4</sub>(s). D) Pb(s)is oxidized to PbSO<sub>4</sub>(s). E) H<sup>+</sup> is oxidized to H<sub>2</sub>O(   ). <div style=padding-top: 35px> ).
Question
Given the following two half-reactions,write the overall reaction in the direction in which it is product-favored,and calculate the standard cell potential. Pb2+(aq)+ 2 e- → Pb(s)
E° = -0.126 V
Fe3+(aq)+ e- → Fe2+(s)
E° = +0.771 V

A) Pb2+(aq)+ 2 Fe2+(s)→ Pb(s)+ 2 Fe3+(aq); <strong>Given the following two half-reactions,write the overall reaction in the direction in which it is product-favored,and calculate the standard cell potential. Pb<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Pb(s) E° = -0.126 V Fe<sup>3+</sup>(aq)+ e<sup>-</sup> → Fe<sup>2+</sup>(s) E° = +0.771 V</strong> A) Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s)→ Pb(s)+ 2 Fe<sup>3+</sup>(aq);   = +0.897 V B) Pb<sup>2+</sup>(aq)+ Fe<sup>2+</sup>(s)→ Pb(s)+ Fe<sup>3+</sup>(aq);   = +0.645 V C) Pb(s)+ 2 Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s);   = +1.416 V D) Pb(s)+ 2 Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s);   = +0.897 V E) Pb(s)+ Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ Fe<sup>2+</sup>(s);   = +0.645 V <div style=padding-top: 35px> = +0.897 V
B) Pb2+(aq)+ Fe2+(s)→ Pb(s)+ Fe3+(aq); <strong>Given the following two half-reactions,write the overall reaction in the direction in which it is product-favored,and calculate the standard cell potential. Pb<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Pb(s) E° = -0.126 V Fe<sup>3+</sup>(aq)+ e<sup>-</sup> → Fe<sup>2+</sup>(s) E° = +0.771 V</strong> A) Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s)→ Pb(s)+ 2 Fe<sup>3+</sup>(aq);   = +0.897 V B) Pb<sup>2+</sup>(aq)+ Fe<sup>2+</sup>(s)→ Pb(s)+ Fe<sup>3+</sup>(aq);   = +0.645 V C) Pb(s)+ 2 Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s);   = +1.416 V D) Pb(s)+ 2 Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s);   = +0.897 V E) Pb(s)+ Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ Fe<sup>2+</sup>(s);   = +0.645 V <div style=padding-top: 35px> = +0.645 V
C) Pb(s)+ 2 Fe3+(aq)→ Pb2+(aq)+ 2 Fe2+(s); <strong>Given the following two half-reactions,write the overall reaction in the direction in which it is product-favored,and calculate the standard cell potential. Pb<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Pb(s) E° = -0.126 V Fe<sup>3+</sup>(aq)+ e<sup>-</sup> → Fe<sup>2+</sup>(s) E° = +0.771 V</strong> A) Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s)→ Pb(s)+ 2 Fe<sup>3+</sup>(aq);   = +0.897 V B) Pb<sup>2+</sup>(aq)+ Fe<sup>2+</sup>(s)→ Pb(s)+ Fe<sup>3+</sup>(aq);   = +0.645 V C) Pb(s)+ 2 Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s);   = +1.416 V D) Pb(s)+ 2 Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s);   = +0.897 V E) Pb(s)+ Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ Fe<sup>2+</sup>(s);   = +0.645 V <div style=padding-top: 35px> = +1.416 V
D) Pb(s)+ 2 Fe3+(aq)→ Pb2+(aq)+ 2 Fe2+(s); <strong>Given the following two half-reactions,write the overall reaction in the direction in which it is product-favored,and calculate the standard cell potential. Pb<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Pb(s) E° = -0.126 V Fe<sup>3+</sup>(aq)+ e<sup>-</sup> → Fe<sup>2+</sup>(s) E° = +0.771 V</strong> A) Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s)→ Pb(s)+ 2 Fe<sup>3+</sup>(aq);   = +0.897 V B) Pb<sup>2+</sup>(aq)+ Fe<sup>2+</sup>(s)→ Pb(s)+ Fe<sup>3+</sup>(aq);   = +0.645 V C) Pb(s)+ 2 Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s);   = +1.416 V D) Pb(s)+ 2 Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s);   = +0.897 V E) Pb(s)+ Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ Fe<sup>2+</sup>(s);   = +0.645 V <div style=padding-top: 35px> = +0.897 V
E) Pb(s)+ Fe3+(aq)→ Pb2+(aq)+ Fe2+(s); <strong>Given the following two half-reactions,write the overall reaction in the direction in which it is product-favored,and calculate the standard cell potential. Pb<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Pb(s) E° = -0.126 V Fe<sup>3+</sup>(aq)+ e<sup>-</sup> → Fe<sup>2+</sup>(s) E° = +0.771 V</strong> A) Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s)→ Pb(s)+ 2 Fe<sup>3+</sup>(aq);   = +0.897 V B) Pb<sup>2+</sup>(aq)+ Fe<sup>2+</sup>(s)→ Pb(s)+ Fe<sup>3+</sup>(aq);   = +0.645 V C) Pb(s)+ 2 Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s);   = +1.416 V D) Pb(s)+ 2 Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s);   = +0.897 V E) Pb(s)+ Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ Fe<sup>2+</sup>(s);   = +0.645 V <div style=padding-top: 35px> = +0.645 V
Question
Write a balanced net ionic equation for the overall reaction represented by the following cell notation. Al(s)| Al3+(aq)|| Cl2(g)| Cl-(aq)| Pt(s)

A) 2 Al(s)+ 3 Cl2(g)→ 2 Al3+(aq)+ 6 Cl-(aq)
B) Al(s)+ Al3+(aq)→ Cl-(aq)+ Cl2(g)
C) 2 Al3+(aq)+ 6 Cl-(aq)→ 2 Al(s)+ 3 Cl2(g)
D) Al(s)+ 3 Cl2(g)→ Al3+(s)+ 2 Cl-(aq)
E) Al(s)+ 2 Cl-(aq)→ Cl2(g)+ Al3+(aq)
Question
An SHE electrode has been assigned a standard reduction potential,E°,of 0.00 volts.Which of the following reactions will occur at this electrode?

A) 2 H2O( <strong>An SHE electrode has been assigned a standard reduction potential,E°,of 0.00 volts.Which of the following reactions will occur at this electrode?</strong> A) 2 H<sub>2</sub>O(   )+ 2 e<sup>-</sup> → H<sub>2</sub>(g)+ 2 OH<sup>-</sup>(aq) B) O<sub>2</sub>(g)+ 4 e<sup>-</sup> → 2 O<sup>2-</sup>(aq) C) Hg<sub>2</sub>Cl<sub>2</sub>(s)+ 2 e<sup>-</sup> → 2 Hg(   )+ 2 Cl<sup>-</sup>(aq) D) Li<sup>+</sup>(aq)+ e<sup>-</sup> → Li(s) E) 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → H<sub>2</sub>(g) <div style=padding-top: 35px> )+ 2 e- → H2(g)+ 2 OH-(aq)
B) O2(g)+ 4 e- → 2 O2-(aq)
C) Hg2Cl2(s)+ 2 e- → 2 Hg( <strong>An SHE electrode has been assigned a standard reduction potential,E°,of 0.00 volts.Which of the following reactions will occur at this electrode?</strong> A) 2 H<sub>2</sub>O(   )+ 2 e<sup>-</sup> → H<sub>2</sub>(g)+ 2 OH<sup>-</sup>(aq) B) O<sub>2</sub>(g)+ 4 e<sup>-</sup> → 2 O<sup>2-</sup>(aq) C) Hg<sub>2</sub>Cl<sub>2</sub>(s)+ 2 e<sup>-</sup> → 2 Hg(   )+ 2 Cl<sup>-</sup>(aq) D) Li<sup>+</sup>(aq)+ e<sup>-</sup> → Li(s) E) 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → H<sub>2</sub>(g) <div style=padding-top: 35px> )+ 2 Cl-(aq)
D) Li+(aq)+ e- → Li(s)
E) 2 H+(aq)+ 2 e- → H2(g)
Question
Calculate the standard cell potential ( <strong>Calculate the standard cell potential (   )for the reaction 2 Ag(s)+ Co<sup>2+</sup>(aq)→ 2 Ag<sup>+</sup>(aq)+ Co(s). The standard reduction potentials are as follows: Ag<sup>+</sup>(aq)+ e<sup>−</sup>→ Ag(s) E° = 0.8 V Co<sup>2+</sup>(aq)+2 e<sup>−</sup>→ Co(s) E° = -0.277 V</strong> A) -1.077 V B) 1.077 V C) -1.877 V D) 1.877 V E) 1.323 V <div style=padding-top: 35px> )for the reaction 2 Ag(s)+ Co2+(aq)→ 2 Ag+(aq)+ Co(s). The standard reduction potentials are as follows:
Ag+(aq)+ e→ Ag(s)
E° = 0.8 V
Co2+(aq)+2 e→ Co(s)
E° = -0.277 V

A) -1.077 V
B) 1.077 V
C) -1.877 V
D) 1.877 V
E) 1.323 V
Question
In the given electrochemical cell,which of the following is the cathode half-reaction? Zn(s)| Zn2+(aq)|| Fe3+(aq),Fe2+(aq)| Pt(s)

A) Fe3+(aq)+ e− → Fe2+(aq)
B) Fe2+(aq)+ e− → Fe3+(aq)
C) Fe2+(aq)+ Pt(s)→ Fe3+(aq)+ e−
D) Zn2+(aq)→ Zn(s)+ 2 e−
E) Zn(s)→ Zn2+(aq)+ 2 e−
Question
Calculate E°cell for the cell for the reaction 2 Ga(s)+ 3 Sn4+(aq)→ 3 Sn2+(aq)+2 Ga3+(aq). The standard reduction potentials are as follows:
Ga3+(aq)+ 3 e− → Ga(s)
E° = -0.55 V
Sn4+(aq)+ 2 e− → Sn2+(aq)
E° = 0.15V

A) 0.70 V
B) 0.40 V
C) 0.33 V
D) -0.40 V
E) −0.70 V
Question
Use the following standard reduction potentials to determine which species is the strongest reducing agent. 2 H+(aq)+ 2 e- → H2(g); 0.00 V
K+(aq)+ e- → K(s); -2.93 V
F2(g)+ 2 e- → 2 F-(aq); 2.87 V
Al3+(aq)+ 3 e- → Al(s); -1.66 V
Pb2+(aq)+ 2 e- → Pb(s); -0.13 V

A) Al3+
B) H+
C) Pb2+
D) F-
E) K
Question
Which of the following species are likely to behave as oxidizing agents? Li(s),H2(g),MnO4-(aq),and Cl-(aq)

A) Li(s)
B) MnO4-(aq)
C) H2(g)and Cl-(aq)
D) Li(s)and MnO4-(aq)
E) Cl-(aq)
Question
The unit for electromotive force,emf,is the volt.1 volt is equal to _____.

A) one joule per second
B) one coulomb per joule
C) one joule per coulomb
D) one coulomb per second
E) one calorie per second
Question
According to the cell notation below,which of the following species is undergoing reduction? Ni | Ni2+(aq)|| Mn2+(aq)| MnO2(s)| Pt(s)

A) Mn2+(aq)
B) Ni2+(aq)
C) Ni(s)
D) MnO2(s)
E) Pt(s)
Question
Use the following standard reduction potentials to determine which species is the strongest oxidizing agent. Fe2+(aq)+ 2 e- → Fe(s); E° = -0.41 V
Pt2+(aq)+ 2 e- → Pt(s); E° = 1.18 V
Cr2O72-(aq)+ 14 H+(aq)+ 6 e- → 2 Cr3+(aq)+ 7 H2O( <strong>Use the following standard reduction potentials to determine which species is the strongest oxidizing agent. Fe<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Fe(s); E° = -0.41 V Pt<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Pt(s); E° = 1.18 V Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ 14 H<sup>+</sup>(aq)+ 6 e<sup>-</sup> → 2 Cr<sup>3+</sup>(aq)+ 7 H<sub>2</sub>O(   ); E° = 1.33 V</strong> A) Fe B) Pt C) Cr<sup>3+</sup> D) Fe<sup>2+</sup> E) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup> <div style=padding-top: 35px> ); E° = 1.33 V

A) Fe
B) Pt
C) Cr3+
D) Fe2+
E) Cr2O72-
Question
Which of the following is the balanced overall reaction and standard cell potential of an electrochemical cell constructed from half-cells with the given half reactions? Pt2+(aq)+ 2 e− → Pt(s); E° = 1.180 V
Pb2+(aq)+ 2 e− → Pb(s); E° = -0.130 V

A) Pt2+(aq)+ Pb(s)→ Pt(s)+ Pb2+(aq); <strong>Which of the following is the balanced overall reaction and standard cell potential of an electrochemical cell constructed from half-cells with the given half reactions? Pt<sup>2+</sup>(aq)+ 2 e− → Pt(s); E° = 1.180 V Pb<sup>2+</sup>(aq)+ 2 e− → Pb(s); E° = -0.130 V</strong> A) Pt<sup>2+</sup>(aq)+ Pb(s)→ Pt(s)+ Pb<sup>2+</sup>(aq);   = 1.310 V B) Pt(s)+ Pb<sup>2+</sup>(aq)→ Pt<sup>2+</sup>(aq)+ Pb(s);   = −1.310 V C) Pt<sup>2+</sup>(aq)+ Pb<sup>2+</sup>(aq)→ Pt(s)+ Pb(s);   = 1.050 V D) Pt<sup>2+</sup>(aq)+ Pb(s)→ Pt(s)+ Pb<sup>2+</sup>(aq);   =0.655 V E) Pt(s)+ Pb<sup>2+</sup>(aq)→ Pt<sup>2+</sup>(aq)+ Pb(s);   = −0.655 V <div style=padding-top: 35px> = 1.310 V
B) Pt(s)+ Pb2+(aq)→ Pt2+(aq)+ Pb(s); <strong>Which of the following is the balanced overall reaction and standard cell potential of an electrochemical cell constructed from half-cells with the given half reactions? Pt<sup>2+</sup>(aq)+ 2 e− → Pt(s); E° = 1.180 V Pb<sup>2+</sup>(aq)+ 2 e− → Pb(s); E° = -0.130 V</strong> A) Pt<sup>2+</sup>(aq)+ Pb(s)→ Pt(s)+ Pb<sup>2+</sup>(aq);   = 1.310 V B) Pt(s)+ Pb<sup>2+</sup>(aq)→ Pt<sup>2+</sup>(aq)+ Pb(s);   = −1.310 V C) Pt<sup>2+</sup>(aq)+ Pb<sup>2+</sup>(aq)→ Pt(s)+ Pb(s);   = 1.050 V D) Pt<sup>2+</sup>(aq)+ Pb(s)→ Pt(s)+ Pb<sup>2+</sup>(aq);   =0.655 V E) Pt(s)+ Pb<sup>2+</sup>(aq)→ Pt<sup>2+</sup>(aq)+ Pb(s);   = −0.655 V <div style=padding-top: 35px> = −1.310 V
C) Pt2+(aq)+ Pb2+(aq)→ Pt(s)+ Pb(s); <strong>Which of the following is the balanced overall reaction and standard cell potential of an electrochemical cell constructed from half-cells with the given half reactions? Pt<sup>2+</sup>(aq)+ 2 e− → Pt(s); E° = 1.180 V Pb<sup>2+</sup>(aq)+ 2 e− → Pb(s); E° = -0.130 V</strong> A) Pt<sup>2+</sup>(aq)+ Pb(s)→ Pt(s)+ Pb<sup>2+</sup>(aq);   = 1.310 V B) Pt(s)+ Pb<sup>2+</sup>(aq)→ Pt<sup>2+</sup>(aq)+ Pb(s);   = −1.310 V C) Pt<sup>2+</sup>(aq)+ Pb<sup>2+</sup>(aq)→ Pt(s)+ Pb(s);   = 1.050 V D) Pt<sup>2+</sup>(aq)+ Pb(s)→ Pt(s)+ Pb<sup>2+</sup>(aq);   =0.655 V E) Pt(s)+ Pb<sup>2+</sup>(aq)→ Pt<sup>2+</sup>(aq)+ Pb(s);   = −0.655 V <div style=padding-top: 35px> = 1.050 V
D) Pt2+(aq)+ Pb(s)→ Pt(s)+ Pb2+(aq); <strong>Which of the following is the balanced overall reaction and standard cell potential of an electrochemical cell constructed from half-cells with the given half reactions? Pt<sup>2+</sup>(aq)+ 2 e− → Pt(s); E° = 1.180 V Pb<sup>2+</sup>(aq)+ 2 e− → Pb(s); E° = -0.130 V</strong> A) Pt<sup>2+</sup>(aq)+ Pb(s)→ Pt(s)+ Pb<sup>2+</sup>(aq);   = 1.310 V B) Pt(s)+ Pb<sup>2+</sup>(aq)→ Pt<sup>2+</sup>(aq)+ Pb(s);   = −1.310 V C) Pt<sup>2+</sup>(aq)+ Pb<sup>2+</sup>(aq)→ Pt(s)+ Pb(s);   = 1.050 V D) Pt<sup>2+</sup>(aq)+ Pb(s)→ Pt(s)+ Pb<sup>2+</sup>(aq);   =0.655 V E) Pt(s)+ Pb<sup>2+</sup>(aq)→ Pt<sup>2+</sup>(aq)+ Pb(s);   = −0.655 V <div style=padding-top: 35px> =0.655 V
E) Pt(s)+ Pb2+(aq)→ Pt2+(aq)+ Pb(s); <strong>Which of the following is the balanced overall reaction and standard cell potential of an electrochemical cell constructed from half-cells with the given half reactions? Pt<sup>2+</sup>(aq)+ 2 e− → Pt(s); E° = 1.180 V Pb<sup>2+</sup>(aq)+ 2 e− → Pb(s); E° = -0.130 V</strong> A) Pt<sup>2+</sup>(aq)+ Pb(s)→ Pt(s)+ Pb<sup>2+</sup>(aq);   = 1.310 V B) Pt(s)+ Pb<sup>2+</sup>(aq)→ Pt<sup>2+</sup>(aq)+ Pb(s);   = −1.310 V C) Pt<sup>2+</sup>(aq)+ Pb<sup>2+</sup>(aq)→ Pt(s)+ Pb(s);   = 1.050 V D) Pt<sup>2+</sup>(aq)+ Pb(s)→ Pt(s)+ Pb<sup>2+</sup>(aq);   =0.655 V E) Pt(s)+ Pb<sup>2+</sup>(aq)→ Pt<sup>2+</sup>(aq)+ Pb(s);   = −0.655 V <div style=padding-top: 35px> = −0.655 V
Question
Consider the following half-reactions. Ag+(aq)+ e- → Ag(s)
E° = +0.80 V
Cu2+(aq)+ 2 e- → Cu(s)
E° = +0.34 V
Pb2+(aq)+ 2 e- → Pb(s)
E° = -0.13 V
Fe2+(aq)+ 2 e- → Fe(s)
E° = -0.44 V
Al3+(aq)+ 3 e- → Al(s)
E° = -1.66 V
Which of the following species will oxidize lead,Pb(s)?

A) Ag+(aq)and Cu2+(aq)
B) Ag(s)and Cu(s)
C) Fe2+(aq)and Al3+(aq)
D) Fe(s)and Al(s)
E) Cu2+(aq)and Fe2+(aq)
Question
Use the following standard reduction potentials to determine which species is the best oxidizing agent. O2(g)+ 4 H+(aq)+ 4 e- → 2 H2O( <strong>Use the following standard reduction potentials to determine which species is the best oxidizing agent. O<sub>2</sub>(g)+ 4 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) E° = +1.229 V Hg<sub>2</sub><sup>2+</sup>(aq)+ 2 e<sup>-</sup> → 2 Hg(   ) E° = +0.789 V I<sub>2</sub>(s)+ 2 e<sup>-</sup> → 2 I<sup>-</sup>(aq) E° = +0.535 V</strong> A) I<sub>2</sub>(s) B) O<sub>2</sub>(g) C) I<sup>-</sup>(aq) D) Hg<sub>2</sub><sup>2+</sup>(aq) E) H<sub>2</sub>O(   ) <div style=padding-top: 35px> )
E° = +1.229 V
Hg22+(aq)+ 2 e- → 2 Hg( <strong>Use the following standard reduction potentials to determine which species is the best oxidizing agent. O<sub>2</sub>(g)+ 4 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) E° = +1.229 V Hg<sub>2</sub><sup>2+</sup>(aq)+ 2 e<sup>-</sup> → 2 Hg(   ) E° = +0.789 V I<sub>2</sub>(s)+ 2 e<sup>-</sup> → 2 I<sup>-</sup>(aq) E° = +0.535 V</strong> A) I<sub>2</sub>(s) B) O<sub>2</sub>(g) C) I<sup>-</sup>(aq) D) Hg<sub>2</sub><sup>2+</sup>(aq) E) H<sub>2</sub>O(   ) <div style=padding-top: 35px> )
E° = +0.789 V
I2(s)+ 2 e- → 2 I-(aq)
E° = +0.535 V

A) I2(s)
B) O2(g)
C) I-(aq)
D) Hg22+(aq)
E) H2O( <strong>Use the following standard reduction potentials to determine which species is the best oxidizing agent. O<sub>2</sub>(g)+ 4 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) E° = +1.229 V Hg<sub>2</sub><sup>2+</sup>(aq)+ 2 e<sup>-</sup> → 2 Hg(   ) E° = +0.789 V I<sub>2</sub>(s)+ 2 e<sup>-</sup> → 2 I<sup>-</sup>(aq) E° = +0.535 V</strong> A) I<sub>2</sub>(s) B) O<sub>2</sub>(g) C) I<sup>-</sup>(aq) D) Hg<sub>2</sub><sup>2+</sup>(aq) E) H<sub>2</sub>O(   ) <div style=padding-top: 35px> )
Question
Consider the following half-reactions. Cl2(g)+ 2 e- → 2 Cl-(aq)
E° = +1.36 V
Ag+(aq)+ e- → Ag(s)
E° = +0.80 V
Cu2+(aq)+ 2 e- → Cu(s)
E° = +0.34 V
Sn2+(aq)+ 2 e- → Sn(s)
E° = -0.14 V
Al3+(aq)+ 3 e- → Al(s)
E° = -1.66 V
Which of the following species will reduce Cu2+(aq)ion?

A) Ag(s)and Sn2+(aq)
B) Cl-(aq)and Ag(s)
C) Cl2(g)and Ag+(aq)
D) Sn(s)and Al(s)
E) Sn2+(aq)and Al3+(aq)
Question
Which of the following is the cell notation for a voltaic cell based on the following reaction? Cu2+(aq)+ Pb(s)+ SO42-(aq)→ Cu(s)+ PbSO4(s)

A) Pb(s)| PbSO4(s)|| Cu2+(aq)|| Cu(s)
B) Cu(s)| Cu2+(aq)|| SO42-(aq)| PbSO4(s)| Pb(s)
C) Cu(s)| Cu2+(aq),SO42-(aq)| PbSO4(s)| Pb(s)
D) Cu(s)| Cu2+(aq),SO42-(aq)|| PbSO4(s)|| Pb(s)
E) Pb(s)| PbSO4(s)| SO42-(aq)|| Cu2+(aq)|| Cu(s)
Question
Write a balanced net ionic equation for the overall reaction represented by the following cell notation. Cu(s)| Cu2+(aq)|| Mn2+(aq)| Mn(s)

A) 2 Cu(s)+ Mn2+(aq)→ Mn(s)+ 2 Cu2+(aq)
B) Cu(s)+ Cu2+(aq)→ Mn(s)+ Mn2+(aq)
C) Cu(s)+ Mn2+(aq)→ Mn(s)+ Cu2+(aq)
D) 2 Mn(s)+ Cu2+(aq)→ Cu(s)+ 2Mn2+(aq)
E) Mn(s)+ Cu2+(aq)→ Cu(s)+ Mn2+(aq)
Question
Which of the following is the cell notation for a cell in which the hydrogen electrode is the anode and the cathode half-reaction is Co3+(aq)+ e− → Co2+(aq)?

A) Pt(s)| H2(g)| H+(aq)|| Co3+(aq),Co2+(aq)| Pt(s)
B) Pt(s)| H2(g)| H+(aq)|| Co3+(aq),Co2+(aq)
C) Co2+(aq),Co3+(aq)|| H+(aq)| H2(g)| Pt(s)
D) Pt(s)| Co2+(aq),Co3+(aq)|| H+(aq)| H2(g)| Pt(s)
E) H2(g)| H+(aq)|| Co2+(aq),Co3+(aq)
Question
Which of the following overall chemical equations is responsible for generating electricity in fuel cells used in NASA's Space Shuttle programs?

A) Fe2O3(s)+ Al(s)→ Fe(s)+ Al2O3(s)
B) Pb(s)+ PbO2(s)+ 2 H2SO4(aq)→ 2 PbSO4(s)+ 2 H2O( <strong>Which of the following overall chemical equations is responsible for generating electricity in fuel cells used in NASA's Space Shuttle programs?</strong> A) Fe<sub>2</sub>O<sub>3</sub>(s)+ Al(s)→ Fe(s)+ Al<sub>2</sub>O<sub>3</sub>(s) B) Pb(s)+ PbO<sub>2</sub>(s)+ 2 H<sub>2</sub>SO<sub>4</sub>(aq)→ 2 PbSO<sub>4</sub>(s)+ 2 H<sub>2</sub>O(   ) C) 2 NiO(OH)(s)+ Cd(s)+ H<sub>2</sub>O(   )→ 2 Ni(OH)<sub>2</sub>(s)+ Cd(OH)<sub>2</sub>(s) D) 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g)→ 2 H<sub>2</sub>O(   ) E) N<sub>2</sub>H<sub>4</sub>(   )+ O<sub>2</sub>(g)→ N<sub>2</sub>(g)+ 2 H<sub>2</sub>O(   ) <div style=padding-top: 35px> )
C) 2 NiO(OH)(s)+ Cd(s)+ H2O( <strong>Which of the following overall chemical equations is responsible for generating electricity in fuel cells used in NASA's Space Shuttle programs?</strong> A) Fe<sub>2</sub>O<sub>3</sub>(s)+ Al(s)→ Fe(s)+ Al<sub>2</sub>O<sub>3</sub>(s) B) Pb(s)+ PbO<sub>2</sub>(s)+ 2 H<sub>2</sub>SO<sub>4</sub>(aq)→ 2 PbSO<sub>4</sub>(s)+ 2 H<sub>2</sub>O(   ) C) 2 NiO(OH)(s)+ Cd(s)+ H<sub>2</sub>O(   )→ 2 Ni(OH)<sub>2</sub>(s)+ Cd(OH)<sub>2</sub>(s) D) 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g)→ 2 H<sub>2</sub>O(   ) E) N<sub>2</sub>H<sub>4</sub>(   )+ O<sub>2</sub>(g)→ N<sub>2</sub>(g)+ 2 H<sub>2</sub>O(   ) <div style=padding-top: 35px> )→ 2 Ni(OH)2(s)+ Cd(OH)2(s)
D) 2 H2(g)+ O2(g)→ 2 H2O( <strong>Which of the following overall chemical equations is responsible for generating electricity in fuel cells used in NASA's Space Shuttle programs?</strong> A) Fe<sub>2</sub>O<sub>3</sub>(s)+ Al(s)→ Fe(s)+ Al<sub>2</sub>O<sub>3</sub>(s) B) Pb(s)+ PbO<sub>2</sub>(s)+ 2 H<sub>2</sub>SO<sub>4</sub>(aq)→ 2 PbSO<sub>4</sub>(s)+ 2 H<sub>2</sub>O(   ) C) 2 NiO(OH)(s)+ Cd(s)+ H<sub>2</sub>O(   )→ 2 Ni(OH)<sub>2</sub>(s)+ Cd(OH)<sub>2</sub>(s) D) 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g)→ 2 H<sub>2</sub>O(   ) E) N<sub>2</sub>H<sub>4</sub>(   )+ O<sub>2</sub>(g)→ N<sub>2</sub>(g)+ 2 H<sub>2</sub>O(   ) <div style=padding-top: 35px> )
E) N2H4( <strong>Which of the following overall chemical equations is responsible for generating electricity in fuel cells used in NASA's Space Shuttle programs?</strong> A) Fe<sub>2</sub>O<sub>3</sub>(s)+ Al(s)→ Fe(s)+ Al<sub>2</sub>O<sub>3</sub>(s) B) Pb(s)+ PbO<sub>2</sub>(s)+ 2 H<sub>2</sub>SO<sub>4</sub>(aq)→ 2 PbSO<sub>4</sub>(s)+ 2 H<sub>2</sub>O(   ) C) 2 NiO(OH)(s)+ Cd(s)+ H<sub>2</sub>O(   )→ 2 Ni(OH)<sub>2</sub>(s)+ Cd(OH)<sub>2</sub>(s) D) 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g)→ 2 H<sub>2</sub>O(   ) E) N<sub>2</sub>H<sub>4</sub>(   )+ O<sub>2</sub>(g)→ N<sub>2</sub>(g)+ 2 H<sub>2</sub>O(   ) <div style=padding-top: 35px> )+ O2(g)→ N2(g)+ 2 H2O( <strong>Which of the following overall chemical equations is responsible for generating electricity in fuel cells used in NASA's Space Shuttle programs?</strong> A) Fe<sub>2</sub>O<sub>3</sub>(s)+ Al(s)→ Fe(s)+ Al<sub>2</sub>O<sub>3</sub>(s) B) Pb(s)+ PbO<sub>2</sub>(s)+ 2 H<sub>2</sub>SO<sub>4</sub>(aq)→ 2 PbSO<sub>4</sub>(s)+ 2 H<sub>2</sub>O(   ) C) 2 NiO(OH)(s)+ Cd(s)+ H<sub>2</sub>O(   )→ 2 Ni(OH)<sub>2</sub>(s)+ Cd(OH)<sub>2</sub>(s) D) 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g)→ 2 H<sub>2</sub>O(   ) E) N<sub>2</sub>H<sub>4</sub>(   )+ O<sub>2</sub>(g)→ N<sub>2</sub>(g)+ 2 H<sub>2</sub>O(   ) <div style=padding-top: 35px> )
Question
The standard reduction potentials are as follows: Cr3+(aq)+ 3 e- → Cr(s); E° = -0.74 V
Fe2+(aq)+ 2 e- → Fe(s); E° = -0.41 V
Calculate the standard Gibbs free energy change for the following reaction.
2 Cr(s)+ 3 Fe2+ → 3 Fe(s)+ 2 Cr3+(aq)

A) 191 kJ
B) 63.7 kJ
C) -504 kJ
D) -191 kJ
E) 1060 kJ
Question
Which of the following equations represents the Nernst equation?

A) <strong>Which of the following equations represents the Nernst equation?</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px>
B) <strong>Which of the following equations represents the Nernst equation?</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px>
C) <strong>Which of the following equations represents the Nernst equation?</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px>
D) <strong>Which of the following equations represents the Nernst equation?</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px>
E) <strong>Which of the following equations represents the Nernst equation?</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px>
Question
Which of the following statements is true concerning the electrochemical cell Zn(s)| Zn2+(aq,1.0 M)|| Ca2+(aq,1.0 M)| Ca(s) The standard reduction potentials are given follows:
Zn2+(aq)+ 2 e− → Zn(s); E° = -0.76 V
Ca2+(aq)+ 2 e− → Ca(s); E° = -2.87 V

A) The standard cell potential is -3.63 V.
B) The standard cell potential is -2.11 V.
C) The standard cell potential is -2.11 V..
D) The standard cell potential is -3.63 V..
E) The standard cell potential is 2.11 V.
Question
The following electrochemical cell has a potential of +0.326 V at 25 °C. Pt | H2(g,1.00 atm)| H+(aq,1.00 M)|| Cl-(aq)| AgCl(s)| Ag
The standard reduction potential,E°,of AgCl(s)is +0.222 V.Calculate the Cl-(aq)ion concentration.

A) 1.9 × 10-19 M
B) 5.5 × 10-10 M
C) 0.018 M
D) 1.03 M
E) 1.8 × 109 M
Question
Calculate the copper(II)ion concentration at 25 °C in the cell Zn(s)| Zn2+(aq,1.0 M)|| Cu2+(aq)| Cu(s)if the measured cell potential is 1.06 V.The standard cell potential is 1.10 V.

A) 0.21 M
B) 1.0 M
C) 0.045 M
D) 1.0 M..
E) 1.0 M.
Question
If the <strong>If the   = -0.362 V for a given electrochemical cell at 25 °C,calculate the pH of the solution at the cathode. Pt | H<sub>2</sub>(g,1.0 atm)| H<sup>+</sup>(aq,1.00 M)|| H<sup>+</sup>(aq)| H<sub>2</sub>(g,1.0 atm)| Pt</strong> A) 1.77 B) 3.06 C) 6.11 D) 7.89 E) 12.23 <div style=padding-top: 35px> = -0.362 V for a given electrochemical cell at 25 °C,calculate the pH of the solution at the cathode. Pt | H2(g,1.0 atm)| H+(aq,1.00 M)|| H+(aq)| H2(g,1.0 atm)| Pt

A) 1.77
B) 3.06
C) 6.11
D) 7.89
E) 12.23
Question
The standard cell potential of the given electrochemical cell is 0.19 V. Pt | Sn4+(aq,1.0 M),Sn2+(aq,1.0 M)|| Cu2+(aq,0.200 M)| Cu
Which of the following factors will increase the measured cell potential of the given electrochemical cell?

A) Switching from a platinum to a graphite anode
B) Increasing the size of the anode
C) Decreasing the concentration of Cu2+
D) Increasing the concentration of Sn4+
E) Decreasing the temperature of the cell
Question
The cell potential of the following electrochemical cell is determined by using an unspecified concentration of acid.Calculate the pH of the acid solution,given that the measured cell potential is -0.431 V and the anode reduction potential (E°)is 0.222 V at 25 °C. ​
Ag(s)| AgCl(s)| Cl−(aq,1.0 M)|| H+(aq)| H2(g,1.0 atm)| Pt(s)

A) 3.53
B) 11.0
C) 4.03
D) 7.06
E) 3.47
Question
Calculate the value of the reaction quotient,Q,for the voltaic cell constructed from the following two half-reactions when the Zn2+ion concentration is 0.0110 M and the Ag+ ion concentration is 1.27 M? <strong>Calculate the value of the reaction quotient,Q,for the voltaic cell constructed from the following two half-reactions when the Zn<sup>2+</sup>ion concentration is 0.0110 M and the Ag<sup>+</sup> ion concentration is 1.27 M?   →     →  </strong> A) 8.66 × 10<sup>-3</sup> B) 6.82 × 10<sup>-3</sup> C) 115 D) 1.25 × 10<sup>-2</sup> E) 147 <div style=padding-top: 35px> <strong>Calculate the value of the reaction quotient,Q,for the voltaic cell constructed from the following two half-reactions when the Zn<sup>2+</sup>ion concentration is 0.0110 M and the Ag<sup>+</sup> ion concentration is 1.27 M?   →     →  </strong> A) 8.66 × 10<sup>-3</sup> B) 6.82 × 10<sup>-3</sup> C) 115 D) 1.25 × 10<sup>-2</sup> E) 147 <div style=padding-top: 35px> <strong>Calculate the value of the reaction quotient,Q,for the voltaic cell constructed from the following two half-reactions when the Zn<sup>2+</sup>ion concentration is 0.0110 M and the Ag<sup>+</sup> ion concentration is 1.27 M?   →     →  </strong> A) 8.66 × 10<sup>-3</sup> B) 6.82 × 10<sup>-3</sup> C) 115 D) 1.25 × 10<sup>-2</sup> E) 147 <div style=padding-top: 35px> <strong>Calculate the value of the reaction quotient,Q,for the voltaic cell constructed from the following two half-reactions when the Zn<sup>2+</sup>ion concentration is 0.0110 M and the Ag<sup>+</sup> ion concentration is 1.27 M?   →     →  </strong> A) 8.66 × 10<sup>-3</sup> B) 6.82 × 10<sup>-3</sup> C) 115 D) 1.25 × 10<sup>-2</sup> E) 147 <div style=padding-top: 35px>

A) 8.66 × 10-3
B) 6.82 × 10-3
C) 115
D) 1.25 × 10-2
E) 147
Question
Calculate Ecell for the following electrochemical cell at 25 °C Pt(s)| H2(g,1.00 atm)| H+(aq,1.00 M)|| Sn2+(aq,0.350 M),Sn4+(aq,0.020 M)| Pt(s)
The standard reduction potentials are as follows:
Sn4+(aq)+ 2 e- → Sn2+(s)
E° = +0.15 V
2 H+(aq)+ 2 e- → H2(g)
E° = 0.00 V

A) -0.19 V
B) +0.08 V
C) +0.11 V
D) +0.19 V
E) +0.22 V
Question
The following has a potential of 0.34 V. <strong>The following has a potential of 0.34 V.   If the concentrations of each of the ions is 1.0 M and the pressure of H<sub>2</sub> is 1.0 atm,then E° for the half-reaction   is _____.</strong> A) 0.17 V B) -0.17 V C) 0.34 V D) -0.34 V E) None of these <div style=padding-top: 35px> If the concentrations of each of the ions is 1.0 M and the pressure of H2 is 1.0 atm,then E° for the half-reaction <strong>The following has a potential of 0.34 V.   If the concentrations of each of the ions is 1.0 M and the pressure of H<sub>2</sub> is 1.0 atm,then E° for the half-reaction   is _____.</strong> A) 0.17 V B) -0.17 V C) 0.34 V D) -0.34 V E) None of these <div style=padding-top: 35px> is _____.

A) 0.17 V
B) -0.17 V
C) 0.34 V
D) -0.34 V
E) None of these
Question
Calculate the cell potential at 25 °C for the following overall reaction. Zn2+(aq)+ 2 Fe2+(aq)→ Zn(s)+ 2 Fe3+(aq)
[Zn2+] = 1.50 × 10-4 M,[Fe3+] = 0.0200 M,and [Fe2+] = 0.0100 M.The standard reduction potentials are as follows:
Zn2+(aq)+ 2 e- → Zn(s)
E° = -0.763 V
Fe3+(aq)+ e- → Fe2+(aq)
E° = +0.771 V

A) -1.665 V
B) -1.534 V
C) -1.439 V
D) -0.008 V
E) +0.008 V
Question
For the following cell reaction,the standard cell potential is 1.34 V.To determine the cell potential at nonstandard conditions,what is the value that should be used for n in the Nernst equation? <strong>For the following cell reaction,the standard cell potential is 1.34 V.To determine the cell potential at nonstandard conditions,what is the value that should be used for n in the Nernst equation?  </strong> A) 8 B) 10 C) 5 D) 2 E) 6 <div style=padding-top: 35px>

A) 8
B) 10
C) 5
D) 2
E) 6
Question
Calculate <strong>Calculate   for the electrochemical cell Pb(s)|PbCl<sub>2</sub>(s)| Cl<sup>-</sup>(aq,1.0 M)|| Fe<sup>3+</sup>(aq,1.0 M),Fe<sup>2+</sup>(aq,1.0 M)| Pt(s). The standard reduction potentials are as follows: Pb<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Pb(s) E° = -0.126 V PbCl<sub>2</sub>(s)+ 2 e<sup>-</sup> → Pb(s)+ 2 Cl<sup>-</sup>(aq) E° = -0.267 V Fe<sup>3+</sup>(aq)+ e<sup>-</sup> → Fe<sup>2+</sup>(aq) E° = +0.771 V Fe<sup>2+</sup>(aq)+ e<sup>-</sup> → Fe(s) E° = -0.44 V</strong> A) -0.504 V B) -0.062 V C) +0.504 V D) +1.038 V E) +1.604 V <div style=padding-top: 35px> for the electrochemical cell Pb(s)|PbCl2(s)| Cl-(aq,1.0 M)|| Fe3+(aq,1.0 M),Fe2+(aq,1.0 M)| Pt(s). The standard reduction potentials are as follows:
Pb2+(aq)+ 2 e- → Pb(s)
E° = -0.126 V
PbCl2(s)+ 2 e- → Pb(s)+ 2 Cl-(aq)
E° = -0.267 V
Fe3+(aq)+ e- → Fe2+(aq)
E° = +0.771 V
Fe2+(aq)+ e- → Fe(s)
E° = -0.44 V

A) -0.504 V
B) -0.062 V
C) +0.504 V
D) +1.038 V
E) +1.604 V
Question
Calculate Ecell for the following electrochemical cell at 25°C.The standard cell potential,E°cell,is 0.460 V. Cu(s)| Cu2+(aq,0.016 M)|| Ag+(aq,0.11 M)| Ag(s)

A) 0.456 V
B) 0.282 V
C) 0.460 V
D) 0.485 V
E) 0.452 V
Question
Calculate Ecell for the following electrochemical cell at 25 °C. Pt(s)| Fe3+(aq,0.100 M),Fe2+(aq,0.040 M)|| Cl-(aq,0.50 M)| AgCl(s)| Ag(s)
The standard reduction potentials are as follows:
AgCl(s)+ e- → Ag(s)+ Cl-(aq)
E° = +0.222 V
Fe3+(aq)+ e- → Fe2+(aq)
E° = +0.771 V

A) -1.034 V
B) -0.590 V
C) -0.508 V
D) -0.555 V
E) +1.034 V
Question
For the electrochemical cell Cu(s)| Cu2+ || Ag+ | Ag(s),the standard cell potential is 0.46 V.A cell using these reagents was made,and the observed potential was 0.26 V at 25 oC.Which of the following is a possible explanation for the observed voltage?

A) The volume of the Cu2+ solution was larger than the volume of the Ag+ solution.
B) The volume of the Ag+ solution was larger than the volume of the Cu2+ solution.
C) The Cu2+ concentration was larger than the Ag+ concentration.
D) The Ag electrode was twice as large as the Cu electrode.
E) The Ag+ concentration was larger than the Cu2+ concentration.
Question
Which of the following is true for a product-favored reaction at equilibrium?

A) ΔrG° < 0; E°cell > 0
B) ΔrG° > 0; E°cell < 0
C) ΔrG° < 0; E°cell < 0
D) ΔrG° > 0; E°cell > 0
E) ΔrG° > 0; E°cell = 0
Question
One Faraday is defined as the:

A) charge on a single electron.
B) quantity of electric charge carried by one mole of electrons.
C) voltage required to reduce one mole of reactant.
D) number of moles of electrons required to reduce one mole of a reactant.
E) charge passed by one ampere of current in one second.
Question
Calculate <strong>Calculate   for the electrochemical cell Ag(s)| AgCl(s)| Cl<sup>-</sup>(aq,1.0 M)|| Cu<sup>2+</sup>(aq,1.0 M)| Cu(s). The standard reduction potentials are as follows: Cu<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Cu(s) E° = +0.337 V AgCl(s)+ e<sup>-</sup> → Ag(s)+ Cl<sup>-</sup>(aq) E° = +0.222 V</strong> A) -0.115 V B) -0.107 V C) +0.115 V D) +0.452 V E) +0.559 V <div style=padding-top: 35px> for the electrochemical cell Ag(s)| AgCl(s)| Cl-(aq,1.0 M)|| Cu2+(aq,1.0 M)| Cu(s). The standard reduction potentials are as follows:
Cu2+(aq)+ 2 e- → Cu(s)
E° = +0.337 V
AgCl(s)+ e- → Ag(s)+ Cl-(aq)
E° = +0.222 V

A) -0.115 V
B) -0.107 V
C) +0.115 V
D) +0.452 V
E) +0.559 V
Question
Calculate the standard reduction potential for the given reaction at 25 °C. AuCl4-(aq)+ 3 e- → Au(s)+ 4 Cl-(aq)
The thermodynamic information is as follows:
Au3+(aq)+ 3 e- → Au(s)
E° = +1.50 V
Au3+(aq)+ 4 Cl-(aq)→ AuCl4-(aq)
Kf = 2.3 × 1025

A) -1.28 V
B) -0.50 V
C) +1.00 V
D) +1.28 V
E) +3.85 V
Question
Claculate the mass of chromium that can be deposited by electrolysis of an aqueous solution of chromium(III)sulfate,Cr2(SO4)3,for 180 min using a constant current of 11.0 A.Assume 100% current efficiency.(F = 96485 C/mol)

A) 0.356 g
B) 21.3 g
C) 192.1 g
D) 0.187 g
E) 32.0 g
Question
The value of E°cell is <strong>The value of E°<sub>cell</sub> is   for the following reaction: Cl<sub>2</sub>(g)+ 2 Fe<sup>2+</sup>(aq)→ 2 Fe<sup>3+</sup>(aq)+ 2 Cl<sup>-</sup>(aq) Calculate the value of E°<sub>cell</sub> for the reaction below. Cl<sup>-</sup>(g)+ Fe<sup>3+</sup>(aq)→ Fe<sup>2+</sup>(aq)+ ½ Cl<sub>2</sub>(g)</strong> A) -1.18 V B) -0.30 V C) 0.59 V D) 0.30 V E) -0.59 V <div style=padding-top: 35px> for the following reaction: Cl2(g)+ 2 Fe2+(aq)→ 2 Fe3+(aq)+ 2 Cl-(aq)
Calculate the value of E°cell for the reaction below.
Cl-(g)+ Fe3+(aq)→ Fe2+(aq)+ ½ Cl2(g)

A) -1.18 V
B) -0.30 V
C) 0.59 V
D) 0.30 V
E) -0.59 V
Question
What half-reaction occurs at the cathode during the electrolysis of molten potassium bromide?

A) K(s)→ K+(l)+ e-
B) Br2(l)+ 2 e- → 2 Br-(l)
C) 2 Br-(l)→ Br2(l)+ 2 e-
D) 2 K+(l)+ 2 e- → 2 K(l)
E) 2 H2O(l)+ 2 e- → H2(g)+ 2 OH-(l)
Question
In an electrolytic cell,reduction occurs at the _____ and oxidation occurs at the _____.
Question
Which of the following are the expected products when an aqueous solution of lithium sulfate is electrolyzed? Reduction Half-Reaction
E° (V)
Li+(aq)+ e- → Li(s)
-3)04
2 H2O(l)+ 2 e- → H2(g)+ 2 OH-(aq)
-0)83
2 H+(aq)+ 2 e- → H2(g)
0)00
O2(g)+ 4 H+(aq)+ 4 e- → 2 H2O(l)
1)23
S2O82-(aq)+ 2 e- → 2 SO42-(aq)
2)01

A) Li(s)and H2(g)
B) H2(g),OH-(aq),O2(g),and H+(aq)
C) O2(g),H+(aq),and Li(s)
D) H2(g),OH-(aq),and Li(s)
E) H2(g),OH-(aq),and S2O82-(aq)
Question
If ΔrG° for the following reaction is -22.2 kJ/mol-rxn,calculate <strong>If Δ<sub>r</sub>G° for the following reaction is -22.2 kJ/mol-rxn,calculate   for the following reaction: Cu<sup>2+</sup>(aq)+ 2 Ag(s)+ 2 Cl<sup>-</sup>(aq)→ Cu(s)+ 2 AgCl(s)</strong> A) -0.460 V B) -0.115 V C) +0.115 V D) +0.230 V E) +0.559 V <div style=padding-top: 35px> for the following reaction: Cu2+(aq)+ 2 Ag(s)+ 2 Cl-(aq)→ Cu(s)+ 2 AgCl(s)

A) -0.460 V
B) -0.115 V
C) +0.115 V
D) +0.230 V
E) +0.559 V
Question
Calculate the equilibrium constant for the reaction below at 25 °C. Co(s)+ 2 Cr3+(aq)→ Co2+(aq)+ 2 Cr2+(aq)
The standard reduction potentials are as follows:
Co2+(aq)+ 2 e- → Co(s)
E° = -0.28 V
Cr3+(aq)+ e- → Cr2+(aq)
E° = -0.41 V

A) 4.0 × 10-5
B) 2.5 × 104
C) 1.0 × 105
D) 1.2 × 105
E) 1.3 × 105
Question
A current of 12.0 A is passed through molten magnesium chloride for 14.0 h.How many moles of magnesium metal can be produced from this electrolysis?

A) 0.0522 mol
B) 3.13 mol
C) 0.37 mol
D) 0.22 mol
E) 6.27 mol
Question
Batteries used in watches contain mercury(II)oxide.As the current flows,mercury(II)oxide is reduced to mercury according to the following reaction: HgO(s)+ H2O( <strong>Batteries used in watches contain mercury(II)oxide.As the current flows,mercury(II)oxide is reduced to mercury according to the following reaction: HgO(s)+ H<sub>2</sub>O(   )+ 2 e<sup>-</sup> → Hg(   )+ 2 OH<sup>-</sup>(aq) If 2.3 × 10<sup>-5</sup> amperes flows continuously for 1200 days,calculate the mass of mercury,Hg(   ),produced.</strong> A) 2.5 g B) 5.0 g C) 9.9 g D) 13 g E) 15 g <div style=padding-top: 35px> )+ 2 e- → Hg( <strong>Batteries used in watches contain mercury(II)oxide.As the current flows,mercury(II)oxide is reduced to mercury according to the following reaction: HgO(s)+ H<sub>2</sub>O(   )+ 2 e<sup>-</sup> → Hg(   )+ 2 OH<sup>-</sup>(aq) If 2.3 × 10<sup>-5</sup> amperes flows continuously for 1200 days,calculate the mass of mercury,Hg(   ),produced.</strong> A) 2.5 g B) 5.0 g C) 9.9 g D) 13 g E) 15 g <div style=padding-top: 35px> )+ 2 OH-(aq)
If 2.3 × 10-5 amperes flows continuously for 1200 days,calculate the mass of mercury,Hg( <strong>Batteries used in watches contain mercury(II)oxide.As the current flows,mercury(II)oxide is reduced to mercury according to the following reaction: HgO(s)+ H<sub>2</sub>O(   )+ 2 e<sup>-</sup> → Hg(   )+ 2 OH<sup>-</sup>(aq) If 2.3 × 10<sup>-5</sup> amperes flows continuously for 1200 days,calculate the mass of mercury,Hg(   ),produced.</strong> A) 2.5 g B) 5.0 g C) 9.9 g D) 13 g E) 15 g <div style=padding-top: 35px> ),produced.

A) 2.5 g
B) 5.0 g
C) 9.9 g
D) 13 g
E) 15 g
Question
Calculate the charge,in coulombs,is required to deposit 1.5 g of solid magnesium from a solution of Mg2+(aq)ion.

A) 4.1 × 102 C
B) 6.0 × 103 C
C) 1.2 × 104 C
D) 2.9 × 105 C
E) 3.1 × 106 C
Question
The use of electrical energy to produce chemical change is known as _____.An example of this process is the reduction of sodium chloride,NaCl( The use of electrical energy to produce chemical change is known as _____.An example of this process is the reduction of sodium chloride,NaCl(   ),to produce solid sodium.<div style=padding-top: 35px> ),to produce solid sodium.
Question
Aluminum(III)ion (Al3+)is reduced to solid aluminum at an electrode.If a current of 2.75 amperes is passed for 36 hours,calculate the mass of aluminum deposited at the electrode.(Assume 100% current efficiency.)

A) 9.2 × 10-3 g
B) 3.3 × 101 g
C) 9.9 × 101 g
D) 1.0 × 102 g
E) 3.0 × 102 g
Question
Calculate the value of the equilibrium constant (K)at 25 °C for the following cell reaction: Sn(s)+ Pb2+(aq)→ Sn2+(aq)+ Pb(s); E°cell = 0.014 V

A) 0.014
B) 1.7
C) 0.4
D) 1.0
E) 3
Question
Calculate ΔrG° for the disproportionation reaction of copper(I)ion (Cu+)at 25 °C. 2 Cu+(aq)→ Cu2+(aq)+ Cu(s)
The standard reduction potentials are as follows:
Cu+(aq)+ e- → Cu(s)
E° = +0.518 V
Cu2+(aq)+ 2 e- → Cu(s)
E° = +0.337 V

A) -165 kJ/mol⋅rxn
B) -135 kJ/mol⋅rxn
C) -34.9 kJ/mol⋅rxn
D) +17.5 kJ/mol⋅rxn
E) +135 kJ/mol⋅rxn
Question
The standard reduction potentials for a reaction are as follows: Pb2+(aq)+ 2 e- → Pb(s)
E° = -0.126 V
PbSO4(s)+ 2 e- → Pb(s)+ SO42-(aq)
E° = -0.355 V
Calculate the Ksp for lead(II)sulfate (PbSO4)at 25 °C.

A) 3.4 × 10-28
B) 1.8 × 10-8
C) 5.6 × 10-5
D) 5.6 × 107
E) 2.9 × 1037
Question
A voltaic cell or galvanic cell converts chemical energy to electrical energy. <strong>A voltaic cell or galvanic cell converts chemical energy to electrical energy.   for the given galvanic cell is -1.80 V. Fe<sup>2+</sup>(aq)+ 2 Cl<sup>-</sup>(aq)→ Fe(s)+ Cl<sub>2</sub>(g) Calculate the value of Δ<sub>r</sub>G° for the reaction.</strong> A) -174 kJ B) 86.8 kJ C) 107 kJ D) 174 kJ E) 347 kJ <div style=padding-top: 35px> for the given galvanic cell is -1.80 V. Fe2+(aq)+ 2 Cl-(aq)→ Fe(s)+ Cl2(g)
Calculate the value of ΔrG° for the reaction.

A) -174 kJ
B) 86.8 kJ
C) 107 kJ
D) 174 kJ
E) 347 kJ
Question
Gold and platinum are commonly used as inert electrodes in laboratory experiments.In commercial applications,such as batteries,_____ is more commonly used as an inert electrodes because it is far less expensive.
Question
Calculate the equilibrium constant for the following reaction at 25 °C, 2 IO3-(aq)+ 5 Hg( <strong>Calculate the equilibrium constant for the following reaction at 25 °C, 2 IO<sub>3</sub><sup>-</sup>(aq)+ 5 Hg(   )+ 12 H<sup>+</sup>(aq)→ I<sub>2</sub>(s)+ 5 Hg<sup>2+</sup>(aq)+ 6 H<sub>2</sub>O(   ) The standard reduction potentials are as follows: IO<sub>3</sub><sup>-</sup>(aq)+ 6 H<sup>+</sup>(aq)+ 5 e<sup>-</sup> → I<sub>2</sub>(s)+ 3 H<sub>2</sub>O(   ) E° = +1.20 V Hg<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Hg(   ) E° = +0.86 V</strong> A) 3 × 10<sup>-58</sup> B) 6 × 10<sup>5</sup> C) 3 × 10<sup>11</sup> D) 6 × 10<sup>28</sup> E) 3 × 10<sup>57</sup> <div style=padding-top: 35px> )+ 12 H+(aq)→ I2(s)+ 5 Hg2+(aq)+ 6 H2O( <strong>Calculate the equilibrium constant for the following reaction at 25 °C, 2 IO<sub>3</sub><sup>-</sup>(aq)+ 5 Hg(   )+ 12 H<sup>+</sup>(aq)→ I<sub>2</sub>(s)+ 5 Hg<sup>2+</sup>(aq)+ 6 H<sub>2</sub>O(   ) The standard reduction potentials are as follows: IO<sub>3</sub><sup>-</sup>(aq)+ 6 H<sup>+</sup>(aq)+ 5 e<sup>-</sup> → I<sub>2</sub>(s)+ 3 H<sub>2</sub>O(   ) E° = +1.20 V Hg<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Hg(   ) E° = +0.86 V</strong> A) 3 × 10<sup>-58</sup> B) 6 × 10<sup>5</sup> C) 3 × 10<sup>11</sup> D) 6 × 10<sup>28</sup> E) 3 × 10<sup>57</sup> <div style=padding-top: 35px> )
The standard reduction potentials are as follows:
IO3-(aq)+ 6 H+(aq)+ 5 e- → I2(s)+ 3 H2O( <strong>Calculate the equilibrium constant for the following reaction at 25 °C, 2 IO<sub>3</sub><sup>-</sup>(aq)+ 5 Hg(   )+ 12 H<sup>+</sup>(aq)→ I<sub>2</sub>(s)+ 5 Hg<sup>2+</sup>(aq)+ 6 H<sub>2</sub>O(   ) The standard reduction potentials are as follows: IO<sub>3</sub><sup>-</sup>(aq)+ 6 H<sup>+</sup>(aq)+ 5 e<sup>-</sup> → I<sub>2</sub>(s)+ 3 H<sub>2</sub>O(   ) E° = +1.20 V Hg<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Hg(   ) E° = +0.86 V</strong> A) 3 × 10<sup>-58</sup> B) 6 × 10<sup>5</sup> C) 3 × 10<sup>11</sup> D) 6 × 10<sup>28</sup> E) 3 × 10<sup>57</sup> <div style=padding-top: 35px> )
E° = +1.20 V
Hg2+(aq)+ 2 e- → Hg( <strong>Calculate the equilibrium constant for the following reaction at 25 °C, 2 IO<sub>3</sub><sup>-</sup>(aq)+ 5 Hg(   )+ 12 H<sup>+</sup>(aq)→ I<sub>2</sub>(s)+ 5 Hg<sup>2+</sup>(aq)+ 6 H<sub>2</sub>O(   ) The standard reduction potentials are as follows: IO<sub>3</sub><sup>-</sup>(aq)+ 6 H<sup>+</sup>(aq)+ 5 e<sup>-</sup> → I<sub>2</sub>(s)+ 3 H<sub>2</sub>O(   ) E° = +1.20 V Hg<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Hg(   ) E° = +0.86 V</strong> A) 3 × 10<sup>-58</sup> B) 6 × 10<sup>5</sup> C) 3 × 10<sup>11</sup> D) 6 × 10<sup>28</sup> E) 3 × 10<sup>57</sup> <div style=padding-top: 35px> )
E° = +0.86 V

A) 3 × 10-58
B) 6 × 105
C) 3 × 1011
D) 6 × 1028
E) 3 × 1057
Question
How many moles of electrons are produced from a current of 17.0 A in 3.40 hours?

A) 5.99 × 10-4 mol
B) 2.16 mol
C) 57.8 mol
D) 3.35 mol
E) 9.33 × 103 mol
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Deck 20: Principles of Chemical Reactivity: Electron Transfer Reactions
1
Which of the following statements concerning voltaic cells is not true?

A) A salt bridge allows cations and anions to move between half-cells.
B) Electrons flow from a cathode to an anode in the external circuit.
C) Oxidation occurs at an anode.
D) A voltaic cell can be used as a source of energy.
E) A voltaic cell consists of two half-cells.
Electrons flow from a cathode to an anode in the external circuit.
2
Which of the following statements is true concerning the voltaic cell shown below? <strong>Which of the following statements is true concerning the voltaic cell shown below?  </strong> A) The Zn anode mass decreases as the cell discharges. B) The Zn cathode mass increases as the cell discharges. C) The Zn cathode mass decreases as the cell discharges. D) The Zn anode mass increases as the cell discharges. E) The mass of the Zn electrode neither increases nor decreases as the cell discharges.

A) The Zn anode mass decreases as the cell discharges.
B) The Zn cathode mass increases as the cell discharges.
C) The Zn cathode mass decreases as the cell discharges.
D) The Zn anode mass increases as the cell discharges.
E) The mass of the Zn electrode neither increases nor decreases as the cell discharges.
The Zn anode mass decreases as the cell discharges.
3
Balance the following half-reaction occurring in an acidic solution. NO3-(aq)→ NO(aq)

A) NO3-(aq)+ 4 H+(aq)+ 3 e− → NO(g)+ 2 H2O(l)
B) NO3-(aq)+ 2 H2O(l)+ 3 e− → NO(g)+ 4 H+(aq)
C) NO3-(aq)+ 4 H+(aq)→ NO(g)+ 2 H2O(l)+ 3 e−
D) NO3-(aq)+ 3 e− → NO(g)+ 4 H+(aq)+ 2 H2O(l)
E) NO3-(aq)+ 4 H+(aq)→ NO(g)+ 2 H2O(l)
NO3-(aq)+ 4 H+(aq)+ 3 e− → NO(g)+ 2 H2O(l)
4
Write a balanced chemical equation for the oxidation of solid cadmium by concentrated nitric acid,producing nitrogen dioxide gas and Cd2+(aq)ion.

A) HNO3(aq)+ Cd(s)→ Cd2+(aq)+ NO2(g)+ OH-(aq)
B) 2 HNO3(aq)+ Cd(s)→ Cd2+(aq)+ 2 NO2(g)+ 2 OH-(aq)
C) HNO3(aq)+ Cd(s)+ H+(aq)→ Cd2+(aq)+ NO2(g)+ H2O( <strong>Write a balanced chemical equation for the oxidation of solid cadmium by concentrated nitric acid,producing nitrogen dioxide gas and Cd<sup>2+</sup>(aq)ion.</strong> A) HNO<sub>3</sub>(aq)+ Cd(s)→ Cd<sup>2+</sup>(aq)+ NO<sub>2</sub>(g)+ OH<sup>-</sup>(aq) B) 2 HNO<sub>3</sub>(aq)+ Cd(s)→ Cd<sup>2+</sup>(aq)+ 2 NO<sub>2</sub>(g)+ 2 OH<sup>-</sup>(aq) C) HNO<sub>3</sub>(aq)+ Cd(s)+ H<sup>+</sup>(aq)→ Cd<sup>2+</sup>(aq)+ NO<sub>2</sub>(g)+ H<sub>2</sub>O(   ) D) 4 HNO<sub>3</sub>(aq)+ Cd(s)→ Cd<sup>2+</sup>(aq)+ 2 NO<sub>2</sub>(g)+ 2 H<sub>2</sub>O(   )+ 2 NO<sub>3</sub><sup>-</sup>(aq) E) HNO<sub>3</sub>(aq)+ Cd(s)→ Cd<sup>2+</sup>(aq)+ NO<sub>2</sub>(g) )
D) 4 HNO3(aq)+ Cd(s)→ Cd2+(aq)+ 2 NO2(g)+ 2 H2O( <strong>Write a balanced chemical equation for the oxidation of solid cadmium by concentrated nitric acid,producing nitrogen dioxide gas and Cd<sup>2+</sup>(aq)ion.</strong> A) HNO<sub>3</sub>(aq)+ Cd(s)→ Cd<sup>2+</sup>(aq)+ NO<sub>2</sub>(g)+ OH<sup>-</sup>(aq) B) 2 HNO<sub>3</sub>(aq)+ Cd(s)→ Cd<sup>2+</sup>(aq)+ 2 NO<sub>2</sub>(g)+ 2 OH<sup>-</sup>(aq) C) HNO<sub>3</sub>(aq)+ Cd(s)+ H<sup>+</sup>(aq)→ Cd<sup>2+</sup>(aq)+ NO<sub>2</sub>(g)+ H<sub>2</sub>O(   ) D) 4 HNO<sub>3</sub>(aq)+ Cd(s)→ Cd<sup>2+</sup>(aq)+ 2 NO<sub>2</sub>(g)+ 2 H<sub>2</sub>O(   )+ 2 NO<sub>3</sub><sup>-</sup>(aq) E) HNO<sub>3</sub>(aq)+ Cd(s)→ Cd<sup>2+</sup>(aq)+ NO<sub>2</sub>(g) )+ 2 NO3-(aq)
E) HNO3(aq)+ Cd(s)→ Cd2+(aq)+ NO2(g)
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5
When the following oxidation-reduction reaction in acidic solution is balanced,what is the lowest whole-number coefficient for Na+(aq)ion? Na(s)+ Ca2+(aq)→ Na+(aq)+ Ca(s)

A) 5
B) 4
C) 1
D) 3
E) 2
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6
Write the balanced oxidation half-reaction for the following overall reaction: 2 H+(aq)+ Ca(s)→ Ca2+(aq)+ H2(g)

A) Ca(s) → 2e- + Ca2+ (aq)
B) 2 H+(aq)→ H2(g)+ 2 e-
C) H2(g)→ 2 H+(aq)+ 2 e-
D) Ca(s)+ 2 e- → Ca2+(aq)
E) 2Ca(s)→ Ca2+(aq)+ 2 e-
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7
Balance the following oxidation-reduction reaction occurring in an acidic solution. MnO4-(aq)+ Cr2+(aq)→ Mn2+(aq)+ Cr3+(aq)

A) MnO4-(aq)+ 8 H+(aq)+ 5 Cr2+(aq)→ Mn2+(aq)+ 4 H2O(l)+ 5 Cr3+(aq)
B) MnO4-(aq)+ 8 H+(aq)+ Cr2+(aq)→ Mn2+(aq)+ 4 H2O(l)+ Cr3+(aq)
C) MnO4-(aq)+ 4 H2(g)+ 5 Cr2+(aq)→ Mn2+(aq)+ 4 H2O(l)+ 5 Cr3+(aq)
D) MnO4-(aq)+ 8 H+(aq)+ 2 Cr2+(aq)→ Mn2+(aq)+ 4 H2O(l)+ 2 Cr3+(aq)
E) MnO4-(aq)+ Cr2+(aq)→ Mn2+(aq)+ 2 O2(g)+ Cr3+(aq)
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8
Balance the following half-reaction occurring in a basic solution. MnO2(s)→ Mn(OH)2(s)

A) MnO2(s)+ 2 H2O(l)+ 2e−→ Mn(OH)2(s)+ 2 OH−(aq)
B) MnO2(s)+ 2 H2O(l)+ 4e−→ Mn(OH)2(s)+ (OH)2−(aq)
C) MnO2(s)+ H22+(aq)+ 2e−→ Mn(OH)2(s)
D) MnO2(s)+ H2(g)→ Mn(OH)2(s)+ 2 e−
E) MnO2(s)+ 2 H2O(l)→ Mn(OH)2(s)+ 2 OH−(aq)
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9
Write a balanced net ionic equation for the reaction below in an acidic solution. Cr2O72-(aq)+ Ni(s)→ Cr3+(aq)+ Ni2+(aq)

A) Cr2O72-(aq)+ 3 Ni(s)+ 14 H+(aq)→ 2 Cr3+(aq)+ 3 Ni2+(aq)+ 7 H2O( <strong>Write a balanced net ionic equation for the reaction below in an acidic solution. Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ Ni(s)→ Cr<sup>3+</sup>(aq)+ Ni<sup>2+</sup>(aq)</strong> A) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ 3 Ni(s)+ 14 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ 3 Ni<sup>2+</sup>(aq)+ 7 H<sub>2</sub>O(   ) B) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ Ni(s)+ 14 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ Ni<sup>2+</sup>(aq)+ 7 H<sub>2</sub>O(   ) C) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ 3 Ni(s)→ 2 Cr<sup>3+</sup>(aq)+ 3 Ni<sup>2+</sup>(aq)+ O<sup>2-</sup>(aq) D) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ Ni(s)+ 14 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ Ni<sup>2+</sup>(aq)+ 7 H<sub>2</sub>O(   ) E) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ 3 Ni(s)+ 7 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ 3 Ni<sup>2+</sup>(aq)+ 7 OH<sup>-</sup>(aq) )
B) Cr2O72-(aq)+ Ni(s)+ 14 H+(aq)→ 2 Cr3+(aq)+ Ni2+(aq)+ 7 H2O( <strong>Write a balanced net ionic equation for the reaction below in an acidic solution. Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ Ni(s)→ Cr<sup>3+</sup>(aq)+ Ni<sup>2+</sup>(aq)</strong> A) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ 3 Ni(s)+ 14 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ 3 Ni<sup>2+</sup>(aq)+ 7 H<sub>2</sub>O(   ) B) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ Ni(s)+ 14 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ Ni<sup>2+</sup>(aq)+ 7 H<sub>2</sub>O(   ) C) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ 3 Ni(s)→ 2 Cr<sup>3+</sup>(aq)+ 3 Ni<sup>2+</sup>(aq)+ O<sup>2-</sup>(aq) D) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ Ni(s)+ 14 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ Ni<sup>2+</sup>(aq)+ 7 H<sub>2</sub>O(   ) E) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ 3 Ni(s)+ 7 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ 3 Ni<sup>2+</sup>(aq)+ 7 OH<sup>-</sup>(aq) )
C) Cr2O72-(aq)+ 3 Ni(s)→ 2 Cr3+(aq)+ 3 Ni2+(aq)+ O2-(aq)
D) Cr2O72-(aq)+ Ni(s)+ 14 H+(aq)→ 2 Cr3+(aq)+ Ni2+(aq)+ 7 H2O( <strong>Write a balanced net ionic equation for the reaction below in an acidic solution. Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ Ni(s)→ Cr<sup>3+</sup>(aq)+ Ni<sup>2+</sup>(aq)</strong> A) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ 3 Ni(s)+ 14 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ 3 Ni<sup>2+</sup>(aq)+ 7 H<sub>2</sub>O(   ) B) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ Ni(s)+ 14 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ Ni<sup>2+</sup>(aq)+ 7 H<sub>2</sub>O(   ) C) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ 3 Ni(s)→ 2 Cr<sup>3+</sup>(aq)+ 3 Ni<sup>2+</sup>(aq)+ O<sup>2-</sup>(aq) D) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ Ni(s)+ 14 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ Ni<sup>2+</sup>(aq)+ 7 H<sub>2</sub>O(   ) E) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ 3 Ni(s)+ 7 H<sup>+</sup>(aq)→ 2 Cr<sup>3+</sup>(aq)+ 3 Ni<sup>2+</sup>(aq)+ 7 OH<sup>-</sup>(aq) )
E) Cr2O72-(aq)+ 3 Ni(s)+ 7 H+(aq)→ 2 Cr3+(aq)+ 3 Ni2+(aq)+ 7 OH-(aq)
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10
Write a balanced half-reaction for the reduction of hydrogen peroxide to water in an acidic solution.

A) 2 H2O2( <strong>Write a balanced half-reaction for the reduction of hydrogen peroxide to water in an acidic solution.</strong> A) 2 H<sub>2</sub>O<sub>2</sub>(   )→ 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) B) 2 H<sub>2</sub>O<sub>2</sub>(   )+ 2e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) C) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) D) H<sub>2</sub>O<sub>2</sub>(   )+ 4 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ H<sub>2</sub>(g) E) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g) )→ 2 H2O( <strong>Write a balanced half-reaction for the reduction of hydrogen peroxide to water in an acidic solution.</strong> A) 2 H<sub>2</sub>O<sub>2</sub>(   )→ 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) B) 2 H<sub>2</sub>O<sub>2</sub>(   )+ 2e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) C) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) D) H<sub>2</sub>O<sub>2</sub>(   )+ 4 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ H<sub>2</sub>(g) E) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g) )+ O2(g)
B) 2 H2O2( <strong>Write a balanced half-reaction for the reduction of hydrogen peroxide to water in an acidic solution.</strong> A) 2 H<sub>2</sub>O<sub>2</sub>(   )→ 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) B) 2 H<sub>2</sub>O<sub>2</sub>(   )+ 2e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) C) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) D) H<sub>2</sub>O<sub>2</sub>(   )+ 4 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ H<sub>2</sub>(g) E) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g) )+ 2e- → 2 H2O( <strong>Write a balanced half-reaction for the reduction of hydrogen peroxide to water in an acidic solution.</strong> A) 2 H<sub>2</sub>O<sub>2</sub>(   )→ 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) B) 2 H<sub>2</sub>O<sub>2</sub>(   )+ 2e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) C) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) D) H<sub>2</sub>O<sub>2</sub>(   )+ 4 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ H<sub>2</sub>(g) E) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g) )+ O2(g)
C) H2O2( <strong>Write a balanced half-reaction for the reduction of hydrogen peroxide to water in an acidic solution.</strong> A) 2 H<sub>2</sub>O<sub>2</sub>(   )→ 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) B) 2 H<sub>2</sub>O<sub>2</sub>(   )+ 2e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) C) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) D) H<sub>2</sub>O<sub>2</sub>(   )+ 4 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ H<sub>2</sub>(g) E) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g) )+ 2 H+(aq)+ 2 e- → 2 H2O( <strong>Write a balanced half-reaction for the reduction of hydrogen peroxide to water in an acidic solution.</strong> A) 2 H<sub>2</sub>O<sub>2</sub>(   )→ 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) B) 2 H<sub>2</sub>O<sub>2</sub>(   )+ 2e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) C) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) D) H<sub>2</sub>O<sub>2</sub>(   )+ 4 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ H<sub>2</sub>(g) E) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g) )
D) H2O2( <strong>Write a balanced half-reaction for the reduction of hydrogen peroxide to water in an acidic solution.</strong> A) 2 H<sub>2</sub>O<sub>2</sub>(   )→ 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) B) 2 H<sub>2</sub>O<sub>2</sub>(   )+ 2e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) C) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) D) H<sub>2</sub>O<sub>2</sub>(   )+ 4 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ H<sub>2</sub>(g) E) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g) )+ 4 H+(aq)+ 2 e- → 2 H2O( <strong>Write a balanced half-reaction for the reduction of hydrogen peroxide to water in an acidic solution.</strong> A) 2 H<sub>2</sub>O<sub>2</sub>(   )→ 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) B) 2 H<sub>2</sub>O<sub>2</sub>(   )+ 2e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) C) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) D) H<sub>2</sub>O<sub>2</sub>(   )+ 4 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ H<sub>2</sub>(g) E) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g) )+ H2(g)
E) H2O2( <strong>Write a balanced half-reaction for the reduction of hydrogen peroxide to water in an acidic solution.</strong> A) 2 H<sub>2</sub>O<sub>2</sub>(   )→ 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) B) 2 H<sub>2</sub>O<sub>2</sub>(   )+ 2e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ O<sub>2</sub>(g) C) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) D) H<sub>2</sub>O<sub>2</sub>(   )+ 4 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → 2 H<sub>2</sub>O(   )+ H<sub>2</sub>(g) E) H<sub>2</sub>O<sub>2</sub>(   )+ 2 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g) )+ 2 H+(aq)+ 4 e- → 2 H2(g)+ O2(g)
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11
Which of the following reactions will require the use of an inert electrode when used in a voltaic cell?

A) Co(s)+ 2 Ag+(aq)→ 2 Ag(s)+ Co2+(aq)
B) 3 Ni(s)+ 2 Au3+(aq)→ 3 Ni2+ + 2 Au(s)
C) Cu(s)+ 2 Ag+(aq)→ Cu2+(aq)+ 2 Ag(s)
D) Zn(s)+ 2 MnO2(s)+ 2 NH4+(aq)→ Zn2+(aq)+ Mn2O3(s)+ 2 NH3(aq)+ H2O(l)
E) 3 Zn2+(aq)+ 2 Al(s)→ 3 Zn(s)+ 2 Al3+(aq)
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12
Write a balanced half-reaction for the reduction of CrO42-(aq)to Cr(OH)3(s)in a basic solution.

A) CrO42-(aq)+ 3 OH-(aq)+ 3 e- → Cr(OH)3(s)+ 2 O2(g)
B) CrO42-(aq)+ 3 H+(aq)+ 3 e- → Cr(OH)3(s)
C) CrO42-(aq)+ 3 H+(aq)→ Cr(OH)3(s)+ 2 e-
D) CrO42-(aq)+ 4 H2O( <strong>Write a balanced half-reaction for the reduction of CrO<sub>4</sub><sup>2-</sup>(aq)to Cr(OH)<sub>3</sub>(s)in a basic solution.</strong> A) CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 OH<sup>-</sup>(aq)+ 3 e<sup>-</sup> → Cr(OH)<sub>3</sub>(s)+ 2 O<sub>2</sub>(g) B) CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 H<sup>+</sup>(aq)+ 3 e<sup>-</sup> → Cr(OH)<sub>3</sub>(s) C) CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 H<sup>+</sup>(aq)→ Cr(OH)<sub>3</sub>(s)+ 2 e<sup>-</sup> D) CrO<sub>4</sub><sup>2-</sup>(aq)+ 4 H<sub>2</sub>O(   )+ 3 e<sup>-</sup> → Cr(OH)<sub>3</sub>(s)+ 5 OH<sup>-</sup>(aq) E) CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 OH<sup>-</sup>(aq)→ Cr(OH)<sub>3</sub>(s)+ 2 O<sub>2</sub>(g) )+ 3 e- → Cr(OH)3(s)+ 5 OH-(aq)
E) CrO42-(aq)+ 3 OH-(aq)→ Cr(OH)3(s)+ 2 O2(g)
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13
Which of the following statements is true for the following reaction,assuming the given reaction proceeds in the forward direction? 3 Sn4+(aq)+ 2 Cr(s)→ 3 Sn2+(aq)+ 2 Cr3+(aq)

A) Sn4+(aq)is the reducing agent and Cr(s)is the oxidizing agent.
B) Cr(s)is the reducing agent and Sn2+(aq)is the oxidizing agent.
C) Sn4+(aq)is the reducing agent and Sn2+(aq)is the oxidizing agent.
D) Cr(s)is the reducing agent and Cr3+(aq)is the oxidizing agent.
E) Cr(s)is the reducing agent and Sn4+(aq)is the oxidizing agent.
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14
Which of the following statements is true for the following reaction,assuming the given reaction proceeds in the forward direction? Fe3+(aq)+ Co(s)→ Fe2+(aq)+ Co2+(aq)

A) Fe3+(aq)is oxidized and Co(s)is reduced.
B) Fe3+(aq)is oxidized and Co2+(aq)is reduced.
C) Co(s)is oxidized and Fe3+(aq)is reduced.
D) Co(s)is oxidized and Co2+(aq)is reduced.
E) Fe2+(aq)is oxidized and Co(s)is reduced.
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15
How many electrons are transferred in the given reaction? Ni + 2 HCl → Ni Cl2 + H2

A) 0
B) 1
C) 2
D) 3
E) 4
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16
When the given oxidation-reduction reaction in an acidic solution is balanced,what is the lowest whole-number coefficient for H+,and on which side of the balanced equation should it appear? MnO4-(aq)+ Br-(aq)→ Mn2+(aq)+ Br2(l)

A) 1; reactant side
B) 2; product side
C) 8; reactant side
D) 16; reactant side
E) 4; product side
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17
Write the balanced reduction half-reaction for the following overall reaction: 2 Fe(s)+ 3 Cl2(aq)→ 2 Fe3+(aq)+ 6 Cl-(aq)

A) Cl2(aq)+ 2 e- → 2 Cl-(aq)
B) Fe(s)+ 3 e- → Fe3+(aq)
C) Fe(s)+ Cl2(aq)→ FeCl3(aq)
D) Cl2(aq)→ 2 Cl-(aq)+ 2 e-
E) 3 Cl2(aq)+ 2 e- → 2 Cl-(aq)
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18
Write a balanced chemical equation for the following reaction in a basic solution. ClO-(aq)+ Cr(OH)3(s)→ Cl-(aq)+ CrO42-(aq)

A) 3 ClO-(aq)+ 2 Cr(OH)3(s)+ 4 OH-(aq)→ 3 Cl-(aq)+ 2 CrO42-(aq)+ 5 H2O( <strong>Write a balanced chemical equation for the following reaction in a basic solution. ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)</strong> A) 3 ClO<sup>-</sup>(aq)+ 2 Cr(OH)<sub>3</sub>(s)+ 4 OH<sup>-</sup>(aq)→ 3 Cl<sup>-</sup>(aq)+ 2 CrO<sub>4</sub><sup>2-</sup>(aq)+ 5 H<sub>2</sub>O(   ) B) ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)+ 3 OH<sup>-</sup>(aq)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 H<sub>2</sub>O(   ) C) 2 ClO<sup>-</sup>(aq)+ 3 Cr(OH)<sub>3</sub>(s)+ 3 OH<sup>-</sup>(aq)→ 2 Cl<sup>-</sup>(aq)+ 3 CrO<sub>4</sub><sup>2-</sup>(aq)+ 6 H<sub>2</sub>O(   ) D) 4 ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)+ 4 OH<sup>-</sup>(aq)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 6 H<sub>2</sub>O(   ) E) ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 H<sup>+</sup>(aq) )
B) ClO-(aq)+ Cr(OH)3(s)+ 3 OH-(aq)→ Cl-(aq)+ CrO42-(aq)+ 3 H2O( <strong>Write a balanced chemical equation for the following reaction in a basic solution. ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)</strong> A) 3 ClO<sup>-</sup>(aq)+ 2 Cr(OH)<sub>3</sub>(s)+ 4 OH<sup>-</sup>(aq)→ 3 Cl<sup>-</sup>(aq)+ 2 CrO<sub>4</sub><sup>2-</sup>(aq)+ 5 H<sub>2</sub>O(   ) B) ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)+ 3 OH<sup>-</sup>(aq)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 H<sub>2</sub>O(   ) C) 2 ClO<sup>-</sup>(aq)+ 3 Cr(OH)<sub>3</sub>(s)+ 3 OH<sup>-</sup>(aq)→ 2 Cl<sup>-</sup>(aq)+ 3 CrO<sub>4</sub><sup>2-</sup>(aq)+ 6 H<sub>2</sub>O(   ) D) 4 ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)+ 4 OH<sup>-</sup>(aq)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 6 H<sub>2</sub>O(   ) E) ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 H<sup>+</sup>(aq) )
C) 2 ClO-(aq)+ 3 Cr(OH)3(s)+ 3 OH-(aq)→ 2 Cl-(aq)+ 3 CrO42-(aq)+ 6 H2O( <strong>Write a balanced chemical equation for the following reaction in a basic solution. ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)</strong> A) 3 ClO<sup>-</sup>(aq)+ 2 Cr(OH)<sub>3</sub>(s)+ 4 OH<sup>-</sup>(aq)→ 3 Cl<sup>-</sup>(aq)+ 2 CrO<sub>4</sub><sup>2-</sup>(aq)+ 5 H<sub>2</sub>O(   ) B) ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)+ 3 OH<sup>-</sup>(aq)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 H<sub>2</sub>O(   ) C) 2 ClO<sup>-</sup>(aq)+ 3 Cr(OH)<sub>3</sub>(s)+ 3 OH<sup>-</sup>(aq)→ 2 Cl<sup>-</sup>(aq)+ 3 CrO<sub>4</sub><sup>2-</sup>(aq)+ 6 H<sub>2</sub>O(   ) D) 4 ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)+ 4 OH<sup>-</sup>(aq)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 6 H<sub>2</sub>O(   ) E) ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 H<sup>+</sup>(aq) )
D) 4 ClO-(aq)+ Cr(OH)3(s)+ 4 OH-(aq)→ Cl-(aq)+ CrO42-(aq)+ 6 H2O( <strong>Write a balanced chemical equation for the following reaction in a basic solution. ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)</strong> A) 3 ClO<sup>-</sup>(aq)+ 2 Cr(OH)<sub>3</sub>(s)+ 4 OH<sup>-</sup>(aq)→ 3 Cl<sup>-</sup>(aq)+ 2 CrO<sub>4</sub><sup>2-</sup>(aq)+ 5 H<sub>2</sub>O(   ) B) ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)+ 3 OH<sup>-</sup>(aq)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 H<sub>2</sub>O(   ) C) 2 ClO<sup>-</sup>(aq)+ 3 Cr(OH)<sub>3</sub>(s)+ 3 OH<sup>-</sup>(aq)→ 2 Cl<sup>-</sup>(aq)+ 3 CrO<sub>4</sub><sup>2-</sup>(aq)+ 6 H<sub>2</sub>O(   ) D) 4 ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)+ 4 OH<sup>-</sup>(aq)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 6 H<sub>2</sub>O(   ) E) ClO<sup>-</sup>(aq)+ Cr(OH)<sub>3</sub>(s)→ Cl<sup>-</sup>(aq)+ CrO<sub>4</sub><sup>2-</sup>(aq)+ 3 H<sup>+</sup>(aq) )
E) ClO-(aq)+ Cr(OH)3(s)→ Cl-(aq)+ CrO42-(aq)+ 3 H+(aq)
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19
In the context of the diagram given below,which of the following statements is true concerning half-cell II? <strong>In the context of the diagram given below,which of the following statements is true concerning half-cell II?  </strong> A) [Cu<sup>2+</sup>] decreases with time,and [K+] increases with time. B) [Cu<sup>2+</sup>] increases with time,and [K+] increases with time. C) [Cu<sup>2+</sup>] decreases with time,and [K+] decreases with time. D) [Cu<sup>2+</sup>] decreases with time,and [SO<sub>4</sub>2-] decreases with time. E) [Cu<sup>2+</sup>] increases with time,and [SO<sub>4</sub>2-] increases with time.

A) [Cu2+] decreases with time,and [K+] increases with time.
B) [Cu2+] increases with time,and [K+] increases with time.
C) [Cu2+] decreases with time,and [K+] decreases with time.
D) [Cu2+] decreases with time,and [SO42-] decreases with time.
E) [Cu2+] increases with time,and [SO42-] increases with time.
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20
Which of the following is the cell notation for a voltaic cell based on the following reaction? Cu2+(aq)+ Fe(s)→ Cu(s)+ Fe2+(aq)

A) Cu(s)| Cu2+(aq)|| Fe2+(aq)| Fe(s)
B) Fe(s)|| Fe2+(aq),Cu2+(aq)| Cu(s)
C) Cu(s)|| Cu2+(aq),Fe2+(aq)|| Fe(s)
D) Cu(s)| Fe2+(aq)|| Cu2+(aq)| Fe(s)
E) Fe(s)| Fe2+(aq)|| Cu2+(aq)| Cu(s)
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21
The electrochemical reaction that powers a lead-acid storage battery is as follows: Pb(s)+ PbO2(s)+ 4 H+(aq)+ 2 SO42-(aq)→ 2 PbSO4(s)+ 2 H2O( <strong>The electrochemical reaction that powers a lead-acid storage battery is as follows: Pb(s)+ PbO<sub>2</sub>(s)+ 4 H<sup>+</sup>(aq)+ 2 SO<sub>4</sub><sup>2-</sup>(aq)→ 2 PbSO<sub>4</sub>(s)+ 2 H<sub>2</sub>O(   ) A single cell of this battery consists of a Pb electrode and a PbO<sub>2</sub> electrode,each submerged in sulfuric acid.Which of the following reactions occurs at the cathode during discharge?</strong> A) Pb(s)is reduced to PbSO<sub>4</sub>(s). B) PbO<sub>2</sub>(s)is reduced to PbSO<sub>4</sub>(s). C) PbO<sub>2</sub>(s)is oxidized to PbSO<sub>4</sub>(s). D) Pb(s)is oxidized to PbSO<sub>4</sub>(s). E) H<sup>+</sup> is oxidized to H<sub>2</sub>O(   ). )
A single cell of this battery consists of a Pb electrode and a PbO2 electrode,each submerged in sulfuric acid.Which of the following reactions occurs at the cathode during discharge?

A) Pb(s)is reduced to PbSO4(s).
B) PbO2(s)is reduced to PbSO4(s).
C) PbO2(s)is oxidized to PbSO4(s).
D) Pb(s)is oxidized to PbSO4(s).
E) H+ is oxidized to H2O( <strong>The electrochemical reaction that powers a lead-acid storage battery is as follows: Pb(s)+ PbO<sub>2</sub>(s)+ 4 H<sup>+</sup>(aq)+ 2 SO<sub>4</sub><sup>2-</sup>(aq)→ 2 PbSO<sub>4</sub>(s)+ 2 H<sub>2</sub>O(   ) A single cell of this battery consists of a Pb electrode and a PbO<sub>2</sub> electrode,each submerged in sulfuric acid.Which of the following reactions occurs at the cathode during discharge?</strong> A) Pb(s)is reduced to PbSO<sub>4</sub>(s). B) PbO<sub>2</sub>(s)is reduced to PbSO<sub>4</sub>(s). C) PbO<sub>2</sub>(s)is oxidized to PbSO<sub>4</sub>(s). D) Pb(s)is oxidized to PbSO<sub>4</sub>(s). E) H<sup>+</sup> is oxidized to H<sub>2</sub>O(   ). ).
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22
Given the following two half-reactions,write the overall reaction in the direction in which it is product-favored,and calculate the standard cell potential. Pb2+(aq)+ 2 e- → Pb(s)
E° = -0.126 V
Fe3+(aq)+ e- → Fe2+(s)
E° = +0.771 V

A) Pb2+(aq)+ 2 Fe2+(s)→ Pb(s)+ 2 Fe3+(aq); <strong>Given the following two half-reactions,write the overall reaction in the direction in which it is product-favored,and calculate the standard cell potential. Pb<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Pb(s) E° = -0.126 V Fe<sup>3+</sup>(aq)+ e<sup>-</sup> → Fe<sup>2+</sup>(s) E° = +0.771 V</strong> A) Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s)→ Pb(s)+ 2 Fe<sup>3+</sup>(aq);   = +0.897 V B) Pb<sup>2+</sup>(aq)+ Fe<sup>2+</sup>(s)→ Pb(s)+ Fe<sup>3+</sup>(aq);   = +0.645 V C) Pb(s)+ 2 Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s);   = +1.416 V D) Pb(s)+ 2 Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s);   = +0.897 V E) Pb(s)+ Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ Fe<sup>2+</sup>(s);   = +0.645 V = +0.897 V
B) Pb2+(aq)+ Fe2+(s)→ Pb(s)+ Fe3+(aq); <strong>Given the following two half-reactions,write the overall reaction in the direction in which it is product-favored,and calculate the standard cell potential. Pb<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Pb(s) E° = -0.126 V Fe<sup>3+</sup>(aq)+ e<sup>-</sup> → Fe<sup>2+</sup>(s) E° = +0.771 V</strong> A) Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s)→ Pb(s)+ 2 Fe<sup>3+</sup>(aq);   = +0.897 V B) Pb<sup>2+</sup>(aq)+ Fe<sup>2+</sup>(s)→ Pb(s)+ Fe<sup>3+</sup>(aq);   = +0.645 V C) Pb(s)+ 2 Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s);   = +1.416 V D) Pb(s)+ 2 Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s);   = +0.897 V E) Pb(s)+ Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ Fe<sup>2+</sup>(s);   = +0.645 V = +0.645 V
C) Pb(s)+ 2 Fe3+(aq)→ Pb2+(aq)+ 2 Fe2+(s); <strong>Given the following two half-reactions,write the overall reaction in the direction in which it is product-favored,and calculate the standard cell potential. Pb<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Pb(s) E° = -0.126 V Fe<sup>3+</sup>(aq)+ e<sup>-</sup> → Fe<sup>2+</sup>(s) E° = +0.771 V</strong> A) Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s)→ Pb(s)+ 2 Fe<sup>3+</sup>(aq);   = +0.897 V B) Pb<sup>2+</sup>(aq)+ Fe<sup>2+</sup>(s)→ Pb(s)+ Fe<sup>3+</sup>(aq);   = +0.645 V C) Pb(s)+ 2 Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s);   = +1.416 V D) Pb(s)+ 2 Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s);   = +0.897 V E) Pb(s)+ Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ Fe<sup>2+</sup>(s);   = +0.645 V = +1.416 V
D) Pb(s)+ 2 Fe3+(aq)→ Pb2+(aq)+ 2 Fe2+(s); <strong>Given the following two half-reactions,write the overall reaction in the direction in which it is product-favored,and calculate the standard cell potential. Pb<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Pb(s) E° = -0.126 V Fe<sup>3+</sup>(aq)+ e<sup>-</sup> → Fe<sup>2+</sup>(s) E° = +0.771 V</strong> A) Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s)→ Pb(s)+ 2 Fe<sup>3+</sup>(aq);   = +0.897 V B) Pb<sup>2+</sup>(aq)+ Fe<sup>2+</sup>(s)→ Pb(s)+ Fe<sup>3+</sup>(aq);   = +0.645 V C) Pb(s)+ 2 Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s);   = +1.416 V D) Pb(s)+ 2 Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s);   = +0.897 V E) Pb(s)+ Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ Fe<sup>2+</sup>(s);   = +0.645 V = +0.897 V
E) Pb(s)+ Fe3+(aq)→ Pb2+(aq)+ Fe2+(s); <strong>Given the following two half-reactions,write the overall reaction in the direction in which it is product-favored,and calculate the standard cell potential. Pb<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Pb(s) E° = -0.126 V Fe<sup>3+</sup>(aq)+ e<sup>-</sup> → Fe<sup>2+</sup>(s) E° = +0.771 V</strong> A) Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s)→ Pb(s)+ 2 Fe<sup>3+</sup>(aq);   = +0.897 V B) Pb<sup>2+</sup>(aq)+ Fe<sup>2+</sup>(s)→ Pb(s)+ Fe<sup>3+</sup>(aq);   = +0.645 V C) Pb(s)+ 2 Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s);   = +1.416 V D) Pb(s)+ 2 Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ 2 Fe<sup>2+</sup>(s);   = +0.897 V E) Pb(s)+ Fe<sup>3+</sup>(aq)→ Pb<sup>2+</sup>(aq)+ Fe<sup>2+</sup>(s);   = +0.645 V = +0.645 V
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23
Write a balanced net ionic equation for the overall reaction represented by the following cell notation. Al(s)| Al3+(aq)|| Cl2(g)| Cl-(aq)| Pt(s)

A) 2 Al(s)+ 3 Cl2(g)→ 2 Al3+(aq)+ 6 Cl-(aq)
B) Al(s)+ Al3+(aq)→ Cl-(aq)+ Cl2(g)
C) 2 Al3+(aq)+ 6 Cl-(aq)→ 2 Al(s)+ 3 Cl2(g)
D) Al(s)+ 3 Cl2(g)→ Al3+(s)+ 2 Cl-(aq)
E) Al(s)+ 2 Cl-(aq)→ Cl2(g)+ Al3+(aq)
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24
An SHE electrode has been assigned a standard reduction potential,E°,of 0.00 volts.Which of the following reactions will occur at this electrode?

A) 2 H2O( <strong>An SHE electrode has been assigned a standard reduction potential,E°,of 0.00 volts.Which of the following reactions will occur at this electrode?</strong> A) 2 H<sub>2</sub>O(   )+ 2 e<sup>-</sup> → H<sub>2</sub>(g)+ 2 OH<sup>-</sup>(aq) B) O<sub>2</sub>(g)+ 4 e<sup>-</sup> → 2 O<sup>2-</sup>(aq) C) Hg<sub>2</sub>Cl<sub>2</sub>(s)+ 2 e<sup>-</sup> → 2 Hg(   )+ 2 Cl<sup>-</sup>(aq) D) Li<sup>+</sup>(aq)+ e<sup>-</sup> → Li(s) E) 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → H<sub>2</sub>(g) )+ 2 e- → H2(g)+ 2 OH-(aq)
B) O2(g)+ 4 e- → 2 O2-(aq)
C) Hg2Cl2(s)+ 2 e- → 2 Hg( <strong>An SHE electrode has been assigned a standard reduction potential,E°,of 0.00 volts.Which of the following reactions will occur at this electrode?</strong> A) 2 H<sub>2</sub>O(   )+ 2 e<sup>-</sup> → H<sub>2</sub>(g)+ 2 OH<sup>-</sup>(aq) B) O<sub>2</sub>(g)+ 4 e<sup>-</sup> → 2 O<sup>2-</sup>(aq) C) Hg<sub>2</sub>Cl<sub>2</sub>(s)+ 2 e<sup>-</sup> → 2 Hg(   )+ 2 Cl<sup>-</sup>(aq) D) Li<sup>+</sup>(aq)+ e<sup>-</sup> → Li(s) E) 2 H<sup>+</sup>(aq)+ 2 e<sup>-</sup> → H<sub>2</sub>(g) )+ 2 Cl-(aq)
D) Li+(aq)+ e- → Li(s)
E) 2 H+(aq)+ 2 e- → H2(g)
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25
Calculate the standard cell potential ( <strong>Calculate the standard cell potential (   )for the reaction 2 Ag(s)+ Co<sup>2+</sup>(aq)→ 2 Ag<sup>+</sup>(aq)+ Co(s). The standard reduction potentials are as follows: Ag<sup>+</sup>(aq)+ e<sup>−</sup>→ Ag(s) E° = 0.8 V Co<sup>2+</sup>(aq)+2 e<sup>−</sup>→ Co(s) E° = -0.277 V</strong> A) -1.077 V B) 1.077 V C) -1.877 V D) 1.877 V E) 1.323 V )for the reaction 2 Ag(s)+ Co2+(aq)→ 2 Ag+(aq)+ Co(s). The standard reduction potentials are as follows:
Ag+(aq)+ e→ Ag(s)
E° = 0.8 V
Co2+(aq)+2 e→ Co(s)
E° = -0.277 V

A) -1.077 V
B) 1.077 V
C) -1.877 V
D) 1.877 V
E) 1.323 V
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26
In the given electrochemical cell,which of the following is the cathode half-reaction? Zn(s)| Zn2+(aq)|| Fe3+(aq),Fe2+(aq)| Pt(s)

A) Fe3+(aq)+ e− → Fe2+(aq)
B) Fe2+(aq)+ e− → Fe3+(aq)
C) Fe2+(aq)+ Pt(s)→ Fe3+(aq)+ e−
D) Zn2+(aq)→ Zn(s)+ 2 e−
E) Zn(s)→ Zn2+(aq)+ 2 e−
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27
Calculate E°cell for the cell for the reaction 2 Ga(s)+ 3 Sn4+(aq)→ 3 Sn2+(aq)+2 Ga3+(aq). The standard reduction potentials are as follows:
Ga3+(aq)+ 3 e− → Ga(s)
E° = -0.55 V
Sn4+(aq)+ 2 e− → Sn2+(aq)
E° = 0.15V

A) 0.70 V
B) 0.40 V
C) 0.33 V
D) -0.40 V
E) −0.70 V
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28
Use the following standard reduction potentials to determine which species is the strongest reducing agent. 2 H+(aq)+ 2 e- → H2(g); 0.00 V
K+(aq)+ e- → K(s); -2.93 V
F2(g)+ 2 e- → 2 F-(aq); 2.87 V
Al3+(aq)+ 3 e- → Al(s); -1.66 V
Pb2+(aq)+ 2 e- → Pb(s); -0.13 V

A) Al3+
B) H+
C) Pb2+
D) F-
E) K
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29
Which of the following species are likely to behave as oxidizing agents? Li(s),H2(g),MnO4-(aq),and Cl-(aq)

A) Li(s)
B) MnO4-(aq)
C) H2(g)and Cl-(aq)
D) Li(s)and MnO4-(aq)
E) Cl-(aq)
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30
The unit for electromotive force,emf,is the volt.1 volt is equal to _____.

A) one joule per second
B) one coulomb per joule
C) one joule per coulomb
D) one coulomb per second
E) one calorie per second
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31
According to the cell notation below,which of the following species is undergoing reduction? Ni | Ni2+(aq)|| Mn2+(aq)| MnO2(s)| Pt(s)

A) Mn2+(aq)
B) Ni2+(aq)
C) Ni(s)
D) MnO2(s)
E) Pt(s)
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32
Use the following standard reduction potentials to determine which species is the strongest oxidizing agent. Fe2+(aq)+ 2 e- → Fe(s); E° = -0.41 V
Pt2+(aq)+ 2 e- → Pt(s); E° = 1.18 V
Cr2O72-(aq)+ 14 H+(aq)+ 6 e- → 2 Cr3+(aq)+ 7 H2O( <strong>Use the following standard reduction potentials to determine which species is the strongest oxidizing agent. Fe<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Fe(s); E° = -0.41 V Pt<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Pt(s); E° = 1.18 V Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>(aq)+ 14 H<sup>+</sup>(aq)+ 6 e<sup>-</sup> → 2 Cr<sup>3+</sup>(aq)+ 7 H<sub>2</sub>O(   ); E° = 1.33 V</strong> A) Fe B) Pt C) Cr<sup>3+</sup> D) Fe<sup>2+</sup> E) Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup> ); E° = 1.33 V

A) Fe
B) Pt
C) Cr3+
D) Fe2+
E) Cr2O72-
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33
Which of the following is the balanced overall reaction and standard cell potential of an electrochemical cell constructed from half-cells with the given half reactions? Pt2+(aq)+ 2 e− → Pt(s); E° = 1.180 V
Pb2+(aq)+ 2 e− → Pb(s); E° = -0.130 V

A) Pt2+(aq)+ Pb(s)→ Pt(s)+ Pb2+(aq); <strong>Which of the following is the balanced overall reaction and standard cell potential of an electrochemical cell constructed from half-cells with the given half reactions? Pt<sup>2+</sup>(aq)+ 2 e− → Pt(s); E° = 1.180 V Pb<sup>2+</sup>(aq)+ 2 e− → Pb(s); E° = -0.130 V</strong> A) Pt<sup>2+</sup>(aq)+ Pb(s)→ Pt(s)+ Pb<sup>2+</sup>(aq);   = 1.310 V B) Pt(s)+ Pb<sup>2+</sup>(aq)→ Pt<sup>2+</sup>(aq)+ Pb(s);   = −1.310 V C) Pt<sup>2+</sup>(aq)+ Pb<sup>2+</sup>(aq)→ Pt(s)+ Pb(s);   = 1.050 V D) Pt<sup>2+</sup>(aq)+ Pb(s)→ Pt(s)+ Pb<sup>2+</sup>(aq);   =0.655 V E) Pt(s)+ Pb<sup>2+</sup>(aq)→ Pt<sup>2+</sup>(aq)+ Pb(s);   = −0.655 V = 1.310 V
B) Pt(s)+ Pb2+(aq)→ Pt2+(aq)+ Pb(s); <strong>Which of the following is the balanced overall reaction and standard cell potential of an electrochemical cell constructed from half-cells with the given half reactions? Pt<sup>2+</sup>(aq)+ 2 e− → Pt(s); E° = 1.180 V Pb<sup>2+</sup>(aq)+ 2 e− → Pb(s); E° = -0.130 V</strong> A) Pt<sup>2+</sup>(aq)+ Pb(s)→ Pt(s)+ Pb<sup>2+</sup>(aq);   = 1.310 V B) Pt(s)+ Pb<sup>2+</sup>(aq)→ Pt<sup>2+</sup>(aq)+ Pb(s);   = −1.310 V C) Pt<sup>2+</sup>(aq)+ Pb<sup>2+</sup>(aq)→ Pt(s)+ Pb(s);   = 1.050 V D) Pt<sup>2+</sup>(aq)+ Pb(s)→ Pt(s)+ Pb<sup>2+</sup>(aq);   =0.655 V E) Pt(s)+ Pb<sup>2+</sup>(aq)→ Pt<sup>2+</sup>(aq)+ Pb(s);   = −0.655 V = −1.310 V
C) Pt2+(aq)+ Pb2+(aq)→ Pt(s)+ Pb(s); <strong>Which of the following is the balanced overall reaction and standard cell potential of an electrochemical cell constructed from half-cells with the given half reactions? Pt<sup>2+</sup>(aq)+ 2 e− → Pt(s); E° = 1.180 V Pb<sup>2+</sup>(aq)+ 2 e− → Pb(s); E° = -0.130 V</strong> A) Pt<sup>2+</sup>(aq)+ Pb(s)→ Pt(s)+ Pb<sup>2+</sup>(aq);   = 1.310 V B) Pt(s)+ Pb<sup>2+</sup>(aq)→ Pt<sup>2+</sup>(aq)+ Pb(s);   = −1.310 V C) Pt<sup>2+</sup>(aq)+ Pb<sup>2+</sup>(aq)→ Pt(s)+ Pb(s);   = 1.050 V D) Pt<sup>2+</sup>(aq)+ Pb(s)→ Pt(s)+ Pb<sup>2+</sup>(aq);   =0.655 V E) Pt(s)+ Pb<sup>2+</sup>(aq)→ Pt<sup>2+</sup>(aq)+ Pb(s);   = −0.655 V = 1.050 V
D) Pt2+(aq)+ Pb(s)→ Pt(s)+ Pb2+(aq); <strong>Which of the following is the balanced overall reaction and standard cell potential of an electrochemical cell constructed from half-cells with the given half reactions? Pt<sup>2+</sup>(aq)+ 2 e− → Pt(s); E° = 1.180 V Pb<sup>2+</sup>(aq)+ 2 e− → Pb(s); E° = -0.130 V</strong> A) Pt<sup>2+</sup>(aq)+ Pb(s)→ Pt(s)+ Pb<sup>2+</sup>(aq);   = 1.310 V B) Pt(s)+ Pb<sup>2+</sup>(aq)→ Pt<sup>2+</sup>(aq)+ Pb(s);   = −1.310 V C) Pt<sup>2+</sup>(aq)+ Pb<sup>2+</sup>(aq)→ Pt(s)+ Pb(s);   = 1.050 V D) Pt<sup>2+</sup>(aq)+ Pb(s)→ Pt(s)+ Pb<sup>2+</sup>(aq);   =0.655 V E) Pt(s)+ Pb<sup>2+</sup>(aq)→ Pt<sup>2+</sup>(aq)+ Pb(s);   = −0.655 V =0.655 V
E) Pt(s)+ Pb2+(aq)→ Pt2+(aq)+ Pb(s); <strong>Which of the following is the balanced overall reaction and standard cell potential of an electrochemical cell constructed from half-cells with the given half reactions? Pt<sup>2+</sup>(aq)+ 2 e− → Pt(s); E° = 1.180 V Pb<sup>2+</sup>(aq)+ 2 e− → Pb(s); E° = -0.130 V</strong> A) Pt<sup>2+</sup>(aq)+ Pb(s)→ Pt(s)+ Pb<sup>2+</sup>(aq);   = 1.310 V B) Pt(s)+ Pb<sup>2+</sup>(aq)→ Pt<sup>2+</sup>(aq)+ Pb(s);   = −1.310 V C) Pt<sup>2+</sup>(aq)+ Pb<sup>2+</sup>(aq)→ Pt(s)+ Pb(s);   = 1.050 V D) Pt<sup>2+</sup>(aq)+ Pb(s)→ Pt(s)+ Pb<sup>2+</sup>(aq);   =0.655 V E) Pt(s)+ Pb<sup>2+</sup>(aq)→ Pt<sup>2+</sup>(aq)+ Pb(s);   = −0.655 V = −0.655 V
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34
Consider the following half-reactions. Ag+(aq)+ e- → Ag(s)
E° = +0.80 V
Cu2+(aq)+ 2 e- → Cu(s)
E° = +0.34 V
Pb2+(aq)+ 2 e- → Pb(s)
E° = -0.13 V
Fe2+(aq)+ 2 e- → Fe(s)
E° = -0.44 V
Al3+(aq)+ 3 e- → Al(s)
E° = -1.66 V
Which of the following species will oxidize lead,Pb(s)?

A) Ag+(aq)and Cu2+(aq)
B) Ag(s)and Cu(s)
C) Fe2+(aq)and Al3+(aq)
D) Fe(s)and Al(s)
E) Cu2+(aq)and Fe2+(aq)
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35
Use the following standard reduction potentials to determine which species is the best oxidizing agent. O2(g)+ 4 H+(aq)+ 4 e- → 2 H2O( <strong>Use the following standard reduction potentials to determine which species is the best oxidizing agent. O<sub>2</sub>(g)+ 4 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) E° = +1.229 V Hg<sub>2</sub><sup>2+</sup>(aq)+ 2 e<sup>-</sup> → 2 Hg(   ) E° = +0.789 V I<sub>2</sub>(s)+ 2 e<sup>-</sup> → 2 I<sup>-</sup>(aq) E° = +0.535 V</strong> A) I<sub>2</sub>(s) B) O<sub>2</sub>(g) C) I<sup>-</sup>(aq) D) Hg<sub>2</sub><sup>2+</sup>(aq) E) H<sub>2</sub>O(   ) )
E° = +1.229 V
Hg22+(aq)+ 2 e- → 2 Hg( <strong>Use the following standard reduction potentials to determine which species is the best oxidizing agent. O<sub>2</sub>(g)+ 4 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) E° = +1.229 V Hg<sub>2</sub><sup>2+</sup>(aq)+ 2 e<sup>-</sup> → 2 Hg(   ) E° = +0.789 V I<sub>2</sub>(s)+ 2 e<sup>-</sup> → 2 I<sup>-</sup>(aq) E° = +0.535 V</strong> A) I<sub>2</sub>(s) B) O<sub>2</sub>(g) C) I<sup>-</sup>(aq) D) Hg<sub>2</sub><sup>2+</sup>(aq) E) H<sub>2</sub>O(   ) )
E° = +0.789 V
I2(s)+ 2 e- → 2 I-(aq)
E° = +0.535 V

A) I2(s)
B) O2(g)
C) I-(aq)
D) Hg22+(aq)
E) H2O( <strong>Use the following standard reduction potentials to determine which species is the best oxidizing agent. O<sub>2</sub>(g)+ 4 H<sup>+</sup>(aq)+ 4 e<sup>-</sup> → 2 H<sub>2</sub>O(   ) E° = +1.229 V Hg<sub>2</sub><sup>2+</sup>(aq)+ 2 e<sup>-</sup> → 2 Hg(   ) E° = +0.789 V I<sub>2</sub>(s)+ 2 e<sup>-</sup> → 2 I<sup>-</sup>(aq) E° = +0.535 V</strong> A) I<sub>2</sub>(s) B) O<sub>2</sub>(g) C) I<sup>-</sup>(aq) D) Hg<sub>2</sub><sup>2+</sup>(aq) E) H<sub>2</sub>O(   ) )
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36
Consider the following half-reactions. Cl2(g)+ 2 e- → 2 Cl-(aq)
E° = +1.36 V
Ag+(aq)+ e- → Ag(s)
E° = +0.80 V
Cu2+(aq)+ 2 e- → Cu(s)
E° = +0.34 V
Sn2+(aq)+ 2 e- → Sn(s)
E° = -0.14 V
Al3+(aq)+ 3 e- → Al(s)
E° = -1.66 V
Which of the following species will reduce Cu2+(aq)ion?

A) Ag(s)and Sn2+(aq)
B) Cl-(aq)and Ag(s)
C) Cl2(g)and Ag+(aq)
D) Sn(s)and Al(s)
E) Sn2+(aq)and Al3+(aq)
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37
Which of the following is the cell notation for a voltaic cell based on the following reaction? Cu2+(aq)+ Pb(s)+ SO42-(aq)→ Cu(s)+ PbSO4(s)

A) Pb(s)| PbSO4(s)|| Cu2+(aq)|| Cu(s)
B) Cu(s)| Cu2+(aq)|| SO42-(aq)| PbSO4(s)| Pb(s)
C) Cu(s)| Cu2+(aq),SO42-(aq)| PbSO4(s)| Pb(s)
D) Cu(s)| Cu2+(aq),SO42-(aq)|| PbSO4(s)|| Pb(s)
E) Pb(s)| PbSO4(s)| SO42-(aq)|| Cu2+(aq)|| Cu(s)
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38
Write a balanced net ionic equation for the overall reaction represented by the following cell notation. Cu(s)| Cu2+(aq)|| Mn2+(aq)| Mn(s)

A) 2 Cu(s)+ Mn2+(aq)→ Mn(s)+ 2 Cu2+(aq)
B) Cu(s)+ Cu2+(aq)→ Mn(s)+ Mn2+(aq)
C) Cu(s)+ Mn2+(aq)→ Mn(s)+ Cu2+(aq)
D) 2 Mn(s)+ Cu2+(aq)→ Cu(s)+ 2Mn2+(aq)
E) Mn(s)+ Cu2+(aq)→ Cu(s)+ Mn2+(aq)
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39
Which of the following is the cell notation for a cell in which the hydrogen electrode is the anode and the cathode half-reaction is Co3+(aq)+ e− → Co2+(aq)?

A) Pt(s)| H2(g)| H+(aq)|| Co3+(aq),Co2+(aq)| Pt(s)
B) Pt(s)| H2(g)| H+(aq)|| Co3+(aq),Co2+(aq)
C) Co2+(aq),Co3+(aq)|| H+(aq)| H2(g)| Pt(s)
D) Pt(s)| Co2+(aq),Co3+(aq)|| H+(aq)| H2(g)| Pt(s)
E) H2(g)| H+(aq)|| Co2+(aq),Co3+(aq)
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40
Which of the following overall chemical equations is responsible for generating electricity in fuel cells used in NASA's Space Shuttle programs?

A) Fe2O3(s)+ Al(s)→ Fe(s)+ Al2O3(s)
B) Pb(s)+ PbO2(s)+ 2 H2SO4(aq)→ 2 PbSO4(s)+ 2 H2O( <strong>Which of the following overall chemical equations is responsible for generating electricity in fuel cells used in NASA's Space Shuttle programs?</strong> A) Fe<sub>2</sub>O<sub>3</sub>(s)+ Al(s)→ Fe(s)+ Al<sub>2</sub>O<sub>3</sub>(s) B) Pb(s)+ PbO<sub>2</sub>(s)+ 2 H<sub>2</sub>SO<sub>4</sub>(aq)→ 2 PbSO<sub>4</sub>(s)+ 2 H<sub>2</sub>O(   ) C) 2 NiO(OH)(s)+ Cd(s)+ H<sub>2</sub>O(   )→ 2 Ni(OH)<sub>2</sub>(s)+ Cd(OH)<sub>2</sub>(s) D) 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g)→ 2 H<sub>2</sub>O(   ) E) N<sub>2</sub>H<sub>4</sub>(   )+ O<sub>2</sub>(g)→ N<sub>2</sub>(g)+ 2 H<sub>2</sub>O(   ) )
C) 2 NiO(OH)(s)+ Cd(s)+ H2O( <strong>Which of the following overall chemical equations is responsible for generating electricity in fuel cells used in NASA's Space Shuttle programs?</strong> A) Fe<sub>2</sub>O<sub>3</sub>(s)+ Al(s)→ Fe(s)+ Al<sub>2</sub>O<sub>3</sub>(s) B) Pb(s)+ PbO<sub>2</sub>(s)+ 2 H<sub>2</sub>SO<sub>4</sub>(aq)→ 2 PbSO<sub>4</sub>(s)+ 2 H<sub>2</sub>O(   ) C) 2 NiO(OH)(s)+ Cd(s)+ H<sub>2</sub>O(   )→ 2 Ni(OH)<sub>2</sub>(s)+ Cd(OH)<sub>2</sub>(s) D) 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g)→ 2 H<sub>2</sub>O(   ) E) N<sub>2</sub>H<sub>4</sub>(   )+ O<sub>2</sub>(g)→ N<sub>2</sub>(g)+ 2 H<sub>2</sub>O(   ) )→ 2 Ni(OH)2(s)+ Cd(OH)2(s)
D) 2 H2(g)+ O2(g)→ 2 H2O( <strong>Which of the following overall chemical equations is responsible for generating electricity in fuel cells used in NASA's Space Shuttle programs?</strong> A) Fe<sub>2</sub>O<sub>3</sub>(s)+ Al(s)→ Fe(s)+ Al<sub>2</sub>O<sub>3</sub>(s) B) Pb(s)+ PbO<sub>2</sub>(s)+ 2 H<sub>2</sub>SO<sub>4</sub>(aq)→ 2 PbSO<sub>4</sub>(s)+ 2 H<sub>2</sub>O(   ) C) 2 NiO(OH)(s)+ Cd(s)+ H<sub>2</sub>O(   )→ 2 Ni(OH)<sub>2</sub>(s)+ Cd(OH)<sub>2</sub>(s) D) 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g)→ 2 H<sub>2</sub>O(   ) E) N<sub>2</sub>H<sub>4</sub>(   )+ O<sub>2</sub>(g)→ N<sub>2</sub>(g)+ 2 H<sub>2</sub>O(   ) )
E) N2H4( <strong>Which of the following overall chemical equations is responsible for generating electricity in fuel cells used in NASA's Space Shuttle programs?</strong> A) Fe<sub>2</sub>O<sub>3</sub>(s)+ Al(s)→ Fe(s)+ Al<sub>2</sub>O<sub>3</sub>(s) B) Pb(s)+ PbO<sub>2</sub>(s)+ 2 H<sub>2</sub>SO<sub>4</sub>(aq)→ 2 PbSO<sub>4</sub>(s)+ 2 H<sub>2</sub>O(   ) C) 2 NiO(OH)(s)+ Cd(s)+ H<sub>2</sub>O(   )→ 2 Ni(OH)<sub>2</sub>(s)+ Cd(OH)<sub>2</sub>(s) D) 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g)→ 2 H<sub>2</sub>O(   ) E) N<sub>2</sub>H<sub>4</sub>(   )+ O<sub>2</sub>(g)→ N<sub>2</sub>(g)+ 2 H<sub>2</sub>O(   ) )+ O2(g)→ N2(g)+ 2 H2O( <strong>Which of the following overall chemical equations is responsible for generating electricity in fuel cells used in NASA's Space Shuttle programs?</strong> A) Fe<sub>2</sub>O<sub>3</sub>(s)+ Al(s)→ Fe(s)+ Al<sub>2</sub>O<sub>3</sub>(s) B) Pb(s)+ PbO<sub>2</sub>(s)+ 2 H<sub>2</sub>SO<sub>4</sub>(aq)→ 2 PbSO<sub>4</sub>(s)+ 2 H<sub>2</sub>O(   ) C) 2 NiO(OH)(s)+ Cd(s)+ H<sub>2</sub>O(   )→ 2 Ni(OH)<sub>2</sub>(s)+ Cd(OH)<sub>2</sub>(s) D) 2 H<sub>2</sub>(g)+ O<sub>2</sub>(g)→ 2 H<sub>2</sub>O(   ) E) N<sub>2</sub>H<sub>4</sub>(   )+ O<sub>2</sub>(g)→ N<sub>2</sub>(g)+ 2 H<sub>2</sub>O(   ) )
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41
The standard reduction potentials are as follows: Cr3+(aq)+ 3 e- → Cr(s); E° = -0.74 V
Fe2+(aq)+ 2 e- → Fe(s); E° = -0.41 V
Calculate the standard Gibbs free energy change for the following reaction.
2 Cr(s)+ 3 Fe2+ → 3 Fe(s)+ 2 Cr3+(aq)

A) 191 kJ
B) 63.7 kJ
C) -504 kJ
D) -191 kJ
E) 1060 kJ
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42
Which of the following equations represents the Nernst equation?

A) <strong>Which of the following equations represents the Nernst equation?</strong> A)   B)   C)   D)   E)
B) <strong>Which of the following equations represents the Nernst equation?</strong> A)   B)   C)   D)   E)
C) <strong>Which of the following equations represents the Nernst equation?</strong> A)   B)   C)   D)   E)
D) <strong>Which of the following equations represents the Nernst equation?</strong> A)   B)   C)   D)   E)
E) <strong>Which of the following equations represents the Nernst equation?</strong> A)   B)   C)   D)   E)
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43
Which of the following statements is true concerning the electrochemical cell Zn(s)| Zn2+(aq,1.0 M)|| Ca2+(aq,1.0 M)| Ca(s) The standard reduction potentials are given follows:
Zn2+(aq)+ 2 e− → Zn(s); E° = -0.76 V
Ca2+(aq)+ 2 e− → Ca(s); E° = -2.87 V

A) The standard cell potential is -3.63 V.
B) The standard cell potential is -2.11 V.
C) The standard cell potential is -2.11 V..
D) The standard cell potential is -3.63 V..
E) The standard cell potential is 2.11 V.
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44
The following electrochemical cell has a potential of +0.326 V at 25 °C. Pt | H2(g,1.00 atm)| H+(aq,1.00 M)|| Cl-(aq)| AgCl(s)| Ag
The standard reduction potential,E°,of AgCl(s)is +0.222 V.Calculate the Cl-(aq)ion concentration.

A) 1.9 × 10-19 M
B) 5.5 × 10-10 M
C) 0.018 M
D) 1.03 M
E) 1.8 × 109 M
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45
Calculate the copper(II)ion concentration at 25 °C in the cell Zn(s)| Zn2+(aq,1.0 M)|| Cu2+(aq)| Cu(s)if the measured cell potential is 1.06 V.The standard cell potential is 1.10 V.

A) 0.21 M
B) 1.0 M
C) 0.045 M
D) 1.0 M..
E) 1.0 M.
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46
If the <strong>If the   = -0.362 V for a given electrochemical cell at 25 °C,calculate the pH of the solution at the cathode. Pt | H<sub>2</sub>(g,1.0 atm)| H<sup>+</sup>(aq,1.00 M)|| H<sup>+</sup>(aq)| H<sub>2</sub>(g,1.0 atm)| Pt</strong> A) 1.77 B) 3.06 C) 6.11 D) 7.89 E) 12.23 = -0.362 V for a given electrochemical cell at 25 °C,calculate the pH of the solution at the cathode. Pt | H2(g,1.0 atm)| H+(aq,1.00 M)|| H+(aq)| H2(g,1.0 atm)| Pt

A) 1.77
B) 3.06
C) 6.11
D) 7.89
E) 12.23
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47
The standard cell potential of the given electrochemical cell is 0.19 V. Pt | Sn4+(aq,1.0 M),Sn2+(aq,1.0 M)|| Cu2+(aq,0.200 M)| Cu
Which of the following factors will increase the measured cell potential of the given electrochemical cell?

A) Switching from a platinum to a graphite anode
B) Increasing the size of the anode
C) Decreasing the concentration of Cu2+
D) Increasing the concentration of Sn4+
E) Decreasing the temperature of the cell
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48
The cell potential of the following electrochemical cell is determined by using an unspecified concentration of acid.Calculate the pH of the acid solution,given that the measured cell potential is -0.431 V and the anode reduction potential (E°)is 0.222 V at 25 °C. ​
Ag(s)| AgCl(s)| Cl−(aq,1.0 M)|| H+(aq)| H2(g,1.0 atm)| Pt(s)

A) 3.53
B) 11.0
C) 4.03
D) 7.06
E) 3.47
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49
Calculate the value of the reaction quotient,Q,for the voltaic cell constructed from the following two half-reactions when the Zn2+ion concentration is 0.0110 M and the Ag+ ion concentration is 1.27 M? <strong>Calculate the value of the reaction quotient,Q,for the voltaic cell constructed from the following two half-reactions when the Zn<sup>2+</sup>ion concentration is 0.0110 M and the Ag<sup>+</sup> ion concentration is 1.27 M?   →     →  </strong> A) 8.66 × 10<sup>-3</sup> B) 6.82 × 10<sup>-3</sup> C) 115 D) 1.25 × 10<sup>-2</sup> E) 147 <strong>Calculate the value of the reaction quotient,Q,for the voltaic cell constructed from the following two half-reactions when the Zn<sup>2+</sup>ion concentration is 0.0110 M and the Ag<sup>+</sup> ion concentration is 1.27 M?   →     →  </strong> A) 8.66 × 10<sup>-3</sup> B) 6.82 × 10<sup>-3</sup> C) 115 D) 1.25 × 10<sup>-2</sup> E) 147 <strong>Calculate the value of the reaction quotient,Q,for the voltaic cell constructed from the following two half-reactions when the Zn<sup>2+</sup>ion concentration is 0.0110 M and the Ag<sup>+</sup> ion concentration is 1.27 M?   →     →  </strong> A) 8.66 × 10<sup>-3</sup> B) 6.82 × 10<sup>-3</sup> C) 115 D) 1.25 × 10<sup>-2</sup> E) 147 <strong>Calculate the value of the reaction quotient,Q,for the voltaic cell constructed from the following two half-reactions when the Zn<sup>2+</sup>ion concentration is 0.0110 M and the Ag<sup>+</sup> ion concentration is 1.27 M?   →     →  </strong> A) 8.66 × 10<sup>-3</sup> B) 6.82 × 10<sup>-3</sup> C) 115 D) 1.25 × 10<sup>-2</sup> E) 147

A) 8.66 × 10-3
B) 6.82 × 10-3
C) 115
D) 1.25 × 10-2
E) 147
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50
Calculate Ecell for the following electrochemical cell at 25 °C Pt(s)| H2(g,1.00 atm)| H+(aq,1.00 M)|| Sn2+(aq,0.350 M),Sn4+(aq,0.020 M)| Pt(s)
The standard reduction potentials are as follows:
Sn4+(aq)+ 2 e- → Sn2+(s)
E° = +0.15 V
2 H+(aq)+ 2 e- → H2(g)
E° = 0.00 V

A) -0.19 V
B) +0.08 V
C) +0.11 V
D) +0.19 V
E) +0.22 V
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51
The following has a potential of 0.34 V. <strong>The following has a potential of 0.34 V.   If the concentrations of each of the ions is 1.0 M and the pressure of H<sub>2</sub> is 1.0 atm,then E° for the half-reaction   is _____.</strong> A) 0.17 V B) -0.17 V C) 0.34 V D) -0.34 V E) None of these If the concentrations of each of the ions is 1.0 M and the pressure of H2 is 1.0 atm,then E° for the half-reaction <strong>The following has a potential of 0.34 V.   If the concentrations of each of the ions is 1.0 M and the pressure of H<sub>2</sub> is 1.0 atm,then E° for the half-reaction   is _____.</strong> A) 0.17 V B) -0.17 V C) 0.34 V D) -0.34 V E) None of these is _____.

A) 0.17 V
B) -0.17 V
C) 0.34 V
D) -0.34 V
E) None of these
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52
Calculate the cell potential at 25 °C for the following overall reaction. Zn2+(aq)+ 2 Fe2+(aq)→ Zn(s)+ 2 Fe3+(aq)
[Zn2+] = 1.50 × 10-4 M,[Fe3+] = 0.0200 M,and [Fe2+] = 0.0100 M.The standard reduction potentials are as follows:
Zn2+(aq)+ 2 e- → Zn(s)
E° = -0.763 V
Fe3+(aq)+ e- → Fe2+(aq)
E° = +0.771 V

A) -1.665 V
B) -1.534 V
C) -1.439 V
D) -0.008 V
E) +0.008 V
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53
For the following cell reaction,the standard cell potential is 1.34 V.To determine the cell potential at nonstandard conditions,what is the value that should be used for n in the Nernst equation? <strong>For the following cell reaction,the standard cell potential is 1.34 V.To determine the cell potential at nonstandard conditions,what is the value that should be used for n in the Nernst equation?  </strong> A) 8 B) 10 C) 5 D) 2 E) 6

A) 8
B) 10
C) 5
D) 2
E) 6
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54
Calculate <strong>Calculate   for the electrochemical cell Pb(s)|PbCl<sub>2</sub>(s)| Cl<sup>-</sup>(aq,1.0 M)|| Fe<sup>3+</sup>(aq,1.0 M),Fe<sup>2+</sup>(aq,1.0 M)| Pt(s). The standard reduction potentials are as follows: Pb<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Pb(s) E° = -0.126 V PbCl<sub>2</sub>(s)+ 2 e<sup>-</sup> → Pb(s)+ 2 Cl<sup>-</sup>(aq) E° = -0.267 V Fe<sup>3+</sup>(aq)+ e<sup>-</sup> → Fe<sup>2+</sup>(aq) E° = +0.771 V Fe<sup>2+</sup>(aq)+ e<sup>-</sup> → Fe(s) E° = -0.44 V</strong> A) -0.504 V B) -0.062 V C) +0.504 V D) +1.038 V E) +1.604 V for the electrochemical cell Pb(s)|PbCl2(s)| Cl-(aq,1.0 M)|| Fe3+(aq,1.0 M),Fe2+(aq,1.0 M)| Pt(s). The standard reduction potentials are as follows:
Pb2+(aq)+ 2 e- → Pb(s)
E° = -0.126 V
PbCl2(s)+ 2 e- → Pb(s)+ 2 Cl-(aq)
E° = -0.267 V
Fe3+(aq)+ e- → Fe2+(aq)
E° = +0.771 V
Fe2+(aq)+ e- → Fe(s)
E° = -0.44 V

A) -0.504 V
B) -0.062 V
C) +0.504 V
D) +1.038 V
E) +1.604 V
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55
Calculate Ecell for the following electrochemical cell at 25°C.The standard cell potential,E°cell,is 0.460 V. Cu(s)| Cu2+(aq,0.016 M)|| Ag+(aq,0.11 M)| Ag(s)

A) 0.456 V
B) 0.282 V
C) 0.460 V
D) 0.485 V
E) 0.452 V
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56
Calculate Ecell for the following electrochemical cell at 25 °C. Pt(s)| Fe3+(aq,0.100 M),Fe2+(aq,0.040 M)|| Cl-(aq,0.50 M)| AgCl(s)| Ag(s)
The standard reduction potentials are as follows:
AgCl(s)+ e- → Ag(s)+ Cl-(aq)
E° = +0.222 V
Fe3+(aq)+ e- → Fe2+(aq)
E° = +0.771 V

A) -1.034 V
B) -0.590 V
C) -0.508 V
D) -0.555 V
E) +1.034 V
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57
For the electrochemical cell Cu(s)| Cu2+ || Ag+ | Ag(s),the standard cell potential is 0.46 V.A cell using these reagents was made,and the observed potential was 0.26 V at 25 oC.Which of the following is a possible explanation for the observed voltage?

A) The volume of the Cu2+ solution was larger than the volume of the Ag+ solution.
B) The volume of the Ag+ solution was larger than the volume of the Cu2+ solution.
C) The Cu2+ concentration was larger than the Ag+ concentration.
D) The Ag electrode was twice as large as the Cu electrode.
E) The Ag+ concentration was larger than the Cu2+ concentration.
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58
Which of the following is true for a product-favored reaction at equilibrium?

A) ΔrG° < 0; E°cell > 0
B) ΔrG° > 0; E°cell < 0
C) ΔrG° < 0; E°cell < 0
D) ΔrG° > 0; E°cell > 0
E) ΔrG° > 0; E°cell = 0
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59
One Faraday is defined as the:

A) charge on a single electron.
B) quantity of electric charge carried by one mole of electrons.
C) voltage required to reduce one mole of reactant.
D) number of moles of electrons required to reduce one mole of a reactant.
E) charge passed by one ampere of current in one second.
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60
Calculate <strong>Calculate   for the electrochemical cell Ag(s)| AgCl(s)| Cl<sup>-</sup>(aq,1.0 M)|| Cu<sup>2+</sup>(aq,1.0 M)| Cu(s). The standard reduction potentials are as follows: Cu<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Cu(s) E° = +0.337 V AgCl(s)+ e<sup>-</sup> → Ag(s)+ Cl<sup>-</sup>(aq) E° = +0.222 V</strong> A) -0.115 V B) -0.107 V C) +0.115 V D) +0.452 V E) +0.559 V for the electrochemical cell Ag(s)| AgCl(s)| Cl-(aq,1.0 M)|| Cu2+(aq,1.0 M)| Cu(s). The standard reduction potentials are as follows:
Cu2+(aq)+ 2 e- → Cu(s)
E° = +0.337 V
AgCl(s)+ e- → Ag(s)+ Cl-(aq)
E° = +0.222 V

A) -0.115 V
B) -0.107 V
C) +0.115 V
D) +0.452 V
E) +0.559 V
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61
Calculate the standard reduction potential for the given reaction at 25 °C. AuCl4-(aq)+ 3 e- → Au(s)+ 4 Cl-(aq)
The thermodynamic information is as follows:
Au3+(aq)+ 3 e- → Au(s)
E° = +1.50 V
Au3+(aq)+ 4 Cl-(aq)→ AuCl4-(aq)
Kf = 2.3 × 1025

A) -1.28 V
B) -0.50 V
C) +1.00 V
D) +1.28 V
E) +3.85 V
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62
Claculate the mass of chromium that can be deposited by electrolysis of an aqueous solution of chromium(III)sulfate,Cr2(SO4)3,for 180 min using a constant current of 11.0 A.Assume 100% current efficiency.(F = 96485 C/mol)

A) 0.356 g
B) 21.3 g
C) 192.1 g
D) 0.187 g
E) 32.0 g
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63
The value of E°cell is <strong>The value of E°<sub>cell</sub> is   for the following reaction: Cl<sub>2</sub>(g)+ 2 Fe<sup>2+</sup>(aq)→ 2 Fe<sup>3+</sup>(aq)+ 2 Cl<sup>-</sup>(aq) Calculate the value of E°<sub>cell</sub> for the reaction below. Cl<sup>-</sup>(g)+ Fe<sup>3+</sup>(aq)→ Fe<sup>2+</sup>(aq)+ ½ Cl<sub>2</sub>(g)</strong> A) -1.18 V B) -0.30 V C) 0.59 V D) 0.30 V E) -0.59 V for the following reaction: Cl2(g)+ 2 Fe2+(aq)→ 2 Fe3+(aq)+ 2 Cl-(aq)
Calculate the value of E°cell for the reaction below.
Cl-(g)+ Fe3+(aq)→ Fe2+(aq)+ ½ Cl2(g)

A) -1.18 V
B) -0.30 V
C) 0.59 V
D) 0.30 V
E) -0.59 V
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64
What half-reaction occurs at the cathode during the electrolysis of molten potassium bromide?

A) K(s)→ K+(l)+ e-
B) Br2(l)+ 2 e- → 2 Br-(l)
C) 2 Br-(l)→ Br2(l)+ 2 e-
D) 2 K+(l)+ 2 e- → 2 K(l)
E) 2 H2O(l)+ 2 e- → H2(g)+ 2 OH-(l)
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65
In an electrolytic cell,reduction occurs at the _____ and oxidation occurs at the _____.
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66
Which of the following are the expected products when an aqueous solution of lithium sulfate is electrolyzed? Reduction Half-Reaction
E° (V)
Li+(aq)+ e- → Li(s)
-3)04
2 H2O(l)+ 2 e- → H2(g)+ 2 OH-(aq)
-0)83
2 H+(aq)+ 2 e- → H2(g)
0)00
O2(g)+ 4 H+(aq)+ 4 e- → 2 H2O(l)
1)23
S2O82-(aq)+ 2 e- → 2 SO42-(aq)
2)01

A) Li(s)and H2(g)
B) H2(g),OH-(aq),O2(g),and H+(aq)
C) O2(g),H+(aq),and Li(s)
D) H2(g),OH-(aq),and Li(s)
E) H2(g),OH-(aq),and S2O82-(aq)
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67
If ΔrG° for the following reaction is -22.2 kJ/mol-rxn,calculate <strong>If Δ<sub>r</sub>G° for the following reaction is -22.2 kJ/mol-rxn,calculate   for the following reaction: Cu<sup>2+</sup>(aq)+ 2 Ag(s)+ 2 Cl<sup>-</sup>(aq)→ Cu(s)+ 2 AgCl(s)</strong> A) -0.460 V B) -0.115 V C) +0.115 V D) +0.230 V E) +0.559 V for the following reaction: Cu2+(aq)+ 2 Ag(s)+ 2 Cl-(aq)→ Cu(s)+ 2 AgCl(s)

A) -0.460 V
B) -0.115 V
C) +0.115 V
D) +0.230 V
E) +0.559 V
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68
Calculate the equilibrium constant for the reaction below at 25 °C. Co(s)+ 2 Cr3+(aq)→ Co2+(aq)+ 2 Cr2+(aq)
The standard reduction potentials are as follows:
Co2+(aq)+ 2 e- → Co(s)
E° = -0.28 V
Cr3+(aq)+ e- → Cr2+(aq)
E° = -0.41 V

A) 4.0 × 10-5
B) 2.5 × 104
C) 1.0 × 105
D) 1.2 × 105
E) 1.3 × 105
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69
A current of 12.0 A is passed through molten magnesium chloride for 14.0 h.How many moles of magnesium metal can be produced from this electrolysis?

A) 0.0522 mol
B) 3.13 mol
C) 0.37 mol
D) 0.22 mol
E) 6.27 mol
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70
Batteries used in watches contain mercury(II)oxide.As the current flows,mercury(II)oxide is reduced to mercury according to the following reaction: HgO(s)+ H2O( <strong>Batteries used in watches contain mercury(II)oxide.As the current flows,mercury(II)oxide is reduced to mercury according to the following reaction: HgO(s)+ H<sub>2</sub>O(   )+ 2 e<sup>-</sup> → Hg(   )+ 2 OH<sup>-</sup>(aq) If 2.3 × 10<sup>-5</sup> amperes flows continuously for 1200 days,calculate the mass of mercury,Hg(   ),produced.</strong> A) 2.5 g B) 5.0 g C) 9.9 g D) 13 g E) 15 g )+ 2 e- → Hg( <strong>Batteries used in watches contain mercury(II)oxide.As the current flows,mercury(II)oxide is reduced to mercury according to the following reaction: HgO(s)+ H<sub>2</sub>O(   )+ 2 e<sup>-</sup> → Hg(   )+ 2 OH<sup>-</sup>(aq) If 2.3 × 10<sup>-5</sup> amperes flows continuously for 1200 days,calculate the mass of mercury,Hg(   ),produced.</strong> A) 2.5 g B) 5.0 g C) 9.9 g D) 13 g E) 15 g )+ 2 OH-(aq)
If 2.3 × 10-5 amperes flows continuously for 1200 days,calculate the mass of mercury,Hg( <strong>Batteries used in watches contain mercury(II)oxide.As the current flows,mercury(II)oxide is reduced to mercury according to the following reaction: HgO(s)+ H<sub>2</sub>O(   )+ 2 e<sup>-</sup> → Hg(   )+ 2 OH<sup>-</sup>(aq) If 2.3 × 10<sup>-5</sup> amperes flows continuously for 1200 days,calculate the mass of mercury,Hg(   ),produced.</strong> A) 2.5 g B) 5.0 g C) 9.9 g D) 13 g E) 15 g ),produced.

A) 2.5 g
B) 5.0 g
C) 9.9 g
D) 13 g
E) 15 g
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71
Calculate the charge,in coulombs,is required to deposit 1.5 g of solid magnesium from a solution of Mg2+(aq)ion.

A) 4.1 × 102 C
B) 6.0 × 103 C
C) 1.2 × 104 C
D) 2.9 × 105 C
E) 3.1 × 106 C
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72
The use of electrical energy to produce chemical change is known as _____.An example of this process is the reduction of sodium chloride,NaCl( The use of electrical energy to produce chemical change is known as _____.An example of this process is the reduction of sodium chloride,NaCl(   ),to produce solid sodium. ),to produce solid sodium.
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73
Aluminum(III)ion (Al3+)is reduced to solid aluminum at an electrode.If a current of 2.75 amperes is passed for 36 hours,calculate the mass of aluminum deposited at the electrode.(Assume 100% current efficiency.)

A) 9.2 × 10-3 g
B) 3.3 × 101 g
C) 9.9 × 101 g
D) 1.0 × 102 g
E) 3.0 × 102 g
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74
Calculate the value of the equilibrium constant (K)at 25 °C for the following cell reaction: Sn(s)+ Pb2+(aq)→ Sn2+(aq)+ Pb(s); E°cell = 0.014 V

A) 0.014
B) 1.7
C) 0.4
D) 1.0
E) 3
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75
Calculate ΔrG° for the disproportionation reaction of copper(I)ion (Cu+)at 25 °C. 2 Cu+(aq)→ Cu2+(aq)+ Cu(s)
The standard reduction potentials are as follows:
Cu+(aq)+ e- → Cu(s)
E° = +0.518 V
Cu2+(aq)+ 2 e- → Cu(s)
E° = +0.337 V

A) -165 kJ/mol⋅rxn
B) -135 kJ/mol⋅rxn
C) -34.9 kJ/mol⋅rxn
D) +17.5 kJ/mol⋅rxn
E) +135 kJ/mol⋅rxn
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76
The standard reduction potentials for a reaction are as follows: Pb2+(aq)+ 2 e- → Pb(s)
E° = -0.126 V
PbSO4(s)+ 2 e- → Pb(s)+ SO42-(aq)
E° = -0.355 V
Calculate the Ksp for lead(II)sulfate (PbSO4)at 25 °C.

A) 3.4 × 10-28
B) 1.8 × 10-8
C) 5.6 × 10-5
D) 5.6 × 107
E) 2.9 × 1037
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77
A voltaic cell or galvanic cell converts chemical energy to electrical energy. <strong>A voltaic cell or galvanic cell converts chemical energy to electrical energy.   for the given galvanic cell is -1.80 V. Fe<sup>2+</sup>(aq)+ 2 Cl<sup>-</sup>(aq)→ Fe(s)+ Cl<sub>2</sub>(g) Calculate the value of Δ<sub>r</sub>G° for the reaction.</strong> A) -174 kJ B) 86.8 kJ C) 107 kJ D) 174 kJ E) 347 kJ for the given galvanic cell is -1.80 V. Fe2+(aq)+ 2 Cl-(aq)→ Fe(s)+ Cl2(g)
Calculate the value of ΔrG° for the reaction.

A) -174 kJ
B) 86.8 kJ
C) 107 kJ
D) 174 kJ
E) 347 kJ
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78
Gold and platinum are commonly used as inert electrodes in laboratory experiments.In commercial applications,such as batteries,_____ is more commonly used as an inert electrodes because it is far less expensive.
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79
Calculate the equilibrium constant for the following reaction at 25 °C, 2 IO3-(aq)+ 5 Hg( <strong>Calculate the equilibrium constant for the following reaction at 25 °C, 2 IO<sub>3</sub><sup>-</sup>(aq)+ 5 Hg(   )+ 12 H<sup>+</sup>(aq)→ I<sub>2</sub>(s)+ 5 Hg<sup>2+</sup>(aq)+ 6 H<sub>2</sub>O(   ) The standard reduction potentials are as follows: IO<sub>3</sub><sup>-</sup>(aq)+ 6 H<sup>+</sup>(aq)+ 5 e<sup>-</sup> → I<sub>2</sub>(s)+ 3 H<sub>2</sub>O(   ) E° = +1.20 V Hg<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Hg(   ) E° = +0.86 V</strong> A) 3 × 10<sup>-58</sup> B) 6 × 10<sup>5</sup> C) 3 × 10<sup>11</sup> D) 6 × 10<sup>28</sup> E) 3 × 10<sup>57</sup> )+ 12 H+(aq)→ I2(s)+ 5 Hg2+(aq)+ 6 H2O( <strong>Calculate the equilibrium constant for the following reaction at 25 °C, 2 IO<sub>3</sub><sup>-</sup>(aq)+ 5 Hg(   )+ 12 H<sup>+</sup>(aq)→ I<sub>2</sub>(s)+ 5 Hg<sup>2+</sup>(aq)+ 6 H<sub>2</sub>O(   ) The standard reduction potentials are as follows: IO<sub>3</sub><sup>-</sup>(aq)+ 6 H<sup>+</sup>(aq)+ 5 e<sup>-</sup> → I<sub>2</sub>(s)+ 3 H<sub>2</sub>O(   ) E° = +1.20 V Hg<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Hg(   ) E° = +0.86 V</strong> A) 3 × 10<sup>-58</sup> B) 6 × 10<sup>5</sup> C) 3 × 10<sup>11</sup> D) 6 × 10<sup>28</sup> E) 3 × 10<sup>57</sup> )
The standard reduction potentials are as follows:
IO3-(aq)+ 6 H+(aq)+ 5 e- → I2(s)+ 3 H2O( <strong>Calculate the equilibrium constant for the following reaction at 25 °C, 2 IO<sub>3</sub><sup>-</sup>(aq)+ 5 Hg(   )+ 12 H<sup>+</sup>(aq)→ I<sub>2</sub>(s)+ 5 Hg<sup>2+</sup>(aq)+ 6 H<sub>2</sub>O(   ) The standard reduction potentials are as follows: IO<sub>3</sub><sup>-</sup>(aq)+ 6 H<sup>+</sup>(aq)+ 5 e<sup>-</sup> → I<sub>2</sub>(s)+ 3 H<sub>2</sub>O(   ) E° = +1.20 V Hg<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Hg(   ) E° = +0.86 V</strong> A) 3 × 10<sup>-58</sup> B) 6 × 10<sup>5</sup> C) 3 × 10<sup>11</sup> D) 6 × 10<sup>28</sup> E) 3 × 10<sup>57</sup> )
E° = +1.20 V
Hg2+(aq)+ 2 e- → Hg( <strong>Calculate the equilibrium constant for the following reaction at 25 °C, 2 IO<sub>3</sub><sup>-</sup>(aq)+ 5 Hg(   )+ 12 H<sup>+</sup>(aq)→ I<sub>2</sub>(s)+ 5 Hg<sup>2+</sup>(aq)+ 6 H<sub>2</sub>O(   ) The standard reduction potentials are as follows: IO<sub>3</sub><sup>-</sup>(aq)+ 6 H<sup>+</sup>(aq)+ 5 e<sup>-</sup> → I<sub>2</sub>(s)+ 3 H<sub>2</sub>O(   ) E° = +1.20 V Hg<sup>2+</sup>(aq)+ 2 e<sup>-</sup> → Hg(   ) E° = +0.86 V</strong> A) 3 × 10<sup>-58</sup> B) 6 × 10<sup>5</sup> C) 3 × 10<sup>11</sup> D) 6 × 10<sup>28</sup> E) 3 × 10<sup>57</sup> )
E° = +0.86 V

A) 3 × 10-58
B) 6 × 105
C) 3 × 1011
D) 6 × 1028
E) 3 × 1057
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80
How many moles of electrons are produced from a current of 17.0 A in 3.40 hours?

A) 5.99 × 10-4 mol
B) 2.16 mol
C) 57.8 mol
D) 3.35 mol
E) 9.33 × 103 mol
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