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Given the Following Two Half-Reactions,write the Overall Reaction in the Direction

Question 84

Multiple Choice

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) ; 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 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) ; 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 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) ; 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 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) ; 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 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) ; 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 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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