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Use the Standard Reduction Potentials Below to Determine Which Element \to

Question 69

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Use the standard reduction potentials below to determine which element or ion is the best oxidizing agent.
O2(g) + 4 H+(aq) + 4 e- \to 2 H2O(  Use the standard reduction potentials below to determine which element or ion is the best oxidizing agent. O<sub>2</sub>(g) + 4 H<sup>+</sup>(aq) + 4 e<sup>-</sup>  \to  2 H<sub>2</sub>O(   )  ~~~~~~~~~~~~~~~~ E<sup> \circ </sup> = +1.229 V Hg<sub>2</sub><sup>2+</sup>(aq) + 2 e<sup>-</sup>  \to  2 Hg(   )  ~~~~~~~~~~~~~~~~~~ ~ ~~~~ ~~~ E<sup> \circ </sup> = +0.789 V I<sub>2</sub>(s) + 2 e<sup>-</sup>  \to 2 I<sup>-</sup>(aq)  ~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~ ~~~~~ E<sup> \circ </sup> = +0.535 V 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 \circ = +1.229 V
Hg22+(aq) + 2 e- \to 2 Hg(  Use the standard reduction potentials below to determine which element or ion is the best oxidizing agent. O<sub>2</sub>(g) + 4 H<sup>+</sup>(aq) + 4 e<sup>-</sup>  \to  2 H<sub>2</sub>O(   )  ~~~~~~~~~~~~~~~~ E<sup> \circ </sup> = +1.229 V Hg<sub>2</sub><sup>2+</sup>(aq) + 2 e<sup>-</sup>  \to  2 Hg(   )  ~~~~~~~~~~~~~~~~~~ ~ ~~~~ ~~~ E<sup> \circ </sup> = +0.789 V I<sub>2</sub>(s) + 2 e<sup>-</sup>  \to 2 I<sup>-</sup>(aq)  ~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~ ~~~~~ E<sup> \circ </sup> = +0.535 V 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 \circ = +0.789 V
I2(s) + 2 e- \to 2 I-(aq)         ~~~~~~~~      ~~ ~~~~        ~~~~~~~~        ~~~~~~~~       ~~ ~~~~~ E \circ = +0.535 V


A) I2(s)
B) O2(g)
C) I-(aq)
D) Hg22+(aq)
E) H2O(  Use the standard reduction potentials below to determine which element or ion is the best oxidizing agent. O<sub>2</sub>(g) + 4 H<sup>+</sup>(aq) + 4 e<sup>-</sup>  \to  2 H<sub>2</sub>O(   )  ~~~~~~~~~~~~~~~~ E<sup> \circ </sup> = +1.229 V Hg<sub>2</sub><sup>2+</sup>(aq) + 2 e<sup>-</sup>  \to  2 Hg(   )  ~~~~~~~~~~~~~~~~~~ ~ ~~~~ ~~~ E<sup> \circ </sup> = +0.789 V I<sub>2</sub>(s) + 2 e<sup>-</sup>  \to 2 I<sup>-</sup>(aq)  ~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~ ~~~~~ E<sup> \circ </sup> = +0.535 V 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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