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Question

Calculate the reduction potential for the following half cell at 25C:
Ag(s)|Ag+(aq, 105 M); E0Ag+/Ag=0.80 V

A
EAg+/Ag=0.5 V
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B
EAg+/Ag=0.25 V
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C
EAg+/Ag=0.8 V
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D
EAg+/Ag=1.5 V
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Solution

The correct option is A EAg+/Ag=0.5 V
Ag(s)|Ag+(aq, 105) M; E0Ag+/Ag=0.80 V


The half call electrode reaction is

Ag+(aq)+eAg(s)

Nernst equation for the half cell is

EAg+/Ag=E0Ag+/Ag2.303RTnFlog[M(s)][Mn+(aq)]

n is the number of electrons involved in the half cell reaction.

n = 1

EAg+/Ag=0.80 0.0591log1[Ag+]

EAg+/Ag=0.80 0.0591log(1/105)

EAg+/Ag=0.80 +0.0591log(105)

EAg+/Ag=0.80 5×0.059=0.50 V




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