What will be the reduction potential for the following half-cell reaction at 298K? (Given: [Ag+]=0.1M and Eocell=+0.80V)
0.741V
0.80V
−0.80V
−0.741V
A
0.80V
B
−0.741V
C
−0.80V
D
0.741V
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Solution
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A galvanic cell or simple battery is made of two electrodes. Each of the electrodes of a galvanic cell is known as a half cell. In a battery, the two half cells form an oxidizing-reducing couple. When two half cells are connected via an electric conductor and salt bridge, an electrochemical reaction is started.
Nernst equation is,
Ecell=E0cell−0.0591nlogQ
Here the reaction is,
Ag++e−→Ag
n=1
E0cell=0.80V
Hence,
Ecell=E0cell−0.0591nlog1[Ag+]
Ecell=0.80−0.05911log10.1
=0.741V
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