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Given:
(i) $Cu^{2+} + 2e^- \rightarrow Cu$, $E^\circ$ = 0.337 V
(ii) $Cu^{2+} + e^- \rightarrow Cu^+$, $E^\circ$ = 0.153 V
Electrode potential $E^\circ$ for the reaction, $Cu^+ + e^- \rightarrow Cu$, will be
(i) $Cu^{2+} + 2e^- \rightarrow Cu$, $E^\circ$ = 0.337 V
(ii) $Cu^{2+} + e^- \rightarrow Cu^+$, $E^\circ$ = 0.153 V
Electrode potential $E^\circ$ for the reaction, $Cu^+ + e^- \rightarrow Cu$, will be
A
0.38 V
B
0.52 V
C
0.90 V
D
0.30 V
Detailed Solution
Electrode potentials are not additive, but Gibbs energies are: $\Delta G^\circ = -nFE^\circ$
(i) $Cu^{2+} + 2e^- \rightarrow Cu$: $\Delta G_1^\circ = -2F(0.337) = -0.674F$
(ii) $Cu^{2+} + e^- \rightarrow Cu^+$: $\Delta G_2^\circ = -1F(0.153) = -0.153F$
Required reaction = (i) − (ii): $Cu^+ + e^- \rightarrow Cu$, so $\Delta G_3^\circ = \Delta G_1^\circ - \Delta G_2^\circ = -0.674F + 0.153F = -0.521F$
$\Delta G_3^\circ = -1\times F\times E^\circ$, so $E^\circ = 0.521$ V
$E^\circ \approx 0.52$ V
(i) $Cu^{2+} + 2e^- \rightarrow Cu$: $\Delta G_1^\circ = -2F(0.337) = -0.674F$
(ii) $Cu^{2+} + e^- \rightarrow Cu^+$: $\Delta G_2^\circ = -1F(0.153) = -0.153F$
Required reaction = (i) − (ii): $Cu^+ + e^- \rightarrow Cu$, so $\Delta G_3^\circ = \Delta G_1^\circ - \Delta G_2^\circ = -0.674F + 0.153F = -0.521F$
$\Delta G_3^\circ = -1\times F\times E^\circ$, so $E^\circ = 0.521$ V
$E^\circ \approx 0.52$ V