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The electrode potentials for $Cu^{2+}_{(aq)} + e^- \rightarrow Cu^+_{(aq)}$ and $Cu^+_{(aq)} + e^- \rightarrow Cu_{(s)}$ are +0.15 V and +0.50 V respectively. The value of $E^\circ_{Cu^{2+}/Cu}$ will be
A
0.150 V
B
0.500 V
C
0.325 V
D
0.650 V
Detailed Solution
Electrode potentials are not additive, but free energies are. Use $\Delta G^\circ = -nFE^\circ$.
(1) $Cu^{2+} + e^- \rightarrow Cu^+$; $\Delta G_1^\circ = -1\times F\times0.15$
(2) $Cu^+ + e^- \rightarrow Cu$; $\Delta G_2^\circ = -1\times F\times0.50$
Adding (1) and (2): $Cu^{2+} + 2e^- \rightarrow Cu$; $\Delta G_3^\circ = -2\times F\times E^\circ$
$\Delta G_3^\circ = \Delta G_1^\circ + \Delta G_2^\circ$
$-2FE^\circ = -0.15F - 0.50F = -0.65F$
$E^\circ = \frac{0.65}{2} = 0.325$ V
(1) $Cu^{2+} + e^- \rightarrow Cu^+$; $\Delta G_1^\circ = -1\times F\times0.15$
(2) $Cu^+ + e^- \rightarrow Cu$; $\Delta G_2^\circ = -1\times F\times0.50$
Adding (1) and (2): $Cu^{2+} + 2e^- \rightarrow Cu$; $\Delta G_3^\circ = -2\times F\times E^\circ$
$\Delta G_3^\circ = \Delta G_1^\circ + \Delta G_2^\circ$
$-2FE^\circ = -0.15F - 0.50F = -0.65F$
$E^\circ = \frac{0.65}{2} = 0.325$ V