For the process: Cu2+ + 2e− → Cu; log[Cu2+] vs. Ered graph is shown in the figure, where OA = 0.34 V

Question

For the process: Cu2+ + 2e− → Cu; log[Cu2+] vs. Ered graph is shown in the figure, where OA = 0.34 V. The electrode potential of the half-cell of Cu|Cu2+ (0.1 M) will be [2.303 RT/F = 0.06]

Accepted Answer

$E_{Cu^{2 +}|Cu} = E_{Cu^{2 +}|Cu}^{o} - \frac{0.06}{2}.\log\frac{1}{\left\lbrack Cu^{2 +} \right\rbrack}$

$= E_{Cu^{2 +}|Cu}^{o} + 0.03\log\left\lbrack Cu^{2 +} \right\rbrack$

$= 0.34 + 0.03 \times \log(0.1) = 0.31\text{ V} $$$\therefore E_{Cu|Cu^{2 +}} = - 0.31\text{ V}$$

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