1. Write the given half-reactions and their standard reduction potentials

Given: $Cu^{2+} + 2e^- \rightarrow Cu \qquad E^\circ = 0.34\,V$

$Cu^+ + e^- \rightarrow Cu \qquad E^\circ = 0.522\,V$

We need: $Cu^{2+} + e^- \rightarrow Cu^+ \qquad E^\circ = ?$

2. Use the relation between $\Delta G^\circ$ and $E^\circ$

Since standard potentials are not directly additive, we use: $\Delta G^\circ = -nFE^\circ$

Let:

• Reaction (1): $Cu^{2+} + 2e^- \rightarrow Cu$ $\Delta G_1^\circ = -2F(0.34) = -0.68F$

• Reaction (2): $Cu^+ + e^- \rightarrow Cu$ $\Delta G_2^\circ = -1F(0.522) = -0.522F$

3. Relate the target reaction to the given reactions

The required reaction is: $Cu^{2+} + e^- \rightarrow Cu^+$

If we add this reaction to reaction (2), we get reaction (1):

$\big(Cu^{2+} + e^- \rightarrow Cu^+\big)$ $\big(Cu^+ + e^- \rightarrow Cu\big)$ $\Rightarrow Cu^{2+} + 2e^- \rightarrow Cu$

Therefore, $\Delta G_1^\circ = \Delta G_\text{req}^\circ + \Delta G_2^\circ$

So, $\Delta G_\text{req}^\circ = \Delta G_1^\circ - \Delta G_2^\circ$

Substitute: $\Delta G_\text{req}^\circ = (-0.68F) - (-0.522F) = -0.158F$

4. Find the required standard potential

For the required reaction, $n=1$, so: $\Delta G_\text{req}^\circ = -1\cdot F \cdot E_\text{req}^\circ$

Thus, $-FE_\text{req}^\circ = -0.158F$

$E_\text{req}^\circ = 0.158\,V$

5. Choose the correct option

$E^\circ(Cu^{2+}/Cu^+) = +0.158\,V$

So the correct option is D.