To find the standard reduction potential for the overall reaction $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$, we need to use standard Gibbs free energy change ($\Delta G^\circ$).

Given half-reactions:

1. $\text{Cu}^{2+}(aq) + e^- \rightarrow \text{Cu}^+(aq) \quad E^\circ_1 = 0.15 \text{ V}$ $\Delta G^\circ_1 = -n_1 F E^\circ_1 = -1 \times F \times 0.15 = -0.15 F$

2. $\text{Cu}^+(aq) + e^- \rightarrow \text{Cu}(s) \quad E^\circ_2 = 0.50 \text{ V}$ $\Delta G^\circ_2 = -n_2 F E^\circ_2 = -1 \times F \times 0.50 = -0.50 F$

Adding equations (1) and (2) gives the required equation: $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$ $\Delta G^\circ_3 = \Delta G^\circ_1 + \Delta G^\circ_2 = -0.15 F + (-0.50 F) = -0.65 F$

For the overall reaction, $\Delta G^\circ_3 = -n_3 F E^\circ_3$, where $n_3 = 2$. $-2 \times F \times E^\circ_3 = -0.65 F$ $E^\circ_3 = \frac{0.65}{2} = +0.325 \text{ V}$.

Therefore, $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.325 \text{ V}$.