Given that the standard potentials (E°) of Cu2+/Cu and Cu+/Cu are 0.34 V and 0.522 V respectively, the E° of Cu2+/Cu+…

Strategy: Relate standard electrode potentials to standard Gibbs free energy ($\Delta G^\circ = -nFE^\circ$). Combine two known half-reactions to find the potential of the target half-reaction.

Step 1: Write known half-reactions and energies

1. $\ce{Cu^{2+} + 2e^- \rightarrow Cu}$: $E^\circ_1 = 0.34 \text{ V}$, $n_1 = 2$ $\Delta G^\circ_1 = -2F(0.34) = -0.68F$
2. $\ce{Cu^+ + e^- \rightarrow Cu}$: $E^\circ_2 = 0.522 \text{ V}$, $n_2 = 1$ $\Delta G^\circ_2 = -1F(0.522) = -0.522F$

Step 2: Define target reaction Target: $\ce{Cu^{2+} + e^- \rightarrow Cu^+}$ This is obtained by: $\text{Reaction (1)} - \text{Reaction (2)}$. $\Delta G^\circ_3 = \Delta G^\circ_1 - \Delta G^\circ_2 = -0.68F - (-0.522F) = -0.158F$

Step 3: Solve for $E^\circ_3$ For target $n_3 = 1$: $E^\circ_3 = \frac{-\Delta G^\circ_3}{1F} = \frac{0.158F}{F} = +0.158 \text{ V}$

$\boxed{\text{Answer: (b) +0.158 V}}$