Given: (i) $Cu^{2+} + 2e^- \rightarrow Cu$, $E^\circ = 0.337 \text{ V}$ (ii) $Cu^{2+} + e^- \rightarrow Cu^+$, $E^\circ = 0.153 \text{ V}$

Electrode potential, $E^\circ$ for the reaction, $Cu^+ + e^- \rightarrow Cu$, will be:

Electrode potentials ($E^\circ$) are intensive properties and cannot be added directly. However, Standard Gibbs Energy ($\Delta G^\circ$) is an extensive property and is additive. The relationship is given by $\Delta G^\circ = -nFE^\circ$ .

1. Analyze the Reactions: Reaction 1: $Cu^{2+} + 2e^- \rightarrow Cu$ ($n_1=2$, $E^\circ_1 = 0.337 \text{ V}$) Reaction 2: $Cu^{2+} + e^- \rightarrow Cu^+$ ($n_2=1$, $E^\circ_2 = 0.153 \text{ V}$)

• Target Reaction 3: $Cu^+ + e^- \rightarrow Cu$ ($n_3=1$, $E^\circ_3 = ?$)

2. Relate the Reactions: Reaction 1 is the sum of Reaction 2 and Target Reaction 3: $(Cu^{2+} + e^- \rightarrow Cu^+) + (Cu^+ + e^- \rightarrow Cu) = (Cu^{2+} + 2e^- \rightarrow Cu)$

3. Apply Gibbs Energy Additivity: $\Delta G^\circ_1 = \Delta G^\circ_2 + \Delta G^\circ_3$ $(-n_1 F E^\circ_1) = (-n_2 F E^\circ_2) + (-n_3 F E^\circ_3)$

4. Substitute Values: $-2F(0.337) = -1F(0.153) + -1F(E^\circ_3)$ Dividing by $-F$: $2(0.337) = 1(0.153) + 1(E^\circ_3)$ $0.674 = 0.153 + E^\circ_3$ $E^\circ_3 = 0.674 - 0.153$ $E^\circ_3 = 0.521 \text{ V}$

Rounding to two decimal places, $E^\circ = 0.52 \text{ V}$.