Given: $E^{\circ}_{Cu^{2+}/Cu} = 0.34\text{ V}$ $E^{\circ}_{Cu^{2+}/Cu^+} = 0.15\text{ V}$

The corresponding half-reactions and their standard Gibbs free energy changes are:

1. $Cu^{2+} + 2e^- \rightarrow Cu \ ; \ \Delta G^{\circ}_1 = -nFE^{\circ} = -2 \times F \times 0.34 = -0.68 F$
2. $Cu^{2+} + e^- \rightarrow Cu^+ \ ; \ \Delta G^{\circ}_2 = -1 \times F \times 0.15 = -0.15 F$

To find the standard reduction potential for $Cu^+/Cu$, subtract reaction 2 from reaction 1: $Cu^+ + e^- \rightarrow Cu \ ; \ \Delta G^{\circ}_3 = \Delta G^{\circ}_1 - \Delta G^{\circ}_2 = -0.68 F - (-0.15 F) = -0.53 F$ Since $\Delta G^{\circ}_3 = -nFE^{\circ}_{Cu^+/Cu} = -1 \times F \times E^{\circ}_{Cu^+/Cu}$, we get $E^{\circ}_{Cu^+/Cu} = 0.53\text{ V}$.

The disproportionation reaction is $2Cu^+ \rightarrow Cu^{2+} + Cu$. The cell potential for this reaction is: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = E^{\circ}_{Cu^+/Cu} - E^{\circ}_{Cu^{2+}/Cu^+} = 0.53\text{ V} - 0.15\text{ V} = 0.38\text{ V}$.