We start with the disproportionation reaction:

$$2\,\mathrm{Cu}^{+} \rightleftharpoons \mathrm{Cu}^{2+} + \mathrm{Cu}$$

This reaction involves one Cu⁺ ion being oxidized to Cu²⁺ and a second Cu⁺ ion being reduced to Cu (metal). The given standard reduction potentials are:

For the redox couple between Cu²⁺ and Cu⁺:

$$\mathrm{Cu}^{2+} + e^- \longrightarrow \mathrm{Cu}^{+}, \quad E^\circ = +0.15\,\mathrm{V}$$

For the redox couple between Cu²⁺ and Cu:

$$\mathrm{Cu}^{2+} + 2e^- \longrightarrow \mathrm{Cu}, \quad E^\circ = +0.34\,\mathrm{V}$$

We can break the disproportionation reaction into two half-reactions:

Oxidation (one Cu⁺ is oxidized):

The oxidation half-reaction is the reverse of the first reaction:

$$\mathrm{Cu}^{+} \longrightarrow \mathrm{Cu}^{2+} + e^-$$

Since the standard reduction potential for

$$\mathrm{Cu}^{2+} + e^- \rightarrow \mathrm{Cu}^{+}$$

is +0.15 V, when reversed, the oxidation potential becomes:

$$E^\circ_{\text{ox}} = -0.15\,\mathrm{V}$$

Reduction (one Cu⁺ is reduced):

We need the half-reaction:

$$\mathrm{Cu}^{+} + e^- \longrightarrow \mathrm{Cu}$$

This potential isn’t given directly, but we can derive it using the overall two‐electron process. Notice that if we add the reduction half-reaction above to the (reverse) of the oxidation half-reaction, we recover:

$$\mathrm{Cu}^{2+} + 2e^- \longrightarrow \mathrm{Cu}$$

whose standard potential is +0.34 V. We can use free energy (ΔG) relationships:

For the reaction

$$\mathrm{Cu}^{2+} + e^- \longrightarrow \mathrm{Cu}^{+}: \quad \Delta G_1 = -F(0.15\,\mathrm{V})$$

For the overall reaction

$$\mathrm{Cu}^{2+} + 2e^- \longrightarrow \mathrm{Cu}: \quad \Delta G_2 = -2F(0.34\,\mathrm{V}) = -0.68F$$

The free energy change for the reduction half-reaction

$$\mathrm{Cu}^{+} + e^- \longrightarrow \mathrm{Cu}$$

is then:

$$\Delta G_{\text{red}} = \Delta G_2 - \Delta G_1 = -0.68F - (-0.15F) = -0.53F$$

Since this half-reaction involves 1 electron, its standard reduction potential is:

$$E^\circ_{\text{red}} = 0.53\,\mathrm{V}$$

Now, to combine the two half-reactions for the overall disproportionation:

Oxidation half: $$\mathrm{Cu}^{+} \longrightarrow \mathrm{Cu}^{2+} + e^- \quad (E^\circ = -0.15\,\mathrm{V})$$

Reduction half: $$\mathrm{Cu}^{+} + e^- \longrightarrow \mathrm{Cu} \quad\,\, (E^\circ = +0.53\,\mathrm{V})$$

When combining redox half-reactions, the overall cell potential is given by adding the cathode (reduction) potential and the anode (oxidation) potential:

$$E^\circ_{\text{overall}} = E^\circ_{\text{red}} + E^\circ_{\text{ox}} = +0.53\,\mathrm{V} + (-0.15\,\mathrm{V}) = +0.38\,\mathrm{V}$$

Thus, the standard electrode potential for the disproportionation of Cu⁺ is:

$$\boxed{+0.38\,\mathrm{V}}$$

This corresponds to Option D.

Electrochemistry

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