1. Write the disproportionation reaction

$2Cu^+(aq) \rightleftharpoons Cu(s) + Cu^{2+}(aq)$

We need the equilibrium constant $K$ at $298\,K$.

2. Identify the two relevant half-reactions

Given standard reduction potentials:

$Cu^{2+} + e^- \rightarrow Cu^+ \qquad E^\circ = 0.16\,V$

$Cu^+ + e^- \rightarrow Cu(s) \qquad E^\circ = 0.52\,V$

For disproportionation, one $Cu^+$ is reduced to $Cu$, and another $Cu^+$ is oxidized to $Cu^{2+}$.

• Cathode (reduction): $Cu^+ + e^- \rightarrow Cu \qquad E^\circ_{\text{red}} = 0.52\,V$

• Anode (oxidation): reverse of $Cu^{2+} + e^- \rightarrow Cu^+ \qquad E^\circ_{\text{red}} = 0.16\,V$

So,

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.52 - 0.16 = 0.36\,V$

3. Relate $E^\circ$ and equilibrium constant

For the reaction,

$\Delta G^\circ = -nFE^\circ$

and

$\Delta G^\circ = -RT\ln K$

Thus,

$RT\ln K = nFE^\circ$

$\ln K = \frac{nFE^\circ}{RT}$

Given:

$\frac{RT}{F} = 0.025$

Hence,

$\ln K = \frac{nE^\circ}{RT/F}$

For this reaction, $n=1$ electron.

So,

$\ln K = \frac{1 \times 0.36}{0.025} = 14.4$

4. Match with the asked format

They ask:

$\ln K = \underline{\hspace{2cm}} \times 10^{-1}$

Since

$14.4 = 144 \times 10^{-1}$

So the required integer is:

$\boxed{144}$

5. Comparison with stored answer

Stored correct answer = $144$.

Our derived answer also = $144$.