For the disproportionation reaction $$2Cu^+(aq) \rightleftharpoons Cu(s) + Cu^{2+}(aq)$$ at 298K, ln K (where K is the equilibrium constant) is ___________ $$\times 10^{-1}$$.
Given: $$(E^°_{Cu^{2+}/Cu^+} = 0.16V, E^°_{Cu^+/Cu} = 0.52V, \frac{RT}{F} = 0.025)$$

We have the disproportionation reaction

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

To find $$\ln K$$, we first need the standard cell potential $$E^\circ_{\text{cell}}$$ for the reaction. The two relevant half-cells are :

Reduction half-cell $$Cu^+(aq)+e^- \;\longrightarrow\;Cu(s),\qquad E^\circ_{\text{red}}=0.52\;{\rm V}$$

Reduction half-cell $$Cu^{2+}(aq)+e^- \;\longrightarrow\;Cu^+(aq),\qquad E^\circ_{\text{red}}=0.16\;{\rm V}$$

In the disproportionation process one $$Cu^+$$ ion is reduced to $$Cu(s)$$ (cathode) and another $$Cu^+$$ ion is oxidised to $$Cu^{2+}$$ (anode). The anode potential is obtained by reversing the second half-reaction, so

$$E^\circ_{\text{ox}} = -\,0.16\;{\rm V}$$

The cell potential is therefore

$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode (red)}} = 0.52\;{\rm V} - 0.16\;{\rm V} = 0.36\;{\rm V}$$

Now we relate $$E^\circ_{\text{cell}}$$ to the equilibrium constant using the thermodynamic relations

$$\Delta G^\circ = -nF\,E^\circ_{\text{cell}}\qquad\text{and}\qquad\Delta G^\circ = -RT\ln K$$

Equating the two expressions gives

$$-nF\,E^\circ_{\text{cell}} = -RT\ln K$$

$$\Rightarrow\;\ln K = \dfrac{nF}{RT}\,E^\circ_{\text{cell}}$$

For the overall reaction one mole of electrons is transferred, so $$n = 1$$. The problem supplies $$\dfrac{RT}{F} = 0.025$$; therefore

$$\frac{F}{RT} = \frac{1}{RT/F} = \frac{1}{0.025} = 40$$

Substituting all the values, we get

$$\ln K = \bigl(1\bigr)\times 40 \times 0.36 = 14.4$$

This can be written as

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

So, the answer is $$144$$.