For the disproportionation reaction $2{\mathrm{Cu}}^{+}\left(\mathrm{aq}\right)\rightleftharpoons \mathrm{Cu}\left(\mathrm{s}\right)+{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)$ at $298\mathrm{K},$ $\ln \left(\mathrm{K}\right)$  (where $\mathrm{K}$ is the equilibrium constant) is _______$\times {10}^{-1}$

Given $:\left(\mathrm{E}{°}_{{\mathrm{Cu}}^{2+}/{\mathrm{Cu}}^{+}}=0.16V E{°}_{{\mathrm{Cu}}^{+}/\mathrm{Cu}}=0.52\mathrm{V}\frac{\mathrm{RT}}{\mathrm{F}}=0.025\right)$

The Gibbs energy change (in $\mathrm{J}$) for the given reaction at $\left[{\mathrm{Cu}}^{2+}\right]=\left[{\mathrm{Sn}}^{2+}\right]=1\text{\hspace{0.17em}M}$ and $298 \mathrm{K}$ is :

$\mathrm{Cu}\left(\mathrm{s}\right)+{\mathrm{Sn}}^{2+}\left(\mathrm{aq}.\right)\to {\mathrm{Cu}}^{2+}\left(\mathrm{aq}.\right)+\mathrm{Sn}\left(\mathrm{s}\right)$; 

$\left({\mathrm{E}}_{{\mathrm{Sn}}^{2+}\left|\mathrm{Sn}\right.}^0=-0.16\mathrm{V},{\mathrm{E}}_{\left.{\mathrm{Cu}}^{2+}\right|\mathrm{Cu}}^0=0.34\mathrm{V},\right.$Take $\left.\mathrm{F}=96500\mathrm{C}{\mathrm{mol}}^{-1}\right)$

The standard electrode potential ${\mathrm{E}}^o$ and its temperature coefficient $\left(\frac{\mathrm{dE}}{\mathrm{dT}}\right)$ for a cell are $2\hspace{0.17em}\mathrm{V}$ and $-5{\times 10}^{-4}\mathrm{\hspace{0.17em}} \mathrm{V} {\mathrm{K}}^{-1}$ at $\mathrm{300\hspace{0.17em}}\mathrm{K}$, respectively. The reaction is $\mathrm{Zn} \left(\mathrm{s}\right)+{\mathrm{Cu}}^{2+} \left(\mathrm{aq}\right)\to {\mathrm{Zn}}^{2+} \left(\mathrm{aq}\right)+\mathrm{Cu} \left(\mathrm{s}\right)$. The standard reaction enthalpy $\left({\mathrm{\Delta }}_{\mathrm{r}}{\mathrm{H}}^{-}\right)$ at $300\hspace{0.17em}\mathrm{K}$ in $\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$ is 

$[$Use $\mathrm{R}=8 \mathrm{J} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1}$ and $\mathrm{F}=96,500 \mathrm{C}\hspace{0.17em}{\mathrm{mol}}^{-1}]$