Cu s + Sn 2 + 0 . 001 M → Cu 2 + 0 . 01 M + Sn s The Gibbs free energy change for the above reaction…

$\mathrm{Cu}\left(\mathrm{s}\right)+{\mathrm{Sn}}^{2+}\left(0.001\mathrm{M}\right)\to {\mathrm{Cu}}^{2+}\left(0.01\mathrm{M}\right)+\mathrm{Sn}\left(\mathrm{s}\right)$

The Gibbs free energy change for the above reaction at $298 \mathrm{K}$ is $x\times {10}^{-1} \mathrm{kJ}{ \mathrm{mol}}^{-1}$. The value of $\mathrm{x}$ is____[nearest integer]

[Given : ${\mathrm{E}}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^{\ominus }=0.34 \mathrm{V};{\mathrm{E}}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn}}^{\ominus }=-0.14 \mathrm{V};\mathrm{F}=96500{\mathrm{Cmol}}^{-1}$]