Cu ( s )+ Sn^2+(0.001 M ) → Cu^2+(0.01 M )+ Sn ( s ) The Gibbs free energy change for the above reaction at 298 K is x × 10^-1 kJ mol^-1. The value…

$$\mathrm{Cu}(\mathrm{~s})+\mathrm{Sn}^{2+}(0.001 \mathrm{M}) \rightarrow \mathrm{Cu}^{2+}(0.01 \mathrm{M})+\mathrm{Sn}(\mathrm{~s}) $$The Gibbs free energy change for the above reaction at $\displaystyle 298$ K is $\displaystyle x \times 10^{-1} \mathrm{~kJ} \mathrm{~mol}^{-1}$. The value of $\displaystyle x$ is \_\_\_\_. [nearest integer] $$\left[\text { Given : } \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\ominus}=0.34 \mathrm{~V} ; \mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\ominus}=-0.14 \mathrm{~V} ; \mathrm{F}=96500 \mathrm{C} \mathrm{~mol}^{-1}\right] $$