JEE Main 2022 · 26 Jun · Shift 2, Chemistry question 87

$$\mathrm{Cu(s)+Sn^{2+}(0.001M)\rightarrow Cu^{2+}(0.01M)+Sn(s)}$$
The Gibbs free energy change for the above reaction at 298 K is $x\times10^{-1}$ kJ mol$^{-1}$. The value of $x$ is ______. [nearest integer]

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