1. Write the reaction and identify electron transfer

Given reaction: $Cu(s) + Sn^{2+}(0.001M) \rightarrow Cu^{2+}(0.01M) + Sn(s)$

Half-reactions are:

• Oxidation: $Cu(s) \rightarrow Cu^{2+} + 2e^-$
• Reduction: $Sn^{2+} + 2e^- \rightarrow Sn(s)$

So, number of electrons transferred: $n=2$

2. Find standard cell potential

Given standard reduction potentials: $E^\circ_{Cu^{2+}/Cu}=0.34\,V$ $E^\circ_{Sn^{2+}/Sn}=-0.14\,V$

In the given reaction:

• $Sn^{2+}$ is reduced at cathode
• $Cu$ is oxidized at anode

Therefore, $E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$ $E^\circ_{cell}=(-0.14)-(0.34)=-0.48\,V$

3. Calculate reaction quotient $Q$

For the reaction $Cu(s)+Sn^{2+}(aq)\rightarrow Cu^{2+}(aq)+Sn(s)$

Solids are omitted, so $Q=\frac{[Cu^{2+}]}{[Sn^{2+}]}=\frac{0.01}{0.001}=10$

4. Apply Nernst equation

At $298\,K$, $E=E^\circ-\frac{0.0591}{n}\log Q$

So, $E=-0.48-\frac{0.0591}{2}\log 10$ $E=-0.48-\frac{0.0591}{2}(1)$ $E=-0.48-0.02955=-0.50955\,V$

5. Calculate Gibbs free energy change

Relation: $\Delta G=-nFE$

Substitute values: $\Delta G=-2\times 96500 \times (-0.50955)$ $\Delta G=98343.15\,J\,mol^{-1}$

Convert to kJ mol$^{-1}$: $\Delta G=98.343\,kJ\,mol^{-1}$

6. Match with the form given

Given: $\Delta G = x\times 10^{-1}\,kJ\,mol^{-1}$

That means $x\times 10^{-1}=98.343$ $x=983.43$

Nearest integer: $x=983$

7. Final Answer

$\boxed{983}$