1. Write the given reaction

$\text{Cu}(s) + \text{Sn}^{2+}(aq) \rightarrow \text{Cu}^{2+}(aq) + \text{Sn}(s)$

Given:

• $E^\circ_{\text{Sn}^{2+}/\text{Sn}} = -0.16\,\text{V}$
• $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\,\text{V}$
• $F = 96500\,\text{C mol}^{-1}$
• $[\text{Cu}^{2+}] = [\text{Sn}^{2+}] = 1\,\text{M}$, so this is standard condition for the ions.

2. Identify oxidation and reduction

From the reaction:

• $\text{Cu}(s) \rightarrow \text{Cu}^{2+} + 2e^-$ (oxidation)
• $\text{Sn}^{2+} + 2e^- \rightarrow \text{Sn}(s)$ (reduction)

3. Calculate the cell emf for the given reaction

Use reduction potentials:

• Cathode (reduction): $\text{Sn}^{2+}/\text{Sn}$, so $E^\circ_{\text{cathode}} = -0.16\,\text{V}$
• Anode corresponds to oxidation of Cu, but using reduction potential of $\text{Cu}^{2+}/\text{Cu}$: $E^\circ_{\text{anode}} = +0.34\,\text{V}$

Therefore, $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ $E^\circ_{\text{cell}} = (-0.16) - (0.34) = -0.50\,\text{V}$

Since concentrations are both $1\,\text{M}$, reaction quotient $Q=1$, hence: $E = E^\circ = -0.50\,\text{V}$

4. Use relation between Gibbs free energy and emf

$\Delta G = -nFE$

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

So, $\Delta G = -2 \times 96500 \times (-0.50)$

$\Delta G = 96500\,\text{J}$

5. Final answer

$\boxed{96500\,\text{J}}$