The Gibbs energy change (in J) for the given reaction at $$[Cu^{2+}] = [Sn^{2+}] = 1$$ M and 298 K is:
$$Cu(s) + Sn^{2+}(aq.) \to Cu^{2+}(aq.) + Sn(s)$$;
$$(E^0_{Sn^{2+}|Sn} = -0.16 \; V, E^0_{Cu^{2+}|Cu} = 0.34 \; V, \text{Take } F = 96500 \; C \; mol^{-1})$$

The relationship between the standard Gibbs energy change and the standard cell potential ($$E^\circ_{\text{cell}}$$) is given by the formula:

$$\Delta G^\circ = -nFE^\circ_{\text{cell}}$$

Where:

• $$n$$ is the number of moles of electrons transferred in the balanced redox equation.
• $$F$$ is the Faraday constant ($$96500 \text{ C mol}^{-1}$$).
• $$E^\circ_{\text{cell}}$$ is the standard electromotive force (EMF) of the cell.

Step-by-Step Calculation:

• Step 1: Determine the Number of Transferred Electrons ($$n$$)

  Let's look at the given cell reaction:

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

  By splitting the process into half-reactions, we can see the total electron exchange:

  • Oxidation (Anode): $$\text{Cu}_{(s)} \rightarrow \text{Cu}^{2+}_{(aq)} + 2e^-$$
  • Reduction (Cathode): $$\text{Sn}^{2+}_{(aq)} + 2e^- \rightarrow \text{Sn}_{(s)}$$

  Thus, exactly $$n = 2$$ moles of electrons are transferred.




• Step 2: Calculate the Standard Cell Potential ($$E^\circ_{\text{cell}}$$)

  Using the standard reduction potentials provided for the half-cells:

  $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ $$E^\circ_{\text{cell}} = E^\circ_{\text{Sn}^{2+}/\text{Sn}} - E^\circ_{\text{Cu}^{2+}/\text{Cu}}$$ $$E^\circ_{\text{cell}} = -0.16 \text{ V} - 0.34 \text{ V} = -0.50 \text{ V}$$



• Step 3: Compute the Gibbs Energy Change ($$\Delta G^\circ$$)

  Substitute the values of $$n$$, $$F$$, and $$E^\circ_{\text{cell}}$$ into our main thermodynamic formula:

  $$\Delta G^\circ = -(2) \times (96500 \text{ C mol}^{-1}) \times (-0.50 \text{ V})$$ $$\Delta G^\circ = +96500 \text{ J}$$

Conclusion:

Because the calculated standard cell potential is negative, the reaction is non-spontaneous in the forward direction, resulting in a positive Gibbs free energy value of $$96500 \text{ J}$$.

Answer: 96500