Free Energy Calculation for Galvanic Cell with Non Standard Concentrations

For the given electrochemical cell: $\text{Zn}(s) \mid \text{ZnSO}_4(aq) \parallel \text{CuSO}_4(aq) \mid \text{Cu}(s)$

The individual half-cell reactions are:

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

The overall cell reaction is: $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightleftharpoons \text{Zn}^{2+}(aq) + \text{Cu}(s)$

The number of moles of electrons transferred in this balanced cell reaction is $n = 2$.

The reaction quotient ($Q$) for the reaction is given by: $Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}$

It is given that the concentration of $\text{Zn}^{2+}$ is $10$ times higher than that of $\text{Cu}^{2+}$, i.e., $[\text{Zn}^{2+}] = 10 [\text{Cu}^{2+}]$. Substituting this into the reaction quotient: $Q = 10$

The relationship between the Gibbs free energy change ($\Delta G$) and the cell potential ($E_{\text{cell}}$) is: $\Delta G = -n F E_{\text{cell}}$

According to the Nernst equation, the cell potential $E_{\text{cell}}$ is expressed as: $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q$

Substituting $E_{\text{cell}}$ from the Nernst equation into the formula for $\Delta G$: $\Delta G = -n F \left( E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q \right)$ $\Delta G = -n F E^\circ_{\text{cell}} + RT \ln Q$

Now, substituting the given values:

• $n = 2$
• $E^\circ_{\text{cell}} = 1.1\text{ V}$
• $Q = 10$
• $\ln(10) = 2.303 \log_{10}(10) = 2.303$

$\Delta G = -2 \times F \times 1.1 + RT (2.303)$ $\Delta G = -2.2 F + 2.303 RT$ $\Delta G = 2.303 RT - 2.2 F$

Thus, the correct expression for $\Delta G$ corresponds to Option B.