Gibbs Free Energy Change for Given Concentration Cell

To calculate the change in Gibbs free energy ($\Delta G$) for the given concentration cell, we use the fundamental thermodynamic relation:

$\Delta G = -nFE_{\text{cell}}$

Step 1: Identify the cell reactions and number of transferred electrons ($n$)

The given cell representation is: $\text{M(s)} \mid \text{M}^{2+}_{\text{anode}} \parallel \text{M}^{2+}_{\text{cathode}} \mid \text{M(s)}$

The half-cell reactions are:

• At Anode (Oxidation): $\text{M(s)} \rightarrow \text{M}^{2+}_{\text{anode}} + 2e^-$
• At Cathode (Reduction): $\text{M}^{2+}_{\text{cathode}} + 2e^- \rightarrow \text{M(s)}$

Net Cell Reaction: $\text{M}^{2+}_{\text{cathode}} \rightarrow \text{M}^{2+}_{\text{anode}}$

From the half-cell reactions, the number of moles of electrons transferred per mole of reaction is $n = 2$.

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Step 2: Calculate $\Delta G$

Given data:

• $n = 2$
• $F = 96500 \text{ C mol}^{-1}$
• $E_{\text{cell}} = 0.059 \text{ V}$

Substituting these values into the formula: $\Delta G = -2 \times 96500 \text{ C mol}^{-1} \times 0.059 \text{ V}$

$\Delta G = -11387 \text{ J mol}^{-1}$

Converting Joules to kilojoules ($\text{kJ}$): $\Delta G = -\frac{11387}{1000} \text{ kJ mol}^{-1} = -11.387 \text{ kJ mol}^{-1} \approx -11.4 \text{ kJ mol}^{-1}$

Thus, the value of $\Delta G$ is $-11.4 \text{ kJ mol}^{-1}$.

Correct Option: D