The standard Gibbs energy change ($\Delta_r G^\circ$) for a cell reaction is given by the relation: $\Delta_r G^\circ = -nFE^\circ_{\text{cell}}$

For the given reaction: $2\text{Fe}^{3+}\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{Fe}^{2+}\text{(aq)} + \text{I}_2\text{(aq)}$ The number of moles of electrons transferred ($n$) is $2$ (since $2\text{I}^- \rightarrow \text{I}_2 + 2e^-$ and $2\text{Fe}^{3+} + 2e^- \rightarrow 2\text{Fe}^{2+}$).

Given: $n = 2$ $F = 96500 \text{ C mol}^{-1}$ $E^\circ_{\text{cell}} = 0.24 \text{ V}$

Substituting the values into the formula: $\Delta_r G^\circ = - (2) \times (96500 \text{ C mol}^{-1}) \times (0.24 \text{ V})$ $\Delta_r G^\circ = - 46320 \text{ J mol}^{-1}$

Converting Joules to kiloJoules: $\Delta_r G^\circ = - 46.32 \text{ kJ mol}^{-1}$