To solve this problem, we will use the relationship between the standard Gibbs energy change and the standard cell potential.$\\$The formula to calculate the standard Gibbs energy change $(\Delta G^\circ)$ is:$\\ \Delta G^\circ = -nFE^\circ_{cell} \\$Where:$\\$• $n$ is the number of moles of electrons transferred in the balanced equation.$\\$• $F$ is the Faraday constant, approximately $96500$ C mol$^{-1}. \\$• $E^\circ_{cell}$ is the standard cell potential, given as $0.24$ V.$\\$For the given cell reaction:$\\ 2Fe^{3+} (aq) + 2I^{-} (aq) \rightarrow 2Fe^{2+} (aq) + I_2 (s) \\$The number of electrons transferred ($n$) is 2, as each $Fe^{3+}$ gains one electron to become $Fe^{2+}. \\$Substitute the known values into the formula:$\\ \Delta G^\circ = -2 \times 96500 \times 0.24 \\$Calculate:$\\ \Delta G^\circ = -46320$ J mol$^{-1} \\$Convert to kJ mol$^{-1}: \\ \Delta G^\circ = -46.32$ kJ mol$^{-1} \\$Therefore, the standard Gibbs energy of the cell reaction is $-46.32$ kJ mol$^{-1}. \\$This corresponds to Option 1.