For the reaction: 

$2{\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right)+2{\mathrm{I}}^{-}\left(\mathrm{aq}\right)\to 2{\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{I}}_2\left(\mathrm{s}\right)$

The magnitude of the standard molar free energy change, 

${\mathrm{\Delta }}_{\mathrm{r}}{\mathrm{G}}_{\mathrm{m}}^{°}=-$ _________________ kJ ( Round off to the Nearest Integer).

$\left[\begin{array}{l}{\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}\left(\mathrm{s}\right)}^{\circ }=-0.440\mathrm{V}; {\mathrm{E}}_{{\mathrm{Fe}}^{3+}/\mathrm{Fe}\left(\mathrm{s}\right)}^{\circ }=-0.036\mathrm{V};\\ {\mathrm{E}}_{{\mathrm{I}}_2/2{\mathrm{I}}^{-}}^{\circ }=0.539\mathrm{V};\mathrm{F}=96500 \mathrm{C}\end{array}\right]$

Considering the given reaction:

$$\begin{gathered} {\mathrm{Fe}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Fe}; {\mathrm{E}}_{{\mathrm{Fe}}^{3+}/\mathrm{Fe}}^{\circ }=-0.036 \mathrm{V} \\ \mathrm{Fe}\to {\mathrm{Fe}}^{2+}+2{\mathrm{e}}^{-}; {\mathrm{E}}_{\mathrm{Fe}/{\mathrm{Fe}}^{2+}}^{\circ }=0.440 \mathrm{V} \\ \overline{{\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-}\to {\mathrm{Fe}}^{2+}, {\mathrm{E}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}^{\circ }=?} \end{gathered}$$

$\begin{array}{l}{\mathrm{\Delta G}}^0=-{\mathrm{nFE}}_{\mathrm{cell}}^0\\ ∆{\mathrm{G}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}^{°}=∆{\mathrm{G}}_{{\mathrm{Fe}}^{3+}/\mathrm{Fe}}^{°}+∆{\mathrm{G}}_{\mathrm{Fe}/{\mathrm{Fe}}^{2+}}^{°}\\ {\mathrm{E}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}^{°}=3\left({\mathrm{E}}_{{\mathrm{Fe}}^{3+}/\mathrm{Fe}}^{°}\right)+2\left({\mathrm{E}}_{\mathrm{Fe}/{\mathrm{Fe}}^{2+}}^{°}\right)\\ 1\left({\mathrm{E}}_{{\mathrm{Fe}}^3/{\mathrm{Fe}}^{+2}}^0\right)=(-0.036)3-(0.44\times 27)\\ {\mathrm{E}}_{{\mathrm{Fe}}^3/{\mathrm{Fe}}^{+2}}^0=0.772\mathrm{V}\end{array}$

${\mathrm{E}}_{\mathrm{cell}}^{°}$ for the given reaction:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}^{°}={\mathrm{E}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}^{°}+{\mathrm{E}}_{{\mathrm{I}}^{-}/{\mathrm{I}}_2}^{°} \\ {\mathrm{E}}_{\mathrm{cell}}^{°}=0.772-0.539=0.233 \mathrm{V} \end{gathered}$$

${\mathrm{\Delta }}_{\mathrm{r}}{\mathrm{G}}_{\mathrm{m}}^{°}$ is calculated as:

$$\begin{gathered} {∆}_{\mathrm{r}}{\mathrm{G}}_{\mathrm{m}}^{°}=-{\mathrm{nFE}}_{\mathrm{cell}}^{\circ } \\ {∆}_{\mathrm{r}}{\mathrm{G}}_{\mathrm{m}}^{°}=-2\times 96500\times 0.233 \\ {∆}_{\mathrm{r}}{\mathrm{G}}_{\mathrm{m}}^{°}=-44969 \mathrm{J}=-44.9 \mathrm{kJ}\cong -45 \mathrm{kJ} \end{gathered}$$