Given the following cell reaction:
 \(\mathrm{2Fe^{3+}(aq) \ + \ 2I^{-}(aq)\rightarrow 2Fe^{2+}(aq) \ + \ I_{2}(aq)}\)

\(E_{cell}^{o} \ = \ 0.24 \ V\) at $298$ $\mathrm{K}$.
The standard Gibbs energy ∆_rG^⊝ of the cell reaction is:

[Given: $\mathrm{F }=\hspace{0.17em}96500$ $\mathrm{C}$ ${\mathrm{mol}}^{-1}$]

1. $23.16$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

2. $-46.32$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

3. $-23.16$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

4. $46.32$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

For a cell involving one electron ${\mathrm{E}}_{\mathrm{cell}}^{⊝}=0.\mathrm{59 }\mathrm{V}$ at 298 K. The equilibrium constant for the cell reaction is :
\(\mathrm{[Given~ that~ \frac {2.303 ~RT}{F} = 0.059 ~V~ at~ T = 298 K]}\)