The cell in which the following reaction occurs:
2Fe³⁺ (aq) + 2I^–(aq) $\to$ 2Fe²⁺ (aq) + I₂(s) has E⁰_cell = 0.236 V at 298 K. Calculate the standerd Gibbs energy and the equilibrium constance of the cell reaction.

Answer:

The cell is
2Fe³⁺ (aq) + 2I^–(aq) $\to$  $2{\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{I}}_2\left(\mathrm{s}\right)$
$$\begin{gathered} {\mathrm{E}}^0 \mathrm{cell} = 0.236 {{\mathrm{V}}^0}_{\mathrm{a}} \\ {∆}_{\mathrm{r}} {\mathrm{G}}^0 =-{{\mathrm{nFE}}^0}_{\mathrm{cell}} \\ =-2\times 0.236\times 96487 \mathrm{C} {\mathrm{mol}}^{-1} \\ = -45541 \mathrm{J} \\ = -45.54 \mathrm{kJ} {\mathrm{mol}}^{-1} \end{gathered}$$

$$\begin{gathered} \log \mathrm{Kc} = \frac{{{\mathrm{nFE}}^0}_{\mathrm{cell}}}{2.303 \times \mathrm{RT}} \\ = \frac{2\times 96487\times 0.236}{2.303\times 8.31\times 298} \\ = \frac{45541.864}{5703.1031} = 7.9854 \\ \mathrm{or} \mathrm{Kc} = \mathrm{Antilog} 7.9854 \\ \mathrm{or} \mathrm{Kc} = 9.62 \times {10}^7 \end{gathered}$$

Λ°_{m (HCOOH)} = Λ°_H+ + Λ°_HCOO^–
= 349.6 + 54.6
= 404.2 s cm²mol^–1
Λ _m= 46.1 s cm² mol^–1