Question

The cell in which the following reaction occurs:
2Fe 3+ (aq) + 21" (aq) →2Fe2+ (aq) +1, (s) has E = 0.236 V at 298 K.
Calculate the standard Gibbs energy and the equilibrium constant of the
cell reaction.​

Answer 1

Given :

E° of the cell = 0.236 V

Temperature (T) = 298 K

To Find :

Standard Gibbs energy

Equilibrium constant of the reaction

Solution :

• The cell reaction is

             $2Fe^{3+}(aq)+2I^-(aq)\rightarrow 2Fe^{2+}(aq)+I_{2}(s)$

• The standard Gibbs energy for the reaction

             $\Delta G_{0}=-nFE^{0}_{cell}$

             $\Delta G_{0}=-2\times0.236 \times96487$

             $\Delta G_{0}= -45.54 kJ/mol$

The  Gibb’s energy for the reaction is -45.54 kJ/mol.

• We know the relation

              $logK_{c}=\frac{nFE^{0}_{cell)}}{2.303RT}$

              $logK_{c}=\frac{2\times96487 \times0.326}{2.303 \times8.31 \times298}$

              $logK_{c}=7.9854$

              $K_{c}=9.62\times10^{7}$

The equilibrium constant for the reaction is 9.62×10⁷.