Thus emf of the cell = 0.91 V. Q.3.6. The cell in which the following reaction occurs :2Fe^(3+) (aq) + 2I^(-)(aq) to 2Fe^(2+0 (aq)+ I_(2)(s) has E_("cell")^(@) = 0.236 V at 298 K Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Thus emf of the cell = 0.91 V. Q.3.6. The cell in which the following

1.	Thus emf of the cell = 0.91 V. Q.3.6. The cell in which the following reaction occurs :2Fe^(3+) (aq) + 2I^(-)(aq) to 2Fe^(2+0 (aq)+ I_(2)(s) has E_("cell")^(@) = 0.236 V at 298 K Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Answer» Solution :The REACTIONS taking place in the cell may be written as under: `2Fe^(3+) + 2e^(-) to 2Fe^(2+)` or `2I^(-) to I_(2) + 2e^(-)` THUS, for the given cell reaction, n = 2. Using the following EQUATION and substituting the values, we get `Delta_(r)G^(@) =-nFE_("cell")^(@) =-2 xx 96500 xx 0.236 J = -45.55 kJ mol^(-1)` Now, using the following relation to calculate cquilibrium CONSTANT `Delta_(r)G^(@) = -2.303 RT log K_( C)` or `K_( C) "antilog" (7.983) = 9.616 xx 10^(7)`