Calculate the gibbs free energy change (dG°) and equlibrium constant (Keq) at 25°c for the reaction - Cu (s)+I 2 (s)-------> Cu 2+ (aq) +2I -(aq).Given E°Cu 2+/Cu=0.34 v and E°I 2/I- =0.54 v.

Q.Calculate the gibbs free energy change  (dG°) and equlibrium constant  (Keq) at 25°c for the reaction - Cu (s)+I 2 (s)-------> Cu 2+ (aq) +2I -(aq).Given E°Cu 2+/Cu=0.34 v and E°I 2/I- =0.54 v.

Answer :

The reaetion is-

cu(s) + I 2 (s)------>Cu2+ ( aq) + 2 I - (aq)

For this reaction cell Potential,

E °cell=E °cu2+/cu - E° I2/I-
=O:54-0:34
=0:20 V

The relation
between standard Gibbs free
energy change and E°ceu is-
dG° = -nF E°cell

n = number of electron involved
and E°cell= 0 · 20 V

We know,
F = 96500 C

::dG°ニ2×96500 × 0.2 J
=38. 600 KJ

:. Standard Gibbs free energy change - 38. 6kJ

Again,The relation between standard Gibbs fee energy change (dG°) and Equilibrium Constant ( Keq) is-

dG °=-RT In Keq

we get dG = - 38 6 00J

Given, T= 25°c=298k
and R= 8.314. J. mol-1k-1

by calculating we get Keq=5.8×10^6