Calculate Gibbs free energy change (dG°) and equlibrium constant (Keq)for the reaction Zn+Fe2+ ------>zn2+ +Fe at 25°C.Given E° Fe2+/Fe=-0.44 and E° Zn2+/Zn=-0.76

Q.Calculate Gibbs free energy change (dG°) and equlibrium constant  (Keq)for the reaction  Zn+Fe2+ ------>zn2+ +Fe.Given E° Fe2+/Fe=-0.44 and E° Zn2+/Zn=-0.76.

Ans.

Given Cell reaction -
Fe2+ + Zn ---->Fe + zn 2+

Now E°cell for this reaction is-
E°cell = E°Fe2+/Fe -E°zn2+/zn
=-0.44 + 0. 76v
= 0:32V

We Know, d G°=- nF E°cell

Where,n= number of electron involved
For this reaction, n= 2

we know, F = 96500 C

E°cell=0.32v

we get dG°=-61.76 kj

Again we know dG°=-RT ln Keq
we get Keq= 6.69×10^10

Gibbs free energy change= - 61.76, KJ and
Equilibrium Constant (keq) =6.69 x10^10