Question

Calculate E0 for the new half-cell reaction

VO2+ + 2H+ + 2e- → V2+ + H2O

And ΔG0 and Keq for the reaction given below at 25 °C

VO2+ + V2+ + 2H+ → 2V3+ + H2O

Assume all reactants and products are at unit activity.

Answer 1

VO+2 + 2H+ + e- → VO+2 + H2O   ---> Eo = 0.991 V        ----> 1

VO+2 + 2H+ + e- ---> V+3 + H2O -----> Eo = 0.337 V   ----> 2

V+3 + e- ----> V+2   ---> Eo = -0.255V   -----> 3

Add 2 and 3

==> VO+2 + 2H+ + 2e- -----> V+2 + H2O

Eo = 0.337 - 0.255 = 0.082 V

Add 2 and reverse of 3rd reaction

VO+2 + 2H+ + e- ---> V+3 + H2O

V+2 -------> V+3 + e-

Net reaction

VO+2 + V2+ + 2H+ → 2V+3 + H2O

Eo = 0.337 + 0.255 = 0.592 V

delta G = -nFE = -1*96500*0.592 = -57.128 KJ

Here n = 1 because number of electrons cancelled in the reaction is 1

delta G = -RT lnK

-57.128KJ = -8.314*298 *ln K

K = 1.0327*10^10