Question

What is the value of the equilibrium constant, K for a redox reaction involving the transfer of 6 mol of electrons if its standard potential is 0.043?

Answer 1

Let us write down the equations first,

G = -RTlnK ........................Eqn (1)

G = -nFE............................Eqn (2)

G = gibbs free energy

R = Gas constant, value of R = 8.3145 J/mol*K

T = Temperature in Kelvin

n = no. of moles transferred

K = Equilibrium constant

E = Standard Electron potential

F = Faradays constants, F = = 96485.34 C/mol

Given data:

n = 6 mols

E = 0.043 V

T = 298 K (Since Standard potential)

K = ?

Solution:

Now combine the above equations (1) & (2),

-nFE = -RTlnK

Therefore,

K = e^(nFE/RT)

K = e^(6 mol * 96485.34 (C/mol) * 0.043 V / (8.3145 J/mol*K)(298 K))

K = e^(10.05) = 23082.28