Question

15. What is the equilibrium constant (K) at 25°C for the following cell reaction?
Fe(s) + Cd2+(aq)
Fe2+(aq) + Cd(s); E°cell = 0.010 V
A) 0.010 B) 1.5 C) 0.25 D) 1.0 E) 2.2

Answer 1

Equilibrium constant and E° are related in the following equation:

E° = RT LnK / nF

Ln K = nFE°/RT Where T = 25 + 273 = 298K and R = 8.314 J/mol K and F=96500 C/mol

In order to get n we have the half reactions:

Fe(s) -------> Fe²⁺ (aq) + 2e-

Cd²⁺ (aq) + 2e- ------> Cd(s)

And the final reaction is:

Fe(s) + Cd²⁺ (aq) -------> Fe²⁺ (aq) + Cd(s) .... and n = 2.

Now, we can get K from the above equation:

Ln K = nFE°/RT = (2)(96,500 Cmol^-1)(0.010JC^-1) / (8.314 Jmol^-1K^-1)(298K)

LnK = 1,930 / 2,478 and Ln K = 0.78

If you apply antilogarithm:

K = 2.2