Question

Use the tabulated half-cell potentials below to calculate the equilibrium constant (K) for the following balanced redox reaction at 25°C. Pb2+(aq) + Cu(s) → Pb(s) + Cu2+(aq)

Answer 1

Pb²⁺(aq) + Cu(s) Pb(s) + Cu²⁺(aq)

The two half cells are:

Pb²⁺(aq) + 2e^-  Pb(s) E° = - 0.126 V

Cu(s) Cu²⁺(aq) + 2e^- E° = - 0.337 V (reversed as oxidation required)

Hence, E° = (-0.126 V) + (-0.337 V) = - 0.463 V.

Now, Using Nerst equation

E° = (RT/nF) x 2.303 log K, [ K is the equilibirium constant at T = 298 K ]

  - 0.463 = [ (8.314 x 298) / (2 x 96500) ] x 2.303 log K

  - 0.463 = (0.0128) x 2.303 log K

log K = - 15.71

K = 10^-15.71

K = 1.95 x 10^-16