In: Chemistry
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)
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