Question

Consider the following reaction:
Cu(s) + 2H+(aq) ↔ Cu2+(aq) + H2(g) E°cell = -0.34V

If you add copper metal to acidic water, do you expect to see H2 evolve? Find the equilibrium constant (K) for this reaction.

Answer 1

Given,

Eo cell = - 0.34 V

We know that,

delta Go = - nFEo

where,

n = number of electrons involved in the reaction equation

F = Faraday constant = 96485

Eo = - 0.34 V

We see that,

Cu ---> Cu2+ + 2e-

and 2H+ + 2e- -----> H2

Therefore,

n = 2

=> delta Go = - 2 x 96485 x (-0.34) = 65609.8 J

Since delta Go > 0, reaction is not feasible

=> H2 will not evolve

delta G = - RT ln K

R = Gas constant = 8.314

T = Temperature = 298 K

=> 65609.8 = - 8.314 x 298 x ln K

=> K = 3.16 x 10^-12 = Equilibrium Constant