Question

Many different ways have been proposed to make batteries. One cell is set up with copper and lead electrodes in contact with CuSO4(aq) and Pb(NO3)2 (aq), respectively. If the Pb2+ and Cu2+ concentrations are each 1.0 M, what is the overall cell potential?

Pb2+ + 2e- ---> Pb E = -0.22 V

Cu2+ + 2e- ----> Cu E = +0.34 V

answer: 0.56 V. how?

Answer 1

The given reduction potentials are

Pb2+ + 2e- ---> Pb E = -0.22 V

Cu2+ + 2e- ----> Cu E = +0.34 V

Since the reduction potential of Cu²⁺ (+0.34V) is positive and more than Pb²⁺ , hence Cu²⁺ will be reduced to Cu and Pb will be oxidised to Pb²⁺ .

Now the half cell reactions can be written as

Cu2+ (aq)+ 2e- ----> Cu(s),    Ered = +0.34 V

Pb(s)    ------------> Pb²⁺ (aq) , Eoxi = - (-0.22V) = +0.22 V

The overall cell reaction is

Cu2+ (aq) + Pb(s) ---------> Cu(s) + Pb²⁺ (aq), Ecell

Hence, Ecell = Eoxi + Ered   = +0.22 V + (+0.34 V) = 0.56V

Given [Cu2+ (aq)] = [Pb²⁺ (aq)] = 1.0 M

Number of electrons transferred during the redox reaction, n = 2

Now the cell potential can be calculated from Nernst equation.

E = E0cell - (0.0591/n) log [Pb²⁺ (aq)] / [Cu2+ (aq)] = 0.56 V - (0.0591/2)xlog( 1.0M / 1.0M) = 0.56 V - (0.03)xlog1

Since log1 = 0

= 0.56 - (0.0591/2) x 0 = 0.56 V    (answer)