Question

What is the calculated value of the cell potential at 298K for an electrochemical cell with the following reaction, when the Pb2+ concentration is 1.20 M and the Mn2+ concentration is 5.43×10-4 M ?

Pb2+(aq) + Mn(s) Pb(s) + Mn2+(aq)

Answer: in V

The cell reaction as written above is spontaneous for the concentrations given: true or false?

Answer 1

Pb+2 (aq) + Mn (s)   --------------> Pb (s) + Mn+2 (aq)

Oxidation : Mn -----------> Mn+2 + 2e-

Reduction : Pb+2 + 2e-   ------------> Pb

Eocell = Eored - Eooxi

           = - 0.13 - (–1.185 )

Eocell = 1.055 V

Ecell = Eo - 0.05916 / n log [Mn+2 / Pb+2]

         = 1.055 - 0.05916 / 2 log [5.43 x 10^-4 / 1.20]

cell potential Ecell = 1.15 V

The cell reaction as written above is spontaneous for the concentrations given : True