Question

3) Calculate the potential in the following cell (8 points) Pb|Pb2+(1.00M)|| Ag+ (1.00M) |Ag

a. right at the beginning when you start using this voltaic cell,

b. when 50% of the Ag+ ions have been consumed,

c. when 99% of the Ag+ ions have been consumed, and

d. when 99.99% of the Ag+ ions have been consumed. Keep in mind that both, the Pb2+ and the Ag+ concentrations are changing

Answer 1

The cell potentials calculations are as given below.

E^ocell = Ecathode - Eanode

           = 1.60 - 0.126

           = 1.726 V

We will use Nernst equation,

Ecell = E^ocell - 0.059/2 log Q

Q = [product]/[reactant]

At the begining,

Q = 1

Ecell = E^ocell = 1.726 V

When 50 % Ag+ is consumed

Ecell = 1.71 V

When 99.99% Ag+ is consumed

Ecell = 1.70 V