Question

A voltaic cell using Pb²⁺/Pb and Ni²⁺/Ni half-cells is set up at standard conditions, and each compartment has a volume of 355 mL. What is the cell potential, in V, after it has delivered 0.150 A for 16.0 hours at 25.0 °C? (E° for Pb²⁺/Pb = -0.130 V and E° for Ni²⁺/Ni = -0.250 V.)

Answer 1

Assume 1M standard conditions to start calculate cell voltage (0.12V).

Now, setup BCA table based on mols, assume constant volume.

Pb2+ --> Ni2+

I 0.35 0.35

C -x +x

E 0.35-x 0.35+x thus Q = (0.35+x)/(0.35-x)

x is mols of Pb2+ consumed = amps x time /96540/2 = mols Ni2+ produced.

since volume is fixed, no need to worry about concentration: mol ratio = molarity ratio

now recalculate E = E0 - RTl/zF lnQ you will note E<<E0, as the cell depletes.