If the concentration of sn2 + in the cathode compartment is 1.50 m and the cell generates an emf of 0.24 v, what is the concentration of pb2 + in the anode compartment?

E° cell = E° cathode - E° anode

=-0.14 + 0.13

= - 0.01V

Q = [pb⁺²]/[Sn⁺²] = Q = [pb⁺²]/1.50]

E = E° = - 0.05916/n log Q

0.24 = - 0.01 = 0.05916/2 log Q

Log Q = - 8.452

Q = + 3.53 * 10⁻⁹

(pb + 2) = 3.53 * 10⁻⁹ * 1.5

=5.297 * 10⁻⁹M

[pb⁺²] = 5.3 * 10⁻⁹M