Question

If the concentration of Sn2+ in the cathode compartment is 1.10 M and the cell generates an emf of 0.22 V , what is the concentration of Pb2+ in the anode compartment?

Answer 1

[Pb] = 1.1*10^-8

Step-by-step explanation:

The Eo values of the half cell reaction of the following are

Pb2+ + 2e- ==> Pb . . .Eo = -0.13 V

Sn2+ + 2e- ==> Sn . . .Eo = -0.14 V

Sn+2 is at cathode(reduction)compartment and Pb+2 is at anode(oxidation) compartment.

the overall reaction will be

Pb ==> Pb2+ + 2e- (oxidation) Eo = +0.13 V

Sn2+ + 2e- ==> Sn (reduction) Eo = -0.14 V

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Pb + Sn2+ ==> Pb2+ + Sn Eo cell = -0.01 V

Qcell = [Pb+2] / [Sn+2] ( pure Sn , Pb do not appear in Q

Ecell = Eo -0.059/n* logQ

0.22 = -0.0.1 - 0.059/2 logQ

0.23 = -0.0295logQ

LogQ= -7.796

Q = 10^-7.796≈ 1/10^8

[Pb+2] / [Sn+2] = 1/10^8

[Pb] = 1.1*10^-8