Question

A voltaic cell consists of a Pb/Pb⁺² half-cell and a Cu/Cu⁺² half-cell at 25°C. The initial concentrations are [Pb⁺²]=0.0500 M and [Cu⁺²]=1.50 M. What is the initial cell potential? What are the concentrations of Pb⁺² and Cu⁺²when the cell potential falls to 0.35 V

Answer 1

The overall cell reaction is: Pb + Cu2+ ==> Pb2+ + Cu . . .Eo cell = +0.47 V.

Q cell = [Pb2+]/[Cu2+] = moles Pb2+/moles Cu2+ since I am assuming that both reactions are going on in the same beaker, or if two beakers are involved, that they both contain the same volume of electrolyte.

A. E cell = Eo cell - 0.059/2 log Q = 0.47 - 0.059/2 log ([Pb2+]/[Cu2+]) = 0.47 - 0.059/2 log (0.0500/1.50) = 0.47 - (0.059/2)(-1.48) = 0.47 - (-0.04) = 0.51 V . . .so we agree.

B. Note that the sum of [Pb2+] and [Cu2+] = 1.55.

E cell = 0.47 - 0.059/2 log Q
0.35 = 0.47 - 0295 loq Q
-0.12 = -0.0295 log Q
4.07 = log Q
Q = 10^4.07 = 11,700

If x = [Pb2+] and y = [Cu2+], then x + y = 1.55 and x/y = 11,700 (or, x = 11,700y).
11,700y + y = 1.55
11.700y = 1.55
y = 0.00013 = [Cu2+] and x ([Pb2+]) = essentially 1.55 M