Question

A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The initial concentrations of Pb2+ and Cu2+ are 5.10×10−2 M and 1.60 M , respectively. The initial potential is 0.51 V and the cell potential is 0.45 V when the concentration of Cu2+ has fallen to 0.230 M. What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.350 V ?

Answer 1

Cu+2 + 2e- ==> Cu E^0=+0.34
Pb+2 + 2e- ==>Pb E^0=-0.13

Pb ==>. Pb+2 + 2e- +0.13
Cu+2 + 2e- =⇒ Cu +0.34
---------------------------------------...
Pb + Cu+2 ==⇒ Pb+2 + Cu E cell =+0.47

Ecell = E^0 + 0.059/2 log [Pb^2+]/[Cu^2+]

0.35 = 0.47 + 0.059/2 log [Pb^2+]/[Cu^2+]

or, [Pb^2+]/[Cu^2+] =  2138

	Pb	Cu^2+	Pb^2+	Cu
inital	--	1.6	5.10×10−2
change	--	-x	+x	--
equilibrium	--	1.6-x	5.10×10−2+x	--

5.10×10−2+x/1.6-x = 2138

or, x = 1.599

So, all the Cu^2+ has been consumed in the reaction.