Question

A voltaic cell is constructed that is based on the following reaction:
Sn2+(aq)+Pb(s)→Sn(s)+Pb2+(aq).

a. If the concentration of Sn2+ in the cathode compartment is 1.50 M and the cell generates an emf of 0.22 V , what is the concentration of Pb2+ in the anode compartment?

b. If the anode compartment contains [SO2−4]= 1.50 M in equilibrium with PbSO4(s), what is the Kspof PbSO4?

Answer 1

a)

Lets find Eo 1st

from data table:

Eo(Pb2+/Pb(s)) = -0.126 V

Eo(Sn2+/Sn(s)) = -0.13 V

As per given reaction/cell notation,

cathode is (Sn2+/Sn(s))

anode is (Pb2+/Pb(s))

Eocell = Eocathode - Eoanode

= (-0.13) - (-0.126)

= -0.004 V

Number of electron being transferred in balanced reaction is 2

So, n = 2

Use:

E = Eo - (0.0592/n) log {[Pb2+]/[Sn2+]}

0.22 = -0.004 - (0.0592/2) log ([Pb2+]/1.5)

log ([Pb2+]/1.5) = -7.568

([Pb2+]/1.5) = 2.707*10^-8

[Pb2+] = 4.06*10^-8 M

Answer: 4.06*10^-8 M

b)

PbSO4 (s)       <—> Pb2+ (aq)   +   SO42- (aq)

Ksp = [Pb2+] [SO4 2-]

= (4.06*10^-8)*(1.50)

= 6.09*10^-8

Answer: 6.09*10^-8