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A voltaic cell is constructed that is based on the following reaction: Sn2+(aq)+Pb(s)→Sn(s)+Pb2+(aq). 1. If the...

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

1. If the concentration of Sn2+ in the cathode compartment is 1.50 M and the cell generates an emf of 0.25 V , what is the concentration of Pb2+ in the anode compartment?

2.If the anode compartment contains [SO2−4]= 1.00 M in equilibrium with PbSO4(s), what is the Ksp of PbSO4?

Expert Solution

1)

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)

= -0.00400 V

Number of electron being transferred in balanced reaction is 2

So, n = 2

Use:

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

0.25 = -0.00400 - (0.0592/2) log ([Pb2+]/1.5)

log ([Pb2+]/1.5) = -8.581

([Pb2+]/1.5) = 2.624*10^-9

[Pb2+] = 3.94*10^-9 M

Answer: 3.94*10^-9 M

2)

PbSO4 (s) <—> Pb2+ (aq) + SO4 2- (aq)

Ksp = [Pb2+][SO42-]

Ksp = (3.94*10^-9)*(1.00)

Ksp = 3.94*10^-9

Answer: 3.94*10^-9

queen_honey_blossom answered 2 years ago

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