A voltaic cell is constructed that is based on the following reaction: Sn^2+ (aq

Question

A voltaic cell is constructed that is based on the following reaction: Sn^2+ (aq) + Pb(s) rightarrow Sn(s) + Pb^2+ (aq). If the concentration of Sn^2+ in the cathode compartment is 1.30 M and the cell generates an emf of 0.16 V, what is the concentration of Pb^2+ in the anode compartment? Express your answer using two significant figures. If the anode compartment contains [SO^2- _4] = 1.00 M in equilibrium with PbSO_4(s), what is the K_sp of PbSO_4 ? Express your answer using two significant figures.

Explanation / Answer

Sn+2 = 1.3

emf = 0.16 V

find Pb+2

Pb2+ + 2 e Pb(s) 0.126

Sn2+ + 2 e Sn(s) 0.13

E°cell = 0.13-0.126 = 0.004 V

so..

Ecell = E°cell - 0.0592/n*log(Pb+2 / Sn+2)

0.16 = 0.004 - 0.0592/2* log([Pb+2] / (1.3) )

solve for Pb+2

-(0.16 -0.004 ) *2 / (0.0592) = log([Pb+2] / (1.3) )

-5.27027 = log([Pb+2] / (1.3) )

10^-5.27027 = Pb+2 / 1.3

[Pb+2] = 1.3* (10^-5.27027 ) = 0.00000697707 M

B)

Ksp =?

Aplpy

dG° = -RT*ln(K)

dG° = -nF*E°cell

ln(K) = nF/(R)*E°cell

PbSO4(s) + 2 e Pb(s) + SO42 0.3588

ln(K) = 2*96500 / (8.314*298) * (-0.3588)

K = exp(-27.950) = 7.268*10^-13

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