Question

A voltaic cell is constructed so its based on the reaction Sn2+(aq)+Pb(s)=Sn(s)+Pb2+(aq) a. If the concentration of Sn2+ in the cathode compartment is 1.00 M and the cell generates an emf of 0.15 V , what is the concentration of Pb2+ in the anode compartment? b. If the anode compartment contains [SO2−4]= 1.20 M in equilibrium with PbSO4(s), what is the Ksp of PbSO4?

Answer 1

Sn²⁺(aq)+ Pb(s) = Sn(s)+Pb²⁺(aq)

The cell potential is given as

E = E^o - 0.059/n log [Pb2+] / [Sn2+]

E = 0.15V, n =2

E^o = E^o_cathode - E^o_anode

E^o = -0.14 - (-0.126) = - 0.014V

0.15 = - 0.014 - 0.059/2 log [Pb2+] / [1.0]

- log [Pb2+] / [1.0] = (0.15 + 0.014 / 0.0295 )

- log [Pb2+] / [1.0] = 5.56

log [Pb2+] / [1.0] = - 5.56

[Pb2+] / [1.0] = 2.75 X 10^-6

[Pb²⁺] = 2.75 X 10^-6M

(b) PbSO₄ <----> Pb²⁺ + SO₄^2-

Ksp = s X s

s = molar solubility. = 1.20

Ksp = s²

Ksp = (1.2)² = 1.44