The voltaic cell, Pb(s) Pb2+, Na2SO4(0.1M),PbSO4(s) // Sn2+(1M) Sn(s), generates voltage of 0.19V. The standard reduction potentials for Pb2+ and Sn2+ are – 0.13V and -0.14V, respectively.

(A)What is the concentration of Pb2+ in the anode compartment?

(B)Calculate the Ksp for PbSO4

Ans: (A)1.75x10-7M (B)1.75*10-8

My calculation:

(A) 0.19 = (0.14 - 0.13) - 0.06/2*log(1/(0.1*[Pb2+]) => [Pb2+] = 4.6*10^6

I wonder where I did wrong. Thanks!