Calculation Error: 0.19 = (0.14 - 0.13) - 0.06/2*log(1/(0.1*[Pb2+]))

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    Calculation Error
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hidemi
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Homework Statement
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
Relevant Equations
E1 = E2 - 0.06/n*log(K)
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!
 
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Write out the equation for the cell reaction. Then write out the correct expression for K.

There appears to be a mistake in your calculation. However, I wonder if the questioner made the same mistake, or there is a mistake in the given data, because I disagree with the given answer.
 
Thank you for pointing out the mistake and I think I got a more reasonable answer.