Chemistry 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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The calculation presented indicates that 0.19 equals (0.14 - 0.13) minus 0.06 divided by 2 times the logarithm of the inverse of (0.1 times [Pb2+]), leading to a calculated [Pb2+] of 4.6*10^6. Participants in the discussion identify a potential error in the calculation and express uncertainty about whether the mistake lies with the questioner or the provided data. There is a call for clarification on the cell reaction equation and the correct expression for K. The conversation highlights discrepancies in the calculations and encourages further examination of the problem. Accurate calculations are crucial for understanding the chemical equilibrium involving lead ions.
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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.
 
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