How to Calculate Ksp for Ag2SO4 Using Electrochemistry?

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Homework Statement

An electrochemical cell is made up of a Pb and an Ag electrode:
Ag+ + e --> Ag(s) E = 0.8V
Pb2+ + 2e --> Pb(s) E = -0.13V

Calculate the Ksp for Ag2SO4(s).
"Note that to obtain ions in the right compartment, excess silver sulfate solid was addded and a small amount of it dissoved"
There is a picture of the cell, showing the Pb electrode on the left and the Ag on the right, a voltage of 0.83V and a concentration of 1.8 M Pb2+

Homework Equations

Ksp = [Ag+]^2 * [SO42-]
E = E(ox) + E(red)
E = Eo + (RT/nF)lnQ

The Attempt at a Solution

lead should be the anode, silver the cathode, so E = +0.93V

0.83V = 0.93V + (8.314*298/(2*96,485) ln(1.8/[Ag+]^2)
*no temperature was given, so 298 K was used*
ln(1.8/[Ag+]^2) = -7.789
[Ag+] = 65.9M

this is way too big, so something must be wrong in the of this solution

making lead the cathode and silver the anode doesn't seem to work

it seems that the reason that there is a voltage is that the system went from being at an equilibrium concentration of Ag+ ions (with no voltage) to a voltage of 0.83V when the solid was added, and the Ag+ concentration went up. there is no data on this initial state, but the ratio of Pb to Ag ions can be found...

 

[Borek]

Write separate Nernst equations for both half cells.