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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...