here we have reduction potentials
reduction potential of Ag+ is greater than Pb+2
so CATHODE = Ag
ANODE= Pb
now we will write reduction half reaction for both anode and cathode
CATHODE = 0.80 - (0.059/1)Log (1/[Ag+]) Ag+ + e- ----------> Ag
ANODE = - 0.13 - (0.059/2)Log (1/[Pb+2]) Pb+2 + 2e- -----------> Pb
E = E cathode - E anode
0.83 = 0.93 + (0.059/2)Log ([Ag+]^2 /[Pb+2])
solving eqn concentration of Ag+ is 0.026457M
Ag2SO4 -------------> 2Ag+ + SO4 -2
2s s
here s is solubality
Ksp = (2s)^2 * s = 4s^3
as we know 2s = 0.026457
s = 0.1322877
Ksp = 4s^3 = 9.26 * 10^ -6
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