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When trying to find the solubility product of a salt in a metal - metal insoluble salt electrode, I find that I am getting a different answer if I consider the cell to be a concentration cell and if I do not.

For example, to find $K_\mathrm{sp}$ of $\ce{PbSO4}$ in the cell: $\ce{Pb}$(s) | $\ce{PbSO4}$(s) | $\ce{Na2SO4}(aq)(0.01M)$ || $\ce{Pb(NO3)2} (0.1M)$| $\ce{Pb}$(s)

Given: $E_\mathrm{cell}$ at $298 K$ = $0.236 V$.

At the anode :
$\ce{Pb -> Pb^2+ + 2e-}$
$\ce{Pb^2+ + SO4^2- -> PbSO4}$

Overall: $\ce{Pb + SO4^2- -> PbSO4 + 2e-}$

So, $E_\mathrm{anode}$ = $E^0_\mathrm{anode}$ - $\frac{0.059}{2}$ $\log\frac{1}{[\ce{SO4^2-}]}$

At the cathode :
$\ce{Pb^2+ + 2e- -> Pb}$
$\ce{Pb(NO3)2 -> Pb^2+ + 2NO3^-}$

Overall: $\ce{Pb(NO3)2 + 2e- -> Pb + 2NO3^-}$

and, $E_\mathrm{cathode}$ = $E^0_\mathrm{cathode}$ - $\frac{0.059}{2}$ $\log{[\ce{NO3^-}]^2}$


$E_\mathrm{cell}$ = $E_\mathrm{cathode}$ - $E_\mathrm{anode}$

$E_\mathrm{cell}$ = $E^0_\mathrm{cell}$ - $\frac{0.059}{2}$ $\log{[\ce{NO3^-}]^2}[\ce{SO4^2-}]$

Thus, $E^0_\mathrm{cell}$ = 0.236 + $\frac{0.059}{2}$ $\log{(0.1)^2}(0.01)$ = 0.118 V

At equilibrium, $E_\mathrm{cell}$ = $0$, so $\log{K_\mathrm{sp}}$ = $\frac{(2)E^0_\mathrm{cell}}{0.059}$

So, $\log{K_\mathrm{sp}}$ = 4

$K_\mathrm{sp}$ = $10^{4}$

However, the answer comes out to be $10^{-11}$ when I consider it to be a concentration cell of $\ce{Pb^2+}$. Where am I going wrong or why does this method not work?

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  • $\begingroup$ E_ cell=RT/(nF)ln([Pb2+][SO4^2-]/Ksp) $\endgroup$
    – Poutnik
    Commented Sep 24, 2023 at 9:25
  • $\begingroup$ @Poutnik could you elaborate? also, my equation does not have Pb2+ in it, so is it wrong that I found Ksp from the E0cell value I got? $\endgroup$
    – algorhythm
    Commented Sep 24, 2023 at 9:30
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    $\begingroup$ It is a concentration cell, where E_ cell=RT/(nF)ln(c1/c2) , where c1 is given by c of Pb^2+ from (dissociated) Pb(NO3)2 and c2 is given by Ksp and c(SO4^2-) from (dissociated) Na2SO4. $\endgroup$
    – Poutnik
    Commented Sep 24, 2023 at 10:38
  • $\begingroup$ @Poutnik Even I thought it was odd, but that was what i got when i added the equations.. $\endgroup$
    – algorhythm
    Commented Sep 24, 2023 at 10:38
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    $\begingroup$ Equations are partly wrong. $\endgroup$
    – Poutnik
    Commented Sep 24, 2023 at 10:38

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