# Calculating the solubility product for a salt in a metal-metal insoluble salt electrode

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?

• E_ cell=RT/(nF)ln([Pb2+][SO4^2-]/Ksp) Commented Sep 24, 2023 at 9:25
• @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? Commented Sep 24, 2023 at 9:30
• 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. Commented Sep 24, 2023 at 10:38
• @Poutnik Even I thought it was odd, but that was what i got when i added the equations.. Commented Sep 24, 2023 at 10:38
• Equations are partly wrong. Commented Sep 24, 2023 at 10:38