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Problem: We want to build a battery employing as electrodes ${\text{PbSO}_4}_{(s)} | {\text{Pb}}_{(s)}\ (E^º=-0.351\ V)$ i ${\text{Cd}}_{(aq)}^{2+} | {\text{Cd}}_{(s)}\ (E^º=-0.4026\ V)$ in a solution of ${\text{CdSO}_4}$ of molar concentration M. At what concentration of ${\text{CdSO}_4}$ will the battery reach equilibrium?

Data: $T=298.15$ K; $P=1$ atm; $F=96500\ C/mole\ e^{-}$.

My attempt:

There's reduction happening on the ${\text{PbSO}_4}_{(s)} | {\text{Pb}}_{(s)}$ electrode and oxidation on ${\text{Cd}}_{(aq)}^{2+} | {\text{Cd}}_{(s)}$. Thus, the global reaction should be

${\text{Cd}}_{(s)} + {\text{PbSO}_4}_{(s)} \rightleftarrows {\text{Pb}}_{(s)} + {\text{Cd}}_{(aq)}^{2+} + {\text{SO}_4}_{(aq)}^{2-}$

I'll use Nernst's equation:

$E=E^º-\dfrac{RT}{nF}\ln{Q}$

At equilibrium:

$0=E^º-\dfrac{RT}{nF}\ln{K}$

So $K=[{\text{CdSO}_4}]=\exp\left({\dfrac{nFE^º}{RT}}\right)=\exp\left({\dfrac{2\cdot 96500\cdot ((-0.351)-(-0.4026))}{8.31\cdot 298.15}}\right)\approx 55.67$ M.

The thing is this is not the actual answer. I don't know what else can I do...

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    $\begingroup$ Take care ! $\ce{K = [Cd^{2+}]·[SO4^{2-}] = [Cd^{2+}]^2}$ $\endgroup$
    – Maurice
    Jan 5, 2023 at 2:12
  • $\begingroup$ That actually makes sense. Thanks. The actual answer should be $7.5$ M approx. $\endgroup$
    – Conreu
    Jan 5, 2023 at 13:10
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    $\begingroup$ Technically $K$ is dimensionless, because the exponential is dimensionless. The way around this is to divide the concentrations by 1 mol/litre each and so $\ce{K=[Cd^{2+}]^2/(1 M^{2} )}$. $\endgroup$
    – porphyrin
    Jan 5, 2023 at 14:20
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    $\begingroup$ Note that at ion concentrations usable for galvanic cells, concentrations are far out from the range where we can afford to consider activity coefficients to be 1 and therefore usual calculations are strongly off the track. $\endgroup$
    – Poutnik
    Jan 5, 2023 at 14:38
  • $\begingroup$ So the activity of $\ce{Cd^{2+}}$ is $\sqrt{55.67} = 7.461$. It is not far from the expected result $7.5$ M. $\endgroup$
    – Maurice
    Jan 5, 2023 at 17:49

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