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:


At equilibrium:


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...

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.