# Concentration of solution when a battery reaches equilibrium

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...

• Take care ! $\ce{K = [Cd^{2+}]·[SO4^{2-}] = [Cd^{2+}]^2}$ Jan 5, 2023 at 2:12
• That actually makes sense. Thanks. The actual answer should be $7.5$ M approx. Jan 5, 2023 at 13:10
• 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 $\ce{K=[Cd^{2+}]^2/(1 M^{2} )}$. Jan 5, 2023 at 14:20
• 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. Jan 5, 2023 at 14:38
• So the activity of $\ce{Cd^{2+}}$ is $\sqrt{55.67} = 7.461$. It is not far from the expected result $7.5$ M. Jan 5, 2023 at 17:49