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...
