Knowing that at 25 °C the following galvanic cell: $$\ce{Pb~|~Pb(NO_3)_2~1M~||~PbS~saturated~|~Pb}$$ shows an $\mathrm{EMF} =0.413~\mathrm{V}$, find the $K_\mathrm{sp}$ of $\ce{PbS}$.
My Approach
This is a concentration cell based on $\ce{Pb^2+}$. Since $\ce{Pb(NO3)2}$ dissociates completely, while $\ce{PbS}$ is a salt with a low solubility, the left semicell will be the cathode and the right one the anode.
So we have the following semireactions:
\begin{align}
\ce{Pb^2+ + 2e- &-> Pb} && \text{(cathode)} \\
\ce{Pb &-> Pb^2+ +2e- } && \text{(anode)}
\end{align}
And for the anode we also have $$\ce{PbS <=> Pb^2+ + S^2- }$$ where $\ce{[Pb^2+]} = \ce{[S^2- ]}= \sqrt{K_\mathrm{sp}}$. So the semicell potentials are: \begin{align} E_\text{cathode} &= E^\circ\\ E_\text{anode} &= E^\circ - \frac{0.059}{2} \log_{10}{[\ce{Pb^2+}]}\\ \end{align}
Thus: $$0.413~\mathrm{V} = E_\text{cathode} - E_\text{anode} = \frac{0.059}{2} \log_{10}\ce{[Pb^2+]} \Rightarrow \ce{[Pb^2+]} = 10^{14} $$ And: $$K_\mathrm{sp} = \ce{[Pb^2+]}^2 = 10^{28}$$
I'm sure that I'm wrong for a sign but I don't understand where is the error.