Electrolysis of PbCl2 and use of Nernst Equation
I've finally read this question. I don't know what the source is, but I disagree with both the answer and the explanation.

Consider first the reduction half reaction for lead:

$$\ce{Pb^{2+} + 2e- -&gt; Pb}, E^{\varnothing}=-0.13\ \mathrm{V}$$

Now, this is for the reduction at standard conditions, which means that the concentration of lead ion is $1\ \mathrm{M}$. If you decrease the concentration of the reactant ions (and you are, if you consider that the concentration of the dissolved lead chloride is less than 1), then the reaction should be less favorable. This means that the potential should be more negative, not less negative.

This is easy to check based on the form of the Nernst equation as the lead ion concentration enters into the denominator of $Q$. Since $\Delta Q&lt;1$, $\Delta \left(\frac{RT}{nF}\ln Q\right) &gt; 0$.