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The answer is -0.09V. And the explanation is that the concentration of $\ce{Pb^2+}$ will decrease.

My question is how will [$\ce{Pb^2+}$] decrease? In my opinion, [$\ce{Pb^2+}$] should remain constant. The cell is a voltaic cell meaning that $\ce{Pb}$ should oxidize to $\ce{Pb^2+}$ to keep the cell running (since $\ce{\frac{Pb^2+}{Pb}}$ = -0.13V). As a result, the concentration of $\ce{Pb^2+}$ will increase. But since the solution is saturated, no more lead can dissolve keeping the value of [$\ce{Pb^2+}$] constant. Can somebody please point out the error in my explanation?

  • $\begingroup$ B and D are presumably Pt electrodes, and B is part of an SHE. What is the composition of C? $\endgroup$ – Zhe Mar 16 '17 at 19:35
  • $\begingroup$ This was to be answered in the first part. The answer is any solution containing H+ ions. $\endgroup$ – mathnoob123 Mar 16 '17 at 19:49

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:

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

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 $\Delta Q<1$, $\Delta \left(\frac{RT}{nF}\ln Q\right) > 0$.

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  • $\begingroup$ Can you please explain a little more on how the Pb2+ concentration is decreased? $\endgroup$ – mathnoob123 Mar 17 '17 at 8:18
  • $\begingroup$ And one more thing if Pb2+ is decreasing and it goes into denominator of Q then, decrease in Pb2+ will lead to increase in Q $\endgroup$ – mathnoob123 Mar 17 '17 at 8:25
  • $\begingroup$ The lead concentration is explicitly provided via the maximum solubility of lead chloride. for the second comment, $\frac{1}{x}$ gets bigger if $x$ is smaller. $\endgroup$ – Zhe Mar 17 '17 at 12:29

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