# How to calculate the solubility product for PbS from galvanic potential?

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.

The galvanic cell must be written like this: $$\ce{Pb~|~PbS~saturated~||~Pb(NO_3)_2~1M~|~Pb}$$ in order to have a positive $\mathrm{EMF} =0,413~\mathrm{V}$.

Your Approach is right with a slight error:
This is a concentration cell based on $\ce{Pb^{2+}}$. Since $\ce{Pb(NO_3)_2}$ dissociates completely, while $\ce{PbS}$ is a salt with a low solubility, the left semicell will be the anode and the right one the cathode.
So we have the following semireactions: \begin{align} \ce{Pb^{2+} + 2e^- &-> Pb} & (cathode) \\ \ce{Pb &-> Pb^{2+} +2e^- } & (anode) \end{align}

And for the anode we also have $$\ce{PbS <=> Pb^{2+} + S^{2-}}$$ where $\ce{[Pb^{2+}]} = \ce{[S^{2-}]}= \sqrt{K_{sp}}$. So the semicell potentials are: \begin{align} E_\text{cathode} &= E^\circ\\ E_\text{anode} &= E^\circ+ \frac{0,059}{2} \log_{10}{[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: $$K_{sp} = \ce{[Pb^{2+}]}^2 = 10^{-28}$$

You swapped the cathode an anode. In a galvanic cell diagram, on the left salt of the bridge is the anode, and on the right side is the cathode. Also, you should have written E(anode)= -E(O)-0.059/2 ... and EMF=E(cathode)+E(anode), or E(anode)=E(0)+0.059/2 ... then EMF=E(cathode)-E(anode) There are two ways of writing the Nernst equation and you are mixing them, so you get something wrong. One way is by using the reaction coefficient Q and the second by using [Ox]/[Red]. What this means you can see here: en.wikipedia.org/wiki/Nernst_equation. Make sure that you first understand these two ways of writing the Nernst equation, then try to derive the EMF by each method. You need to make these things clear and then return to the problem again. The red-ox potential of the cathode must be higher than the one of the anode.

• It should conventionally be. But in this problem, it is asked also to determine which of them is the cathode/anode (and the order in the diagram has no importance). My reasoning brought me to think that the left one is the cathode, even if I thought too it should be swapped. What to think so? – Brontolo Sep 2 '14 at 13:54
• The order in the diagram has big importance, it is always like that, and by looking at it you can easily deduce which electrode is the cathode and which is the anode. The EMF would be negative in the opposite case which is thermodynamically unfavorable. – Marko Sep 2 '14 at 14:28
• Also, as far as I know, it's written Ksp for solubility product. – Skeptical Chymist Sep 2 '14 at 16:19
• Sorry, just forgot to translate it (in italian is Kps). Anyway I'd like to know a qualitative/quantitative method, if there's, to move in this kind of situations without using this kind of conventional rules. – Brontolo Sep 2 '14 at 17:40
• @Marko Could you expand your answer with the tips you gave in the comments. I guess anyone who is not an everyday-trained electrochemist will have trouble to understand the point you made with this wording. – Martin - マーチン Sep 4 '14 at 7:13