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

Question

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.

Answer 1

The galvanic cell must be written like this:

$$\ce{Pb | PbS saturated || Pb(NO3)2 \pu{1 M} | Pb}$$

in order to have a positive $\mathrm{EMF} = \pu{0.413 V}$.

Your approach is right with a slight error. 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 anode and the right one the cathode.
So, we have the following semireactions:

$$ \begin{align} \ce{Pb^2+ + 2 e^- &-> Pb} &\quad &\text{(cathode)} \\ \ce{Pb &-> Pb^2+ + 2 e-} &\quad &\text{(anode)} \end{align} $$

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:

$$\pu{0.413 V} = E_\text{cathode} - E_\text{anode} = -\frac{0.059}{2}\log_{10}[\ce{Pb^2+}] \implies [\ce{Pb^2+}] = 10^{-14}$$

And

$$K_\mathrm{sp} = [\ce{Pb^2+}]^2 = 10^{-28}$$

Answer 2

Actually, I think the question has an error in the EMF of the cell given. It should be $\pu{-0.413 V}$ rather than $\pu{0.413 V}.$ So, actually you are right in saying that the left half cell would be cathode if the EMF of the cell has to be positive.

The mistake that you are doing is when you wrote the reduction potential of anode, you should have written

$$E = E^\circ - \frac{0.059}{2}\log\frac{1}{[\ce{Pb^2+}]},$$

as the reduction reaction would be

$$\ce{Pb^2+ + 2 e- -> Pb},$$

and the reduction potential is given by

$$E = E^\circ - \frac{0.059}{2}\log\frac{[\text{Products}]}{[\text{Reactants}]}.$$

$\ce{Pb^2+}$ is on the reactant side, so it should come in the denominator.

Answer 3

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

$$E(\text{anode})= -E^\circ - 0.059/2 \ldots$$

and

$$\text{EMF} = E(\text{cathode}) + E(\text{anode}),$$

or

$$E(\text{anode}) = E^\circ + 0.059/2\ldots$$

Then

$$\text{EMF} = E(\text{cathode}) - E(\text{anode}).$$

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 $[\text{Ox}]/[\text{Red}].$

Make sure that you first understand these two ways of writing the Nernst equation, then try to derive the $\text{EMF}$ 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.