How does precipitate affect cell potential?

Question

A cell is set up with copper and lead electrodes in contact with $\ce{CuSO4(aq)}$ and $\ce{Pb(NO3)2(aq)},$ respectively, at $\pu{25 °C}.$ The standard reduction potentials are:

$$ \begin{align} \ce{Pb^2+ + 2 e- &-> Pb} &\quad E^\circ &= \pu{-0.13 V} \\ \ce{Cu^2+ + 2 e- &-> Cu} &\quad E^\circ &= \pu{+0.34 V} \end{align} $$

If sulfuric acid is added to the $\ce{Pb(NO3)2}$ solution, forming a precipitate of $\ce{PbSO4},$ what will happen to the cell potential?

I know the lead is the anode while the copper is the cathode:

$$\ce{Pb(s) + Cu^2+ -> Pb^2+ + Cu(s)}$$

The addition of $\ce{H2SO4}$ will cause less $\ce{Pb^2+}$ to be in the solution, thus causing the equilibrium to push to the right creating more $\ce{Pb^2+}.$ With more $\ce{Pb^2+},$ if you were to put it in the Nernst equation, would the greater number in the log expression cause the cell potential to decrease?

Answer 1

The solubility product of $\ce{PbSO4}$ is about $10^{-8}$. Now suppose $[\ce{Pb^{2+}}$] falls down from $\pu{1 M}$ to an arbitrary low value like $\pu{10^{-8} M},$ due to addition of $\ce{SO4^{2-}}$ ions. In this case, Nernst's law can be applied, and the potential of the lead electrode falls from $E^\circ_\ce{Pb} = \pu{-0.13 V}$ down to

$$E_\ce{Pb} = \pu{-0.13 V} + \pu{0.0296 V}\times\log{10^{-8}} = \pu{-0.37 V}\tag{1}$$

As a consequence, the $\ce{Pb}$ potential decreases. But the cell potential increases and goes from the initial value (without sulfate)

$$E^\circ_\mathrm{cell, i} = \pu{0.34 V} - (\pu{-0.13 V}) = \pu{0.47 V}\tag{2}$$

to the final value (after precipitation of lead sulfate)

$$E_\mathrm{cell, f} = \pu{0.34 V} - (\pu{-0.37 V}) = \pu{0.71 V}\tag{3}$$