# How does precipitate affect cell potential?

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?

• [Pb^2+] is determined by [SO4^2-] and PbSO4 solubility product. Commented Feb 23, 2021 at 12:24

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)
$$E^\circ_\mathrm{cell, i} = \pu{0.34 V} - (\pu{-0.13 V}) = \pu{0.47 V}\tag{2}$$
$$E_\mathrm{cell, f} = \pu{0.34 V} - (\pu{-0.37 V}) = \pu{0.71 V}\tag{3}$$