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?