Answer
$[Pb^{2+}]$ reduction at the anode explains the white precipitate formation which is $PbSO_4$$_{(s)}$; this increases the value of $E_{cell}$, so that $E_{cell} > E^{\circ}_{cell}$.
Work Step by Step
$[Pb^{2+}]$ reduction at the anode explains the white precipitate formation which is $PbSO_4$$_{(s)}$; this increases the value of $E_{cell}$, so that $E_{cell} > E^{\circ}_{cell}$.
