What is the change in oxidation number of Pb at positive electrode of lead accumulator acting as galvanic cell ?

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In a lead accumulator acting as a galvanic cell (discharging), the positive electrode acts as the cathode where reduction takes place.

The reaction at the positive electrode is:
$P b O_{2} (s) + S O_{4}^{2 -} (a q) + 4 H^{+} (a q) + 2 e^{-} \to P b S O_{4} (s) + 2 H_{2} O (l)$

The oxidation state of Pb in $P b O_{2}$ is $+ 4$ .
The oxidation state of Pb in $P b S O_{4}$ is $+ 2$ .

The oxidation number of Pb changes from $+ 4$ to $+ 2$ , which means it decreases by $2$ .

Answer: decreases by 2

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