This can be represented by the following half-reaction:
\[ Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^- \]
At the cathode, lead dioxide (PbO2) reacts with the sulfate ion (SO4^2-) and 4 hydrogen ions (H^+) from the sulfuric acid, and 2 electrons to form lead
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