Answer:
Oxidation: Pb → Pb⁴⁺ + 4 e⁻
Reduction: 2 Cl₂ + 4 e⁻ → 4 Cl⁻
Explanation:
Let's consider the following redox reaction.
Pb + 2 Cl₂ ⇄ PbCl₄
We can identify both half-reactions.
Oxidation: Pb → Pb⁴⁺ + 4 e⁻
Reduction: 2 Cl₂ + 4 e⁻ → 4 Cl⁻
Pb is oxidized and its oxidation number increases from 0 to 4+ while Cl₂ is reduced and its oxidation number decreases from 0 to 1-.