Unbalanced chemical reaction: PbO₂ + I₂ → Pb²⁺ +IO₃⁻<span>. First </span><span>determine oxidation numbers of elements: 1) lead on the left side of reaction has oxidation number +4 (x + 2</span>·(-2) = 0) and on the right side +2, so lead is reduced. 2) iodine has neutal charge (0) on left and oxidation number +5 on the right side of chemical reaction (x + 3 · (-2) = -1), so iodine is oxidized. 3) oxygen has oxidation number -2 on both side of chemical reaction. Second, write balanced half reactions, because this reaction is acidic solution, add hydrogen ions on one side and water on another side of half reaction; balance atoms and electrons on both half reactions: Half reaction 1: PbO₂ + 4H⁺ + 2e⁻ → Pb²⁺ + 2H₂O /×5 5PbO₂ + 20H⁺ + 10e⁻ → 5Pb²⁺ + 10H₂O. Half reaction 2: I₂ + 6H₂O → 2IO₃⁻ + 12H⁺ + 10e⁻. Sum both half reactions and short same ions: 5PbO₂ + 20H⁺ + I₂ + 6H₂O → 5Pb²⁺ + 10H₂O + 2IO₃⁻ + 12H⁺. Abbreviate reaction: 5PbO₂ + 8H⁺ + I₂ → 5Pb²⁺ + 4H₂O + 2IO₃⁻..
