Question

Can someone help me solve this? I was given the answer at the end of class but I don't understand how the teacher got it. (Directions: Balance each redox reaction in acid solution using the half reaction method.)
PbO2+I2=>Pb^2+ +IO3^- = I2+8H^+ +5PbO2=>2IO3^- +5Pb^+2 +4H2O

Answer 1

Unbalanced chemical reaction: PbO₂ + I₂ → Pb²⁺ +IO₃⁻.
First determine oxidation numbers of elements:
1) lead on the left side of reaction has oxidation number +4 (x + 2·(-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⁻..