Answer:
NO₃⁻ is reduced; I₂ is oxidized
Explanation:
I think your equation is supposed to be
NO₃⁻ + I₂ ⟶ IO₃⁻ + NO₂
If you break the equation into two half-reactions, you get
NO₃⁻ + 2H⁺ + e⁻ ⟶ NO₂ + H₂O
I₂ + 3H₂O ⟶ IO₃⁻ + 6H⁺ + 5e⁻
We see that NO₃⁻ gains an electron, so it is reduced.
I₂ loses five electrons, so it is oxidized.
