Answer:
Oxidation: Cr → Cr⁴⁺ + 4 e⁻
Reduction: 4 e⁻ + O₂ → 2 O²⁻
Step-by-step explanation:
Let's consider the following redox reaction.
Cr + O₂ → CrO₂
Cr is oxidized. Its oxidation number increases from 0 to +4. The corresponding half-reaction is:
Cr → Cr⁴⁺ + 4 e⁻
O is reduced. Its oxidation number decreases from 0 to -2. The corresponding half-reaction is:
4 e⁻ + O₂ → 2 O²⁻