2l− → l2 + 2e−l− is

Each iodide ion loses an electron as two I⁻ ions form I₂: 2I⁻ → I₂ + 2e⁻. Iodine’s oxidation state increases from −1 in I⁻ to 0 in I₂, so iodide is oxidized.

AI-written and checked against the verified answer.