The disproportionation of chlorine in cold dilute NaOH (3 Cl₂ + 6 OH⁻ → 5 Cl⁻ + ClO₃⁻ + 3 H₂O is the hot reaction). At cold and dilute conditions, Cl₂ + 2 OH⁻ →:

1. In cold dilute NaOH, Cl₂ disproportionates to Cl⁻ (reduced) and ClO⁻ (oxidised). 2. Cl₂ + 2 OH⁻ → Cl⁻ + ClO⁻ + H₂O. 3. Cl in Cl₂ is 0; in Cl⁻ it is −1 (reduction); in ClO⁻ it is +1 (oxidation). 4. The hot and concentrated version produces Cl⁻ and ClO₃⁻ (a +5 state). _Source: NCERT Class 11 Chem Ch 8 §8.2 Disproportionation_