5NaClO + I2 + 2NaOH → 5NaCl + 2NaIO3 + H2O
2I (0) – 10 e → 2I (+5) | ÷ 10 — 1
Cl (+1) + 2 e → Cl (–1) | ÷ 10 — 5
5NaClO + I2 + 2NaOH → 5NaCl + 2NaIO3 + H2O
2I (0) – 10 e → 2I (+5) | ÷ 10 — 1
Cl (+1) + 2 e → Cl (–1) | ÷ 10 — 5