In alkaline medium, MnO4- oxidises I- to
- I2
- IO3-
- IO-
- IO4-
Answer
(2) IO3-
Solution
Neutral / faintly alkaline medium: MnO4- (+7) → MnO2 (+4)
2MnO4- + H2O + I- 2MnO2 + 2OH- + IO3-
∵ I is oxidised from −1 to +5 (iodate), 6 e- lost = 2 × 3 e- gained by Mn
Answer: (2)