The number of moles of KMnO4 reduced by one mole of KI in alkaline medium is
- one
- two
- five
- one fifth
Answer
(2) two
Solution
As per NCERT, in neutral or faintly alkaline medium permanganate oxidises iodide to iodate:
2MnO4- + I- + H2O 2MnO2 + IO3- + 2OH-
I- (-1) IO3- (+5): 6 electrons lost per I-
MnO4- (+7) MnO2 (+4): 3 electrons gained per MnO4-
Moles of KMnO4 reduced by 1 mol KI = 63 = 2
Answer: (2)