The number of moles of `KMnO_(4)` reduced by `1 "mol of" KI` in alkaline medium is

Options

A. one
B. two
C. five
D. one fifth

Answer

B

Explanation

The product of oxidation of `I^(-)` with `MnO_(4)^(-)` in alkaline medium is `IO_(3)^(-)`
`2MnO_(4)^(-) + H_(2)O + I^(-) rarr 2MnO_(2) + 2OH^(-) + IO_(3)^(-)`
Change in oxidation numbers:
`I^(-) rarr IO_(3)^(-)` (increase by 6 units)
Thus, one mole of KI contains 6 equivalents of KI.
`MnO_(4)^(-) rarr MnO_(2)` (decrease by 3 units)
Thus, one mole of `KMnO_(4)` contains 3 equivalents. This implies that 6 equivalents of KI (one mole) will reduce 6 equivalents of `MnO_(4)` (2 moles) in alkaline medium.