The equivalent weight of `KMnO_4` in a redox reaction in a neutral medium is .

Options

A. `M//5`
B. `M`
C. `M//3`
D. `M//4`

Answer

C

Explanation

In neutral medium, ` KMnO_4` is directly reduced to manganese dioxide :
`Koverset(+7)(Mn)O_(4) rarr overset(+4)(Mn)O_(2)`
Thus, the oxidation number of manganese decreases by three units .
Equivalent weight
` =("Formula mass")/(("Total change in oxidation number of element"),("oxidized or reduced per mole of compound"))`
` M/3`
In acidic medium :
`overset(+7)(Mn)O_(4)^(-) rarr overset(+2)(M)n^(2+)`
The oxidation number of Mn decrease by five units, Thus the equivalent weight of ` KMnO_4` is `M//5`
Under alkaline conditions, `KMnO_4` is first reduced to potassium manganate `(K_2MnO_4)` and then to insoluble manganese oxide `(MnO_2)` :
`overset(+7)(M)nO_(4)^(-) rarr overset(+4)(M)nO_(2)`
Thus, the equivalent weight of ` KMnO_4` in alkaline medimum is ` M//3`.
In concentrated alkalies, permanganate gives manganate :
`Koverset(+7)(Mn)O_(4) rarr K_(2)overset(+6)(Mn)O_(4)`
Thus, the equivalent weight of ` KMnO_4` in strongly alkaline med-ium is M.