The equivalent weight of `KMnO_4` in a redox reaction in a neutral med

1.	The equivalent weight of `KMnO_4` in a redox reaction in a neutral medium is .A. `M//5`B. `M`C. `M//3`D. `M//4`
Answer» Correct Answer - C 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.

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The equivalent weight of `KMnO_4` in a redox reaction in a neutral medium is .A. `M//5`B. `M`C. `M//3`D. `M//4`

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