Question

Equivalent weight of $KMn{O}_4$ and its reduced species in different mediums are given

Medium	Equivalent weight	Reduced speciest
Acidic	$\frac{M}{5}$	I
Neutral	$\frac{M}{1}$	II
Concentrated	$\frac{M}{3}$	III
Alkaline

• A. (I)-$MN{O}_2$,(II)-$Mn{O}_4^{2-}$ (III)-$M{n}^{2+}$
• B. (I)-$MN{O}_2$,(II)-$M{n}^{2+}$ (III)-$Mn{O}_4^{2-}$
• C. (I)-$M{N}^{2+}$,(II)-$Mn{O}_2$ (III)-$Mn{O}_4^{2-}$
• D. (I)-$M{N}^{2+}$,(II)-$Mn{O}_4^{2-}$ (III)-$Mn{O}_2$

Answer 1

Answer: (D)

Equivalent weight=$\frac{molarmass}{changeinoxidationnumber}$

	Change in oxidation number	Equivalent weight
$M\underset{7}{n}{O}_4^{-}\to M\underset{2}{n^{2+}}$	5	$\frac{M}{5}$
$M\underset{7}{n}{O}_4^{-}\to M\underset{6}{n}{O}_4^{2-}$	1	$\frac{M}{1}$
$M\underset{7}{n}{O}_4^{-}\to M\underset{4}{n}{O}_2$	3	$\frac{M}{3}$