Question

Equivalent weight of $KMnO_{4}$ 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)-$MNO_{2}$,(II)-$MnO^{2-}_{4}$ (III)-$Mn^{2+}$
• B. (I)-$MNO_{2}$,(II)-$Mn^{2+}$ (III)-$MnO_{4}^{2-}$
• C. (I)-$MN^{2+}$,(II)-$MnO_{2}$ (III)-$MnO_{4}^{2-}$
• D. (I)-$MN^{2+}$,(II)-$MnO^{2-}_{4}$ (III)-$MnO_{2}$

Answer 1

Answer: (D)

Equivalent weight=$\frac{molar mass}{change \,in\, oxidation\, number}$

	Change in oxidation number	Equivalent weight
$M\underset{7}{n}O^{-}_{4} \to M\underset{2}{n^{2+}}$	5	$\frac{M}{5}$
$M\underset{7}{n}O^{-}_{4} \to M\underset{6}{n}O^{2-}_{4}$	1	$\frac{M}{1}$
$M\underset{7}{n}O^{-}_{4} \to M\underset{4}{n}O_2$	3	$\frac{M}{3}$