Question

A solution of $KMnO_4$ is reduced to various products depending upon its $pH$. At $pH < 7$ it is reduced to a colourless solution $(A)$, at $pH = 7$ it forms a brown precipitate $(B)$ and at $pH > 7$ it gives a green solution $(C)$. $(A)$, $(B)$ and $(C)$ are

	(A)	(B)	(C)
(a)	$Mn^{2+}$	$MnO_2$	$MnO_4^{2-}$
(b)	$MnO_2$	$Mn^{2+}$	$MnO_4^{2-}$
(c)	$Mn^{2+}$	$MnO_4^{2-}$	$MnO_2$
(d)	$MnO_4^{2-}$	$Mn^{2+}$	$MnO_2$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (A)

At $p H<7$, in acidic medium
$MnO _{4}^{-}+8 H ^{+}+5 e^{-} \rightarrow \underset{\text { (colourless) }}{ Mn ^{2+}}+4 H _{2} O$
At $p H>7$, in alkaline medium
$MnO _{4}^{-}+e^{-} \rightarrow \underset{(\text { green })}{ MnO _{4}^{2-}}$