A solution of KMnO4 is reduced to various products depending upon its pH. At pH 7 it gives a green solution (C). (A), (B) and (C) are (A) (B) (C) (a) Mn2+ MnO2 MnO42- (b) MnO2 Mn2+ MnO42- (c) Mn2+ MnO42- MnO2 (d) MnO42- Mn2+ MnO2

Question

A solution of KMnO4 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) Mn2+ MnO2 MnO42- (b) MnO2 Mn2+ MnO42- (c) Mn2+ MnO42- MnO2 (d) MnO42- Mn2+ MnO2 (A) a (B) b (C) c (D) d

Tardigrade Admin · Accepted Answer

At p H<7, in acidic medium MnO 4-+8 H ++5 e- arrow underset text (colourless) Mn 2++4 H 2 O At p H>7, in alkaline medium MnO 4-+e- arrow underset( text green ) MnO 42-

	(A)	(B)	(C)
(a)	$M n^{2 +}$	$M n O_{2}$	$M n O_{4}^{2 -}$
(b)	$M n O_{2}$	$M n^{2 +}$	$M n O_{4}^{2 -}$
(c)	$M n^{2 +}$	$M n O_{4}^{2 -}$	$M n O_{2}$
(d)	$M n O_{4}^{2 -}$	$M n^{2 +}$	$M n O_{2}$

a

b

c

d

At pH<7, in acidic medium MnO4−​+8H++5e−→ (colourless) Mn2+​+4H2​O At pH>7, in alkaline medium MnO4−​+e−→( green )MnO42−​​

At $p H < 7$ , in acidic medium
$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to (colourless) M n^{2 +} + 4 H_{2} O$
At $p H > 7$ , in alkaline medium
$M n O_{4}^{-} + e^{-} \to (green) M n O_{4}^{2 -}$