Reduction of the metal centre in aqueous permanganate ion involves :

3

electrons in neutral medium

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

5

electrons in neutral medium

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3

electrons in alkaline medium

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

5

electrons in acidic medium

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct options are
A $3$ electrons in neutral medium
C $3$ electrons in alkaline medium
D $5$ electrons in acidic medium
In acidic medium:
${M n O}_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n 2^{+} + 4 H_{2} O$

In neutral medium:
${M n O}_{4}^{-} + 2 H_{2} O + 3 e^{-} \to M n O_{2} + 4 O H^{-}$

In basic medium:
${M n O}_{4}^{-} + H_{2} O_{\to} 2 M n O_{2} + O_{2}$

${M n O}_{4}^{-}$ to $M n O_{2}$ means Mn from +7 to +4, i.e., 3 electrons are gained.

Hence options A, C and D are correct.

Suggest Corrections

Similar questions

M n O_{4}^{-}

M n O_{3}^{-}

How many mole of electrons are involved in the reduction of one mole of $M n O_{4}^{-}$ ion in alkaline medium to $M n O_{4}^{- 2}$ ?

C O_{2}

C O_{2}

K M n O_{4}

K = 39.09

M n = 54.94

O = 16.0

2 K M n O_{4} + 2 K O H \to 2 K_{2} {M n O}_{4} + H_{2} O + [O]

K M n O_{4}

K M n O_{4}

K M n O_{4}

Join BYJU'S Learning Program