Reduction of the metal centre in aqueous permanganate ion involves

Question

Reduction of the metal centre in aqueous permanganate ion involves (A) 3 electrons in neutral medium (B) 5 electrons in neutral medium (C) 3 electrons in alkaline medium (D) 5 electrons in acidic medium

Tardigrade Admin · Accepted Answer

In acidic medium MnO 4-+8 H ++5 e - longrightarrow Mn 2++4 H 2 O In neutral medium MnO 4-+2 H 2 O +3 e - longrightarrow MnO 2+4 OH - Hence, number of electron loose in acidic and neutral medium 5 and 3 electrons respectively.

Q. Reduction of the metal centre in aqueous permanganate ion involves

3 electrons in neutral medium

5 electrons in neutral medium

3 electrons in alkaline medium

5 electrons in acidic medium

In acidic medium
$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} ⟶ M n^{2 +} + 4 H_{2} O$
In neutral medium
$M n O_{4}^{-} + 2 H_{2} O + 3 e^{-} ⟶ M n O_{2} + 4 O H^{-}$
Hence, number of electron loose in acidic and neutral medium 5 and 3 electrons respectively.

Q. Reduction of the metal centre in aqueous permanganate ion involves

3 electrons in neutral medium

5 electrons in neutral medium

3 electrons in alkaline medium

5 electrons in acidic medium

In acidic medium MnO4−​+8H++5e−⟶Mn2++4H2​O In neutral medium MnO4−​+2H2​O+3e−⟶MnO2​+4OH− Hence, number of electron loose in acidic and neutral medium 5 and 3 electrons respectively.

In acidic medium
$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} ⟶ M n^{2 +} + 4 H_{2} O$
In neutral medium
$M n O_{4}^{-} + 2 H_{2} O + 3 e^{-} ⟶ M n O_{2} + 4 O H^{-}$
Hence, number of electron loose in acidic and neutral medium 5 and 3 electrons respectively.