Question

Reduction of the metal centre in aqueous permanganate ion involves: (This question has multiple correct answers)
(A) 3 electrons in neutral medium
(C) 5 electrons in neutral medium
(D) 1 electron in alkaline medium
(D) 5 electrons in acidic medium

Answer 1

Hint: In acidic medium, the permanganate ion changes to $M{n}^{2+}$, in the neutral medium the permanganate ion changes to $Mn{O}_2$, and in the basic or alkaline medium the permanganate ion changes to $Mn{O}_4^{2-}$ ion.

Complete step by step solution:
The formula of permanganate ion is $Mn{O}_4^{-}$ and it is usually present as potassium permanganate that has a formula $KMn{O}_4$. The potassium permanganate is a deep purple solution that is used for chemical titration. It is used in titration because it can be used for acidic medium, alkaline medium, and neutral medium.
In an acidic medium, when the permanganate ion reacts with hydrogen ion to produce manganese ion which has a $+2$ oxidation state. Water is also produced in this reaction. The reaction is:
$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+2H_{2}O$
In this reaction, 5 electrons are used.
In a neutral medium, the permanganate ion reacts with water to produce manganese dioxide and hydroxyl ions. The reaction is given below:
$Mn{O}_4^{-}+2H_{2}O+3e^{-}\to Mn{O}_2+4O{H}^{-}$
In this reaction, 3 electrons are used.
In a basic or alkaline medium, the permanganate ion reacts with an electron to form a manganate ion. The reaction is given below:
$Mn{O}_4^{-}+e^{-}\to Mn{O}_4^{2-}$
In this reaction, one electron is used.
So, in acidic medium 5 electrons, in neutral medium 3 electrons, an alkaline medium 1 electron is used.

Therefore, the correct answers are options (A), (C), and (D).

Note: In a neutral solution permanganate ion can be used to oxidize manganous sulfate, sodium thiosulfate, hydrogen sulfide. In a basic medium, it can be used to oxidize potassium iodide, olefinic compounds, etc. in an acidic solution, it is used to oxidize hydrogen disulfide, sulfur dioxide, etc.