How many electrons are involved in the oxidation of KMnO4 in basic medium?

One electron is involved in the oxidation of KMnO4 in basic medium.

Explanation

KMnO4 oxidised in acidic, basic as well as in neutral medium. Let us have  a look at it

  • Acidic medium

The oxidation state of the Mn atom in KMnO4  is +7.

If you put it in an acidic medium, you get this:

MnO4¯ + 8H+ + 5e- → Mn2+ + 4H2O

MnO4¯→ Mn2+

Change in oxidation Number = 7-2= +5

As you can see, Mn gives up 5 electrons. (Making it an oxidizing agent.)

  • Basic medium

In a strongly alkaline solution, you get:

MnO4¯ + e- → MnO42-

Change in oxidation Number = 7-6= +1

So, it only gives up one of its electrons. Making it a much weaker oxidizing agent.

  • Neutral medium

MnO4¯ + e- → MnO42-

The oxidation state reduces from +7 to +4

Change in oxidation Number = 7-4= +3

So, it only gives up three of its electrons in neutral medium.

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