Question

How does ${{H}_{2}}{{O}_{2}}$ reacts with $KMn{{O}_{4}}$ in the alkaline medium?

Answer 1

Hint: Try to analyze what is the nature of the reaction of the hydrogen peroxide reacting in the presence of the oxidizing agent in the acidic or the basic media and what are the properties of it understand with a suitable reaction, and then try to answer the above question.

Complete Solution :
In the presence of the strong oxidizing agent, hydrogen peroxide behaves as a reducing agent in acidic as well as the basic media. This means that the component is reacting with the hydrogen peroxide in the acidic or the basic media will get reduced. In these reactions, molecular oxygen is always produced by the combination of hydrogen peroxide with oxygen atom released by the strong oxidizing agent.
- This means when the hydrogen peroxide is treated with the nascent oxygen produces the molecular oxygen and the water. This molecular oxygen in the acidic media or in the basic media produces the free electrons which further react with the strong oxidizing agent to reduce it.
- Hence when we treated ${{H}_{2}}{{O}_{2}}$ reacts with $KMn{{O}_{4}}$ in the alkaline medium the hydrogen peroxide produces the electrons which reduces the $KMn{{O}_{4}}$ as shown below

$2KMn{{O}_{4}}+3{{H}_{2}}{{O}_{2}}\to 2Mn{{O}_{2}}+3{{O}_{2}}+2KOH+4{{H}_{2}}O$

- Hence the potassium permanganate will reduce into manganese dioxide when it is treated with the hydrogen peroxide in the presence of the basic media or the alkaline media.

Note: Not only in the basic media but also in the acidic media the hydrogen peroxide will act as the reducing agent when it is treated with the strong oxidizing agent. The hydrogen peroxide tries to reduce the compounds that react with it and hence the hydrogen peroxide is known as a reducing agent.