Question

How many electrons are involved in the oxidation of $KMn{O_4}$ in the basic medium?

Answer 1

Hint: This problem requires us to understand the different reactions that $KMn{O_4}$ undergoes under different conditions. These conditions include acidic, basic and neutral mediums. $KMn{O_4}$ reacts with different chemical compounds and shows different oxidizing effects, based on the different mediums.

Complete answer:
Commercial potassium permanganate is made by combining a solution of potassium hydroxide and powdered manganese oxide with oxidising agents such as potassium chlorate. The mixture is cooked until it has evaporated, then cooked in iron pans until it acquires a pasty consistency.
The solution created when potassium permanganate ($KMn{O_4}$) crystals are dissolved in water is purple. It's a potent oxidising agent that doesn't produce any hazardous byproducts. Other minerals, such as manganese oxide, are frequently used to make it.
Because potassium permanganate is a powerful oxidising agent, it can be utilised as an oxidant in a wide range of chemical reactions. When executing a redox reaction with potassium permanganate, the dark purple solution turns colourless and eventually into a brown solution, demonstrating its oxidising power.
In basic medium (strongly alkaline medium), the exchange of ions occurs as follows:
So, we can observe that the oxidation of $Mn$ changes from +7 to +6. So, in the basic medium, $KMn{O_4}$ acts as a weak oxidizing agent. The transfer of only one electron takes place.
Potassium permanganate is converted to manganate and oxygen gas is released when heated with alkalies.

Note:
Once all of the permanganate ions have been used up in the process, the solution loses its pink colour. Potassium permanganate is a self-indicator since it serves as an indication in addition to being one of the reactants.