Question

Discuss in detail the oxidising character of potassium permanganate in alkaline solutions. Give at least two reactions in the given case.

Answer 1

Purple potassium permanganate turns into green potassium manganate when an alkali is present.

\[\ce{\underset{(purple)}{\underset{Pot. permanganate}{2KMnO4}} + 2KOH -> \underset{(green)}{\underset{Pot. permanganate}{2K2MnO4}} + H2O + O}\]

In the presence of a reducing agent, the resulting potassium manganate is further reduced to MnO₂.

\[\ce{K2MnO4 + H2O -> MnO2 + 2KOH + O}\]

Therefore, the complete reaction is

\[\ce{2KMnO4 + H2O -> 2MnO2 + 2KOH + 3O}\]

Thus, the reduction of Mn04 ions into Mn02 occurs during the reaction in an alkaline solution. This is how it can be expressed in the ionic form.

\[\ce{MnO^-_4 + e- -> MnO^{2-}_4}\]

\[\ce{MnO^{2-}_4 + 2H2O + 2e- -> MnO2 + 4OH-}\]

\[\ce{\overline{\text{MnO}^-_4 + 2\text{H}_2\text{O} + 3\text{e}^- \leftarrow \text{MnO}_2 + 4\text{OH}^-}}\]

The reaction is identical to that which takes place in a neutral solution. In this case as well, it takes three moles of electrons to convert one mole of MnO₄ ions to MnO₂. Therefore, KMnO₄’s equivalent mass in an alkaline medium is equal to one-third of its molecular mass.