Consider the following reaction,

MnO₂ --(Fused with KOH, oxidized with air)--> A --(Electrolytic oxidation in alkaline solution)--> B

The colour of B is

Solution:

The reaction sequence is as follows:

1. When manganese dioxide (MnO₂) is fused with KOH and oxidized with air, it forms potassium manganate (K₂MnO₄).

   2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O

   Potassium manganate (K₂MnO₄) is green in colour. So, A is K₂MnO₄.

2. Potassium manganate (K₂MnO₄) undergoes electrolytic oxidation in an alkaline solution to form potassium permanganate (KMnO₄).

   2K₂MnO₄ + 2H₂O → 2KMnO₄ + 2KOH + H₂ (at cathode) or more simply, at the anode: 2MnO₄²⁻(aq) → 2MnO₄⁻(aq) + 2e⁻

   Potassium permanganate (KMnO₄) is purple in colour. So, B is KMnO₄.

Therefore, the colour of B is purple.