Question

A black-brown coloured solid (A), when fused with KOH in the presence of air, produces a dark green compound (B), which on electrolytic oxidation in alkaline medium gives a dark purple compound (C). Identify (A), (B) and (C) and write the reactions involved.

Hint:

MnO₂ → (KOH, air) K₂MnO₄ → (electrolytic oxidation) KMnO₄.

Answer 1

Concept: The clues (a black-brown solid that becomes green with KOH and air, then purple on electrolysis) point straight to the laboratory making of potassium permanganate from the ore pyrolusite.

Step 1: Identify the three compounds
The black-brown solid (A) is manganese dioxide, $\mathrm{MnO_2}$ (pyrolusite). The dark green compound (B) is potassium manganate, $\mathrm{K_2MnO_4}$. The dark purple compound (C) is potassium permanganate, $\mathrm{KMnO_4}$.

Step 2: First reaction (A to B)
When $\mathrm{MnO_2}$ is fused with KOH in the presence of air (the air supplies oxygen), it is oxidised to the green manganate:\[ 2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O \]

Step 3: Second reaction (B to C)
The green manganate is then oxidised further by electrolysis in alkaline medium, where each manganate ion loses one electron to become purple permanganate:\[ MnO_4^{2-} \rightarrow MnO_4^- + e^- \]

Answer: (A) is $\mathrm{MnO_2}$, (B) is $\mathrm{K_2MnO_4}$, and (C) is $\mathrm{KMnO_4}$.