Question

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

Hint:

\(MnO_2\) is black-brown, \(K_2MnO_4\) is green and \(KMnO_4\) is purple. Manganate ion disproportionates in acidic medium.

Answer 1

Concept:
Potassium permanganate is prepared from manganese dioxide. Manganese dioxide is first fused with potassium hydroxide in presence of air to form potassium manganate. Potassium manganate is green in colour. It is then oxidised to potassium permanganate, which is purple in colour.

Step 1: Identify black-brown compound A.
The black-brown solid used for preparation of potassium permanganate is manganese dioxide. Therefore: \[ A=MnO_2 \]

Step 2: Fusion with KOH

in presence of air.
When \(MnO_2\) is fused with \(KOH\) in presence of oxygen, potassium manganate is formed. \[ 2MnO_2+4KOH+O_2 \rightarrow 2K_2MnO_4+2H_2O \] Potassium manganate is dark green. Therefore: \[ B=K_2MnO_4 \]

Step 3: Electrolytic oxidation of potassium manganate.
Potassium manganate is oxidised electrolytically in alkaline medium to potassium permanganate. \[ MnO_4^{2-}\rightarrow MnO_4^-+e^- \] Thus: \[ K_2MnO_4 \rightarrow KMnO_4 \] Potassium permanganate is dark purple. Therefore: \[ C=KMnO_4 \]

Step 4: Behaviour of acidic solution of green compound B.
The green compound is potassium manganate: \[ K_2MnO_4 \] In acidic medium, manganate ion undergoes disproportionation. Some \(MnO_4^{2-}\) is oxidised to \(MnO_4^-\), while some is reduced to \(MnO_2\). The reaction is: \[ 3MnO_4^{2-}+4H^+ \rightarrow 2MnO_4^-+MnO_2+2H_2O \]

Step 5: Name of the reaction.
In this reaction, manganese in \(MnO_4^{2-}\) has oxidation state \(+6\). It changes into: \[ MnO_4^- \quad (Mn=+7) \] and \[ MnO_2 \quad (Mn=+4) \] The same element is both oxidised and reduced. Therefore, this reaction is called disproportionation reaction. Hence: \[ A=MnO_2,\quad B=K_2MnO_4,\quad C=KMnO_4 \] \[ 2MnO_2+4KOH+O_2 \rightarrow 2K_2MnO_4+2H_2O \] \[ 3MnO_4^{2-}+4H^+ \rightarrow 2MnO_4^-+MnO_2+2H_2O \] The reaction is disproportionation.