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When 2 Mno is Fused with Koh in the Presence of 3 Kno as an Oxidizing Agent, It Gives a Dark Green Compound (A). Compound (A) Disproportionates in Acidic Solution to Give Purple Compound - Chemistry

Question

Short Note

Answer the following question.
When MnO2 is fused with KOH in the presence of KNO3 as an oxidizing agent, it gives a dark green compound (A). Compound (A) disproportionates in an acidic solution to give a purple compound (B).  An alkaline solution of compound (B) oxidizes KI to compound (C) whereas an acidified solution of compound (B) oxidizes KI to (D). Identify (A), (B), (C), and (D).

Solution

Potassium permanganate is prepared by fusion of MnO2 with an alkali metal hydroxide and an oxidizing agent like KNO3. This produces the dark green K2MnO4 which disproportionates in a neutral or acidic solution to give permanganate.

\[\ce{2MnO2 + 4KOH + O2 -> \underset{\text{(A) Dark Green}}{2K2MnO4}+ 2H2O}\] 

\[\ce{3MnO^2-_4 + 4H^+ ->\underset{\text{(B) purple}}{2MnO^-_4} + MnO2 + 2H2O}\]

In acid solutions:
Iodine is librated from potassium iodine: 

\[\ce{10I^- +\underset{\text{(B)}}{2MnO^-_4}+ 16H^+ -> 2Mn^2+ + 8H2O +\underset{\text{(D)}}{5l2}}\]

In neutral or faintly alkaline solution:
A notable reaction is the oxidation of iodine to iodate: 

\[\ce{\underset{\text{(B)}}{2MnO^-_4}+ H2O + I^- -> 2MnO2 + 2OH^- + \underset{\text{(C)}}{IO^-_3}}\].

  Is there an error in this question or solution?
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