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).

 

 

 

 
 
 
 
 

Answers (1)
S Sumit

2MnO_2 + 4KOH + O_2 \rightarrow\underset{\textup{Potassium manganate}}{\underset{(green)}{\underset{A}{2K_2MnO_4}}} + 2H_2O

\underset{A}{2MnO_4^{2-}} + 4H^+ \rightarrow {\underset{(purple)}{\underset{B}{2MnO_4^-}}} + 2MnO_2 +2H_2O

\underset{B}{2MnO_4^{-}} + H_2O +KI \rightarrow 2MnO_2 + 2KOH + \underset{\textup{Potassium iodate}}{\underset {C}{KIO_3}}

\underset{B}{2MnO_4^{-}} + 16H^+ +10I^- \rightarrow 2Mn^{2+} + 8H_2O + \underset{D}{5I_2}

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