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Karnataka Board PUCPUC Science 2nd PUC Class 12

Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with oxalic acid? Write the ionic equation for the reaction.

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Question

Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with oxalic acid? Write the ionic equation for the reaction.

Chemical Equations/Structures
Long Answer
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Solution

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

\[\ce{2MnO2 + 4KOH + O2 -> 2K2MnO4 + 2H2O}\]

\[\ce{3MnO^{2-}_4 + 4H+ -> 2MnO^-4 + MnO2 + 2H2O}\]

In an acidic medium (dilute sulphuric acid), permagnate ion reacts with oxalate ion \[\ce{(C2O^{2-}_4)}\] or oxalic acid at 333 K to evolve carbon dioxide gas.

Reduction half-reaction: \[\ce{[MnO^-_4 + 8H+ + 5e- -> Mn^{2+} + 4H2O] \times 2}\]

Oxidation half-reaction: \[\ce{[C2O^-_4 -> 2CO2 + 2e-] \times 5}\]

Net ionic equation: \[\ce{2MnO^-_4 + 16H+ + 5C2O^{2-}_4 -> 2Mn^{2+} + 8H2O + 10CO2}\]

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Chapter 4: d-and ƒ-Block Elements - 'NCERT TEXT-BOOK, Exercises [Page 506]

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Nootan Chemistry [English] Class 12 ISC
Chapter 4 d-and ƒ-Block Elements
'NCERT TEXT-BOOK, Exercises | Q 8.16 (iii) | Page 506
NCERT Chemistry Part 1 and 2 [English] Class 12
Chapter 4 The d-block and f-block Elements
Exercises | Q 4.16 (iii) | Page 116

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