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Describe with an example the oxidising actions of permanganate ion in alkaline media. What acids are actually used for? - Chemistry (Theory)

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Describe with an example the oxidising actions of permanganate ion in alkaline media. What alkali are actually used for?

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उत्तर

In the presence of an alkali, purple potassium permanganate transforms to green potassium manganate.

\[\ce{\underset{(purple)}{\underset{Pot. permanganate}{2KMnO4}} + 2KOH -> \underset{(green)}{\underset{Pot. manganate}{2K2MnO4}} + H2O + O}\]

Potassium manganate is reduced to MnO2 in the presence of a reducing agent.

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

Therefore, the complete reaction is:

\[\ce{2KMnO4 + H2O -> 2MnO2 + 2KOH + 3O}\]

In alkaline solutions, the process includes reducing MnO4 ions to MnO2. In ionic form, it is written as follows.

\[\ce{MnO^-_4 + e- -> MnO^{2-}_4}\]

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

\[\ce{\overline{\text{MnO}^-_4 + 2\text{H}_2\text{O} + 3\text{e}^- \leftarrow \text{MnO}_2 + 4\text{OH}^-}}\];

E° = 1.23 V

The reaction is the same as in a neutral solution. Reducing one mole of MnO4 ions to MnO2 requires three moles of electrons. In alkaline conditions, the corresponding mass of KMnO4 is one-third of its molecular mass.

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अध्याय 8: d-and ƒ-Block Elements - REVIEW EXERCISES [पृष्ठ ४८४]

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नूतन Chemistry Part 1 and 2 [English] Class 12 ISC
अध्याय 8 d-and ƒ-Block Elements
REVIEW EXERCISES | Q 8.42 ii. | पृष्ठ ४८४
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