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NVAEasyJEE 2023Oxidation Number & Redox Reactions

JEE Chemistry 2023 Question with Solution

In alkaline medium, the reduction of permanganate anion involves a gain of _____ electrons.

Answer

Correct answer:3

Step-by-step solution

Standard Method

Given: The reduction of permanganate anion in alkaline medium is considered.

Find: The number of electrons gained.

Permanganate ion is MnO4−\text{MnO}_4^-, where the oxidation state of manganese is +7+7.

In alkaline medium, it is reduced to manganese with oxidation state +4+4.

MnO4−→Mn4+\text{MnO}_4^- \rightarrow \text{Mn}^{4+}

The change in oxidation number is:

+7→+4+7 \rightarrow +4

So, manganese gains:

33

electrons.

Therefore, the reduction of permanganate anion involves the gain of 33 electrons.

Oxidation Number Change

Given: Manganese in MnO4−\text{MnO}_4^- is in the +7+7 oxidation state.

Find: How many electrons are required for its reduction in alkaline medium.

Reduction means a decrease in oxidation number. Here manganese changes from +7+7 to +4+4.

Δoxidation number=7−4=3\Delta \text{oxidation number} = 7 - 4 = 3

A decrease by 33 units means gain of 33 electrons.

Thus, the required numerical value is 33.

Common mistakes

  • Assuming permanganate always gains 55 electrons is incorrect because the number of electrons depends on the medium. In alkaline medium, the reduction is different from acidic medium. Always check the reaction medium before counting electrons.

  • Using the oxidation state of manganese incorrectly can lead to a wrong answer. In MnO4−\text{MnO}_4^-, manganese is +7+7, not +4+4 initially. First determine the oxidation state correctly, then find the change.

  • Confusing decrease in oxidation number with loss of electrons is wrong. A decrease from +7+7 to +4+4 indicates reduction, which means gain of electrons. Use oxidation number decrease to count electrons gained.

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