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Redox Reactions question

2023 · 10 Apr · Shift 2 · Q15

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Redox Reactions
2023 · 10 Apr · Shift 2 · Q15

Redox Reactions question

2023 · 10 Apr · Shift 2 · Q15

JEE MainChemistryRedox ReactionsNumerical+4 / −1

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

\underline{\hspace{2cm}}

electrons.

Numerical answer

View written solutionFree

Correct answer: 3

We need the reduction of permanganate ion in alkaline medium.
The permanganate ion is: $\mathrm{MnO_4^-}$ In alkaline (or neutral) medium, it is reduced to manganese dioxide: $\mathrm{MnO_4^- \rightarrow MnO_2}$
Find the oxidation state of manganese on both sides.

In $\mathrm{MnO_4^-}$ : $x + 4(-2) = -1$ $x - 8 = -1$ $x = +7$
In $\mathrm{MnO_2}$ : $x + 2(-2) = 0$ $x - 4 = 0$ $x = +4$

Change in oxidation state of Mn: $+7 \to +4$ So manganese gains: $3\text{ electrons}$
The balanced half-reaction in alkaline medium is: $\mathrm{MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-}$ This confirms that $3$ electrons are gained.

Therefore, the number of electrons gained is: $\boxed{3}$

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