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D and F Block Elements question

2023 · 30 Jan · Shift 1 · Q20

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D and F Block Elements
2023 · 30 Jan · Shift 1 · Q20

D and F Block Elements question

2023 · 30 Jan · Shift 1 · Q20

JEE MainChemistryD and F Block ElementsNumerical+4 / −1

The number of electrons involved in the reduction of permanganate to manganese dioxide in acidic medium is

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.

Numerical answer

View written solutionFree

Correct answer: 3

We need the reduction of permanganate ion, $\mathrm{MnO_4^-}$ , to manganese dioxide, $\mathrm{MnO_2}$ , in acidic medium.
Find the oxidation state of manganese:

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

Thus manganese is reduced from $+7$ to $+4$ .
Change in oxidation state: $+7 \to +4$ So, electrons gained per Mn atom: $7 - 4 = 3$
The balanced half-reaction in acidic medium is: $\mathrm{MnO_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O}$ This confirms that 3 electrons are involved.
Therefore, the required integer answer is: $\boxed{3}$

Comparison with stored correct answer:

Stored correct answer = $3$
Derived answer = $3$
They agree.

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