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D and F Block Elements question

2023 · 15 Apr · Shift 1 · Q21

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D and F Block Elements
2023 · 15 Apr · Shift 1 · Q21

D and F Block Elements question

2023 · 15 Apr · Shift 1 · Q21

JEE MainChemistryD and F Block ElementsNumerical+4 / −1

The total change in the oxidation state of manganese involved in the reaction of

\mathrm{KMnO}_{4}

and potassium iodide in the acidic medium is

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 5

Identify the relevant redox reaction

In acidic medium, permanganate ion oxidizes iodide to iodine while manganese gets reduced.

The ionic reaction is:

2\text{MnO}_4^- + 10\text{I}^- + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{I}_2 + 8\text{H}_2\text{O}

Find oxidation state of Mn in reactant and product

In $\mathrm{KMnO_4}$ , manganese is in the oxidation state $+7$ .
In acidic medium, $\mathrm{MnO_4^-}$ is reduced to $\mathrm{Mn^{2+}}$ , so manganese becomes $+2$ .

Calculate the change in oxidation state of manganese

For one manganese atom:

+7 \to +2

So the decrease is:

7 - 2 = 5

Thus, the total change in oxidation state of manganese involved is 5 per Mn atom reduced.

Final answer

\boxed{5}

Comparison with stored correct answer

Stored correct answer = $5$ .

My derived answer also equals $5$ , so they agree.

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