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MHTCET 25 Apr 2025 Shift 2 Paper

Question : 53 of 150

Marks: +1, -0

2 \mathrm{molKMnO}_{4}

\mathrm{MnSO}_{4}

Solution:

Concept:

Reduction is the gain of electrons, and the charge needed in Faradays equals the total moles of electrons required.

Explanation:

\mathrm{KMnO}_{4}

, the oxidation state of Mn is

+7

\mathrm{MnSO}_{4}

, the oxidation state of Mn is

+2

So, each Mn ion undergoes:

\mathrm{Mn}^{+7} + 5e^- \longrightarrow \mathrm{Mn}^{+2}

Thus, 1 mole of

\mathrm{KMnO}_{4}

requires 5 moles of electrons, i.e.

5\ \mathrm{F}

of charge.

For 2 moles of

\mathrm{KMnO}_{4}

, the charge required is

2 \times 5\ \mathrm{F} = 10\ \mathrm{F}

Answer:

10\ \mathrm{F}

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