JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Redox Reactions and Electrochemistry

In the chemical reaction between stoichiometric quantities of

KMnO_{4}

and KI in weakly basic solution, what is the number of moles of

I_{2}

released for 4 moles of

KMnO_{4}

consumed?

Answer: 6

Solution

In weakly basic medium,

KMnO_{4}

is reduced to

MnO_{2}

KMnO_{4} + KI ⟶ MnO_{2} + I_{2}

n-factor calculation:

$KMnO_{4}$ : Mn goes from +7 to +4, so n-factor = 3
$I_{2}$ : Each I goes from -1 to 0, so for $I_{2}$ , n-factor = 2

Using equivalents:

Equivalents of KMnO_{4} = Equivalents of I_{2}

4 \times 3 = n \times 2

n = \frac{12}{2} = 6

Therefore, 6 moles of

I_{2}

are released.

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Integer answer

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