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JEE Main 201910 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 10 January
Shift: Evening Shift
Subject: Chemistry
Chapter: Redox Reactions

JEE Main 2019 · 10 January · Evening Shift · Chemistry · Redox Reactions

In the reaction of oxalate with permanganate in acidic medium, the...

In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of

{CO}_{2}

is:

Options

A $2$
B $10$
C $1$
D $5$

Correct answer

C. $1$

Step-by-step solution

$2{MnO}_{4}^{-} + 5{C}_{2}{O}_{4}^{2 -} + 16{H}^{+} \to 10{CO}_{2} + 2{Mn}^{2 +} + 8{H}_{2}O$

$C$ goes from $+ 3$ oxidation state to $+ 4$ oxidation state.

Total $10{e}^{-}$ are transferred for $10$ molecules of ${CO}_{2}$ .

Hence, for $1$ molecule of ${CO}_{2}$ , only one ${e}^{-}$ is transferred.
So, number of electrons involved for $1$ molecule of ${CO}_{2}$ is $1$ .

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