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QJEEM19Che927: JEE Main 2019 Chemistry - Class 11 JEEM19Che927 Solution: In the reaction of oxalate with permanganate in acidic medium,…

In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of CO2 is :

Multiple Choice Options

  • A 10
  • B 2
  • C 1
  • D 5

Step-by-Step Verified Solution

The balanced reaction of oxalate with permanganate in acidic medium is :

5 C2O42−+2 MnO4−+16 H+→10 CO2+2 Mn2++8 H2O5 \, \text{C}_2\text{O}_4^{2-} + 2 \, \text{MnO}_4^- + 16 \, \text{H}^+ \rightarrow 10 \, \text{CO}_2 + 2 \, \text{Mn}^{2+} + 8 \, \text{H}_2\text{O}

In this redox reaction, the oxalate ion C2O42−\text{C}_2\text{O}_4^{2-} is oxidized to carbon dioxide CO2\text{CO}_2 , and the permanganate ion MnO4−\text{MnO}_4^- is reduced to manganese(II) ion Mn2+\text{Mn}^{2+} .

The oxidation half-reaction, representing the change for the oxalate ion, is :

C2O42−→2 CO2+2 e−\text{C}_2\text{O}_4^{2-} \rightarrow 2 \, \text{CO}_2 + 2 \, e^-

From this half-reaction, it's clear that each oxalate ion produces two CO2 molecules and releases two electrons in the process.

So, the number of electrons involved in producing one molecule of CO2 is :

2 e−2 CO2=1 e−/CO2\frac{2 \, e^-}{2 \, \text{CO}_2} = 1 \, e^-/\text{CO}_2

Therefore, the correct answer is Option C: One electron is involved in the production of one molecule of CO2.

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