JEE Main 2019 — Redox Reactions Question with Solution

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

$$5{\text{C}}_2{\text{O}}_4^{2-}+2{\text{MnO}}_4^{-}+16{\text{H}}^{+}\to 10{\text{CO}}_2+2{\text{Mn}}^{2+}+8{\text{H}}_2\text{O}$$

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

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

$${\text{C}}_2{\text{O}}_4^{2-}\to 2{\text{CO}}_2+2e^{-}$$

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 :

$$\frac{2e^{-}}{2{\text{CO}}_2}=1e^{-}/{\text{CO}}_2$$

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