JEE Main 2024 — Electrochemistry Question with Solution

Statement I is indeed true. When $${\mathrm{MnO}}_2$$ (manganese dioxide) is fused with $$\mathrm{KOH}$$ (potassium hydroxide) and an oxidising agent such as $${\mathrm{KNO}}_3$$ (potassium nitrate), it forms dark green potassium manganate ($${\mathrm{K}}_2{\mathrm{MnO}}_4$$). The reaction can be represented as follows:

$${2MnO}_2+4KOH+{\mathrm{O}}_2\to {2K}_2{\mathrm{MnO}}_4+2H_2\mathrm{O}$$

This process involves the oxidation of manganese dioxide to manganate ion ($${\mathrm{MnO}}_4^{2-}$$) in an alkaline medium.

Statement II is also true. The manganate ion ($${\mathrm{MnO}}_4^{2-}$$), when subjected to electrolytic oxidation in an alkaline medium, can indeed be converted into permanganate ion ($${\mathrm{MnO}}_4^{-}$$), which is characterized by a deep purple color. This conversion is an example of an electron-loss (oxidation) process at the anode of an electrolytic cell. The reaction can be represented as follows:

$${\mathrm{MnO}}_4^{2-}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{MnO}}_4^{-}+2{\mathrm{OH}}^{-}+2e^{-}$$

Thus, both statements I and II are accurate, making the correct answer:

Option D: Both Statement I and Statement II are true.