JEE MAIN - Chemistry (2024 - 8th April Evening Shift - No. 14)
Given below are two statements :
Statement (I) : Fusion of $$\mathrm{MnO}_2$$ with $$\mathrm{KOH}$$ and an oxidising agent gives dark green $$\mathrm{K}_2 \mathrm{MnO}_4$$.
Statement (II) : Manganate ion on electrolytic oxidation in alkaline medium gives permanganate ion.
In the light of the above statements, choose the correct answer from the options given below :
Explanation
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:
$$\mathrm{2MnO}_2 + 4KOH + \mathrm{O}_2 \rightarrow \mathrm{2K}_2\mathrm{MnO}_4 + 2H_2\mathrm{O}$$
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:
$$\mathrm{MnO}_4^{2-} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{MnO}_4^- + 2\mathrm{OH}^- + 2e^-$$
Thus, both statements I and II are accurate, making the correct answer:
Option D: Both Statement I and Statement II are true.
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