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 :

<p>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:</p> <p>$$\mathrm{2MnO}_2 + 4KOH + \mathrm{O}_2 \rightarrow \mathrm{2K}_2\mathrm{MnO}_4 + 2H_2\mathrm{O}$$</p> <p>This process involves the oxidation of manganese dioxide to manganate ion ($\mathrm{MnO}_4^{2-}$) in an alkaline medium.</p> <p>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:</p> <p>$$\mathrm{MnO}_4^{2-} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{MnO}_4^- + 2\mathrm{OH}^- + 2e^-$$</p> <p>Thus, both statements I and II are accurate, making the correct answer:</p> <p>Option D: <i>Both Statement I and Statement II are true</i>.</p>