What happens when acidic solution of potassium permanganate is allowed to stand for sometime ? Give the equation involved. What is this type of reaction called ?

Question

Statement-I : Consider the following pairs of ions $(\text{Sc}^{3+}, \text{Ti}^{3+}), (\text{Ti}^{4+}, \text{Ni}^{2+}), (\text{Cu}^{2+}, \text{Zn}^{2+})$ and $(\text{Cr}^{3+}, \text{Mn}^{3+})$. Out of these pairs three pairs consist of ions that are both coloured :
Statement-II : Among the lanthanide ions $\text{Eu}^{2+}$, $\text{Gd}^{3+}$, $\text{Ce}^{4+}$ and $\text{Tb}^{4+}$, the ion $\text{Tb}^{4+}$ is the strongest reducing agent.
Choose the correct option.

Answer 1

When an acidic solution of potassium permanganate ($ KMnO_4 $) is allowed to stand for some time, it undergoes a reduction reaction where $ KMnO_4 $ is reduced to manganese dioxide ($ MnO_2 $) and the purple color of $ KMnO_4 $ disappears due to the reduction of $ Mn^{7+} $ to $ Mn^{4+} $. In the acidic medium, the potassium permanganate is typically reduced to $ Mn^{2+} $ (manganese(II) ions), and the hydrogen ions from the acid help to facilitate the reaction. The equation involved is: \[ 2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O \] The reaction is known as a **disproportionation reaction**, where a species is both reduced and oxidized simultaneously. In this case, the $ MnO_4^- $ ions are reduced to $ Mn^{2+} $ while being oxidized to $ MnO_2 $ (which occurs under different conditions or with different stoichiometries). This type of reaction is called a **disproportionation reaction** because one molecule of potassium permanganate undergoes both oxidation and reduction, resulting in the formation of two different oxidation states of manganese.

What happens when acidic solution of potassium permanganate is allowed to stand for sometime ? Give the equation involved. What is this type of reaction called ?

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