What happens when an acidic solution of the green compound (B) is allowed to stand for some time ? Write the equation and name the type of reaction.
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Solution and Explanation
Step 1: What happens on standing in acid
When an acidic solution of the manganate ion $\mathrm{MnO_4^{2-}}$ is left for some time, some manganese goes up to +7 (purple permanganate) and some goes down to +4 (brown $\mathrm{MnO_2}$).
Step 2: Write the equation
\[ 3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O \]
Step 3: Name the reaction type
Here the same element, manganese, is both oxidised (+6 to +7) and reduced (+6 to +4) in one reaction. A reaction where one element does both at once is called a disproportionation reaction.
Answer: The manganate disproportionates: $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$, which is a disproportionation reaction.
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