Step 1: Understanding the Concept:
Disproportionation is a specific class of redox reaction where a single chemical element contained within a reactant undergoes simultaneous oxidation and reduction, splitting into two distinct products with higher and lower oxidation states. The manganate ion ($\text{MnO}_4^{2-}$) is stable only in highly alkaline solutions; introducing an acidic environment causes it to instantly disproportionate.
Step 2: Key Formula or Approach:
1. Determine the baseline oxidation state of Manganese ($\text{Mn}$) in the manganate reactant ion ($\text{MnO}_4^{2-}$).
2. Balance the corresponding chemical reaction steps in an acidic medium.
3. Identify the oxidation states of manganese in the resulting products to confirm that one is oxidized and the other is reduced.
Step 3: Detailed Explanation:
Let's find the initial oxidation state of $\text{Mn}$ in $\text{MnO}_4^{2-}$:
\[ x + 4(-2) = -2 \implies x - 8 = -2 \implies x = +6 \]
In an acidic medium, the $\text{Mn(VI)}$ state is thermodynamically unstable and undergoes a rapid auto-redox rearrangement:
- Oxidation component: $\text{Mn(VI)}$ loses an electron to form the permangante ion ($\text{MnO}_4^-$), where the oxidation state climbs to $+7$:
\[ x + 4(-2) = -1 \implies x = +7 \]
- Reduction component: $\text{Mn(VI)}$ gains electrons to form manganese dioxide ($\text{MnO}_2$), where the oxidation state drops down to $+4$:
\[ x + 2(-2) = 0 \implies x = +4 \]
The fully balanced ionic chemical equation for this disproportionation process in an acidic medium is written as:
\[ 3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O} \]
Here, we can see clearly that out of three moles of $\text{MnO}_4^{2-}$, two moles are oxidized to $\text{MnO}_4^-$ and one mole is reduced to $\text{MnO}_2$. This confirms that the products are $\text{MnO}_4^-$ and $\text{MnO}_2$.
Step 4: Final Answer:
$\text{MnO}_4^{2-}$ in an acidic medium disproportionates to $\text{MnO}_4^-$ and $\text{MnO}_2$.