Remember the stability of different oxidation states of manganese. Mn⁺⁷ (in MnO₄⁻) and Mn⁺⁴ (in MnO₂) are relatively stable products. In strongly acidic conditions, Mn⁺² is the most stable state, but for the disproportionation of MnO₄²⁻, MnO₂ is the typical reduction product.
The question asks for the products formed when the manganate ion, MnO₄²⁻, undergoes a disproportionation reaction in an acidic medium.
A disproportionation reaction is one in which the same species is simultaneously oxidized and reduced.
Step 2: Oxidation State Analysis:
First, determine the oxidation state of manganese in MnO₄²⁻.
Let the oxidation state of Mn be \(x\).
The oxidation state of oxygen is −2.
\[
x + 4(-2) = -2
\]
\[
x - 8 = -2
\]
\[
x = +6
\]
So, manganese is in the +6 oxidation state in the manganate ion.
Step 3: Identifying Disproportionation Products:
In a disproportionation reaction, Mn(+6) will be both oxidized and reduced.
Oxidation:
A common higher oxidation state of manganese is +7, which corresponds to the permanganate ion, MnO₄⁻.
Reduction:
A common lower oxidation state of manganese in acidic or neutral medium is +4, which corresponds to manganese dioxide, MnO₂.
Thus, the expected products are MnO₄⁻ and MnO₂.
Step 4: Balancing the Reaction:
Unbalanced skeletal equation:
\[
\text{MnO}_4^{2-} \rightarrow \text{MnO}_4^- + \text{MnO}_2
\]
Here:
Mn(+6) → Mn(+7) involves loss of 1 electron (oxidation).
Mn(+6) → Mn(+4) involves gain of 2 electrons (reduction).
To balance electrons, we take three manganate ions:
\[
3\text{MnO}_4^{2-} \rightarrow 2\text{MnO}_4^- + \text{MnO}_2
\]
Now balance charge by adding H⁺ ions (acidic medium):
Left side charge = \(3 \times -2 = -6\)
Right side charge = \(2 \times -1 = -2\)
Add 4 H⁺ ions to the reactant side:
\[
3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2
\]
Now balance oxygen and hydrogen atoms by adding water molecules:
\[
3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O}
\]
The equation is now fully balanced.
Step 5: Final Answer:
The disproportionation of manganate ion (MnO₄²⁻) in an acidic medium produces:
Permanganate ion (MnO₄⁻) and
Manganese dioxide (MnO₂)
