Question

Which of the following reactions are disproportionation reactions?
\(\text{(A) } \text{Cu}^+ \rightarrow \text{Cu}^{2+} + \text{Cu}\)
\(\text{(B) } 3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O}\)
\(\text{(C) } 2\text{KMnO}_4 \rightarrow \text{K}_2\text{MnO}_4 + \text{MnO}_2 + \text{O}_2\)
\(\text{(D) } 2\text{MnO}_4^- + 3\text{Mn}^{2+} + 2\text{H}_2\text{O} \rightarrow 5\text{MnO}_2 + 4\text{H}^+\)
Choose the correct answer from the options given below.

• A. (A), (B)
• B. (B), (C), (D)
• C. (A), (B), (C)
• D. (A), (D)

Answer 1

The Correct Option is A

The problem asks to identify which of the given chemical reactions are disproportionation reactions. A disproportionation reaction is a specific type of redox reaction where a single element from a single reactant is simultaneously oxidized and reduced, forming at least two different products containing that element in different oxidation states.

Concept Used:

To identify a disproportionation reaction, we must perform the following checks for each reaction:

1. Identify an element in a reactant that appears in multiple products.
2. Determine the oxidation state of this element in the reactant and in each of the products.
3. Verify if the oxidation state of the element in the reactant is intermediate to its oxidation states in the products.
4. Confirm that the element is both oxidized (its oxidation state increases) and reduced (its oxidation state decreases) in the reaction.

The general form for disproportionation is: \( \text{Element (Oxidation State N)} \rightarrow \text{Element (Oxidation State > N)} + \text{Element (Oxidation State < N)} \)

Step-by-Step Solution:

Step 1: Analyze Reaction (A)

The reaction is: \( \text{Cu}^+ \rightarrow \text{Cu}^{2+} + \text{Cu} \)

We determine the oxidation states of copper (Cu) in the reactant and products:

• Oxidation state of Cu in reactant \( \text{Cu}^+ \) is +1.
• Oxidation state of Cu in product \( \text{Cu}^{2+} \) is +2.
• Oxidation state of Cu in product \( \text{Cu} \) (elemental form) is 0.

Here, \( \text{Cu}^+ \) is oxidized to \( \text{Cu}^{2+} \) (oxidation state increases from +1 to +2), and it is also reduced to \( \text{Cu} \) (oxidation state decreases from +1 to 0). Since the same species \( \text{Cu}^+ \) undergoes both oxidation and reduction, this is a disproportionation reaction.

Step 2: Analyze Reaction (B)

The reaction is: \( 3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O} \)

We determine the oxidation states of manganese (Mn):

• In reactant \( \text{MnO}_4^{2-} \): Let oxidation state of Mn be \(x\). Then \( x + 4(-2) = -2 \Rightarrow x = +6 \).
• In product \( \text{MnO}_4^{-} \): Let oxidation state of Mn be \(y\). Then \( y + 4(-2) = -1 \Rightarrow y = +7 \).
• In product \( \text{MnO}_2 \): Let oxidation state of Mn be \(z\). Then \( z + 2(-2) = 0 \Rightarrow z = +4 \).

Manganese in \( \text{MnO}_4^{2-} \) (oxidation state +6) is oxidized to \( \text{MnO}_4^{-} \) (+7) and is also reduced to \( \text{MnO}_2 \) (+4). Since the same reactant \( \text{MnO}_4^{2-} \) undergoes both oxidation and reduction, this is a disproportionation reaction.

Step 3: Analyze Reaction (C)

The reaction is: \( 2\text{KMnO}_4 \rightarrow \text{K}_2\text{MnO}_4 + \text{MnO}_2 + \text{O}_2 \)

We determine the oxidation states of the elements involved:

• In reactant \( \text{KMnO}_4 \): Oxidation state of Mn is +7, and O is -2.
• In product \( \text{K}_2\text{MnO}_4 \): Oxidation state of Mn is +6.
• In product \( \text{MnO}_2 \): Oxidation state of Mn is +4.
• In product \( \text{O}_2 \): Oxidation state of O is 0.

In this reaction, Manganese (Mn) is reduced from +7 to +6 and +4. Oxygen (O) is oxidized from -2 to 0. Since two different elements (Mn and O) from the same reactant are undergoing redox changes, this is an intramolecular redox reaction, but not a disproportionation reaction.

Step 4: Analyze Reaction (D)

The reaction is: \( 2\text{MnO}_4^- + 3\text{Mn}^{2+} + 2\text{H}_2\text{O} \rightarrow 5\text{MnO}_2 + 4\text{H}^+ \)

We determine the oxidation states of manganese (Mn):

• In reactant \( \text{MnO}_4^- \): Oxidation state of Mn is +7.
• In reactant \( \text{Mn}^{2+} \): Oxidation state of Mn is +2.
• In product \( \text{MnO}_2 \): Oxidation state of Mn is +4.

Here, Mn from \( \text{MnO}_4^- \) (+7) is reduced to \( \text{MnO}_2 \) (+4), and Mn from \( \text{Mn}^{2+} \) (+2) is oxidized to \( \text{MnO}_2 \) (+4). Although the same element is involved, it comes from two different reactants. This type of reaction, where an element from two different oxidation states forms a product with an intermediate oxidation state, is called a comproportionation (or synproportionation) reaction, which is the reverse of disproportionation.

Final Result:

Based on the analysis, reactions (A) and (B) are disproportionation reactions.

Therefore, the correct options are (A) and (B).