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 objective is to identify which of the provided chemical reactions exemplify disproportionation. A disproportionation reaction is a redox reaction wherein a single element, originating from a single reactant, is concurrently oxidized and reduced. This process yields at least two distinct products containing that element, each at a different oxidation state.

Concept Used:

To ascertain if a reaction is a disproportionation reaction, the following criteria must be assessed for each reaction:

1. Identify an element present in a reactant that also appears in multiple products.
2. Determine the oxidation state of this identified element in the reactant and in each of the products.
3. Confirm that the oxidation state of the element in the reactant lies between its oxidation states in the products.
4. Verify that the element undergoes both oxidation (increase in oxidation state) and reduction (decrease in oxidation state) within the reaction.

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

Step-by-Step Solution:

Step 1: Analysis of Reaction (A)

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

The oxidation states of copper (Cu) in the reactant and products are as follows:

• Reactant \( \text{Cu}^+ \): Cu oxidation state is +1.
• Product \( \text{Cu}^{2+} \): Cu oxidation state is +2.
• Product \( \text{Cu} \) (elemental state): Cu oxidation state is 0.

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

Step 2: Analysis of Reaction (B)

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

The oxidation states of manganese (Mn) are determined as follows:

• Reactant \( \text{MnO}_4^{2-} \): Let the oxidation state of Mn be \(x\). The equation is \( x + 4(-2) = -2 \), yielding \( x = +6 \).
• Product \( \text{MnO}_4^{-} \): Let the oxidation state of Mn be \(y\). The equation is \( y + 4(-2) = -1 \), yielding \( y = +7 \).
• Product \( \text{MnO}_2 \): Let the oxidation state of Mn be \(z\). The equation is \( z + 2(-2) = 0 \), yielding \( z = +4 \).

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

Step 3: Analysis of Reaction (C)

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

The oxidation states of the involved elements are determined:

• Reactant \( \text{KMnO}_4 \): Mn is +7, O is -2.
• Product \( \text{K}_2\text{MnO}_4 \): Mn is +6.
• Product \( \text{MnO}_2 \): Mn is +4.
• Product \( \text{O}_2 \): 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) originating from the same reactant undergo redox changes, this is classified as an intramolecular redox reaction, not a disproportionation reaction.

Step 4: Analysis of Reaction (D)

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

The oxidation states of manganese (Mn) are determined:

• Reactant \( \text{MnO}_4^- \): Mn oxidation state is +7.
• Reactant \( \text{Mn}^{2+} \): Mn oxidation state is +2.
• Product \( \text{MnO}_2 \): Mn oxidation state 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 originates from two distinct reactants. This type of reaction, where an element from two different oxidation states converges to form a product with an intermediate oxidation state, is termed a comproportionation (or synproportionation) reaction, which is the inverse of disproportionation.

Final Result:

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

Consequently, the correct classifications are for reactions (A) and (B).