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.
\(\text{(A) } \text{Cu}^+ \rightarrow \text{Cu}^{2+} + \text{Cu}\)
\(\text{(B) } 3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O}\)
\(\text{(C) } 2\text{KMnO}_4 \rightarrow \text{K}_2\text{MnO}_4 + \text{MnO}_2 + \text{O}_2\)
\(\text{(D) } 2\text{MnO}_4^- + 3\text{Mn}^{2+} + 2\text{H}_2\text{O} \rightarrow 5\text{MnO}_2 + 4\text{H}^+\)
Choose the correct answer from the options given below.
- (A), (B)
- (B), (C), (D)
- (A), (B), (C)
- (A), (D)
The Correct Option is A
Solution and Explanation
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:
- Identify an element in a reactant that appears in multiple products.
- Determine the oxidation state of this element in the reactant and in each of the products.
- Verify if the oxidation state of the element in the reactant is intermediate to its oxidation states in the products.
- 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).
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