Question:

Chlorine undergoes disproportionation in alkaline medium as shown below:
\(a \, \text{Cl}_2(g) + b \, \text{OH}^-(\text{aq}) \rightarrow c \, \text{ClO}^-(\text{aq}) + d \, \text{Cl}^-(\text{aq}) + e \, \text{H}_2\text{O}(\text{l})\)
The values of \( a \), \( b \), \( c \), and \( d \) in a balanced redox reaction are respectively:

Updated On: Nov 3, 2025
  • 1, 2, 1, and 1
  • 2, 2, 1, and 3
  • 2, 2, 1, and 3
  • 2, 2, 1, and 3
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The Correct Option is A

Approach Solution - 1

To find the values of \(a\), \(b\), \(c\), and \(d\) in the balanced equation of chlorine disproportionation in an alkaline medium, let's follow these steps: 

  1. Disproportionation involves the simultaneous oxidation and reduction of the same element. In this case, chlorine (\(\text{Cl}_2\)) is both reduced to chloride ions (\(\text{Cl}^-\)) and oxidized to hypochlorite ions (\(\text{ClO}^-\)).
  2. First, write the half-reactions for oxidation and reduction:
  • Reduction: \(\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-\)
  • Oxidation: \(\text{Cl}_2 + 2\text{OH}^- \rightarrow 2\text{ClO}^- + 2\text{H}_2\text{O} + 2e^-\)
  1. Combine the half-reactions. Make sure the number of electrons lost is equal to the number of electrons gained to maintain electron balance:
  • Overall equation after canceling electrons: \( \text{Cl}_2 + 2\text{OH}^- \rightarrow \text{ClO}^- + \text{Cl}^- + \text{H}_2\text{O} \)
  1. Balancing the atoms and charges, we have:

From the combined reaction:

  • The coefficient of \(\text{Cl}_2\) (\(a\)) is 1.
  • The coefficient of \(\text{OH}^-\) (\(b\)) is 2.
  • The coefficient of \(\text{ClO}^-\) (\(c\)) is 1.
  • The coefficient of \(\text{Cl}^-\) (\(d\)) is 1.
  1. Thus, the balanced redox reaction coefficients are \(a = 1\), \(b = 2\), \(c = 1\), and \(d = 1\).

Therefore, the correct set of coefficients is 1, 2, 1, and 1, which matches with the given option: 1, 2, 1, and 1.

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Approach Solution -2

The disproportionation reaction of chlorine in alkaline medium can be balanced as follows. Chlorine (Cl2) is both reduced and oxidized:

Cl2 + 2OH → Cl + ClO + H2O

Thus, the values of a, b, c, and d are 1, 2, 1, and 1 respectively.

So, the correct answer is: Option (1)

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