JEE Main · 2024 · Shift-IeasyRDX-040

Chlorine undergoes disproportionation in alkaline medium as shown below: {aCl_2(g) + bOH^{-}(aq) cClO^{-}(aq) +…

Redox Reactions · Class 11 · JEE Main Previous Year Question

Question

Chlorine undergoes disproportionation in alkaline medium as shown below:

aCl2(g)+bOH−(aq)→cClO−(aq)+dCl−(aq)+eH2O(l)\mathrm{aCl_2(g) + bOH^{-}(aq) \rightarrow cClO^{-}(aq) + dCl^{-}(aq) + eH_2O(l)}

The values of aa, bb, cc and dd in a balanced redox reaction are respectively:

Options
  1. a

    1, 2, 1 and 1

    ✓
  2. b

    2, 2, 1 and 3

  3. c

    3, 4, 4 and 2

  4. d

    2, 4, 1 and 3

Correct Answera

1, 2, 1 and 1

Detailed Solution

Step 1 — Identify the half-reactions:

Oxidation: Cl2→ClO−\mathrm{Cl_2 \rightarrow ClO^-} Cl: 0→+10 \rightarrow +1 (loss of 1e⁻ per Cl)

Reduction: Cl2→Cl−\mathrm{Cl_2 \rightarrow Cl^-} Cl: 0→−10 \rightarrow -1 (gain of 1e⁻ per Cl)

Step 2 — Balance electrons: Each Cl2\mathrm{Cl_2} provides one Cl that gets oxidised (+1) and one that gets reduced (-1) → already balanced with 1 molecule of Cl2\mathrm{Cl_2}.

Step 3 — Balance atoms: Cl2+2OH−→ClO−+Cl−+H2O\mathrm{Cl_2 + 2OH^- \rightarrow ClO^- + Cl^- + H_2O}

Check:

  • Cl: 2 = 1 + 1 ✓
  • O: 2 = 1 + 1 ✓
  • H: 2 = 2 ✓
  • Charge: −2=−1+(−1)+0=−2-2 = -1 + (-1) + 0 = -2 ✓

Values: a=1a = 1, b=2b = 2, c=1c = 1, d=1d = 1

Answer: Option (1) — 1, 2, 1 and 1

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