JEE Main 2026 — Redox Reactions Question with Solution
JEE Main 2026 (24 January Shift 2)
Question
One mole of was passed into 2 L of cold 2 M KOH solution. After the reaction, the concentrations of and are respectively (assume volume remains constant)
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Show full solutionCorrect option: A
Correct answer
A
Step-by-step explanation
Cl₂ undergoes disproportionation in cold KOH:
Initial: 1 mole Cl₂, 2 L × 2 M KOH = 4 moles OH⁻
Molar ratio Cl₂:OH⁻ = 1:2 (stoichiometric)
Products formed: 1 mole Cl⁻, 1 mole ClO⁻; Excess OH⁻ = 4 - 2 = 2 moles
In 2 L solution: [Cl⁻] = 0.5 M, [ClO⁻] = 0.5 M, [OH⁻] = 1 M
Initial: 1 mole Cl₂, 2 L × 2 M KOH = 4 moles OH⁻
Molar ratio Cl₂:OH⁻ = 1:2 (stoichiometric)
Products formed: 1 mole Cl⁻, 1 mole ClO⁻; Excess OH⁻ = 4 - 2 = 2 moles
In 2 L solution: [Cl⁻] = 0.5 M, [ClO⁻] = 0.5 M, [OH⁻] = 1 M
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This is a previous-year question from JEE Main 2026, covering the Redox Reactions chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.