JEE Main 202624 JanuaryEvening Shift
- Exam
- JEE Main
- Year
- 2026
- Date
- 24 January
- Shift
- Evening Shift
- Subject
- Chemistry
- Chapter
- Redox Reactions
JEE Main 2026 · 24 January · Evening Shift · Chemistry · Redox Reactions
One mole of Cl_ 2 (~g) was passed into 2 L of cold 2 M KOH...
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)
Options
- A
- B
- C
- D
Correct answer
A.
Step-by-step solution
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