One mole of $\mathrm{Cl}_{2}(\mathrm{~g})$ was passed into 2 L of cold 2 M KOH solution. After the reaction, the concentrations of $\mathrm{Cl}^{-}, \mathrm{ClO}^{-}$ and $\mathrm{OH}^{-}$ are respectively (assume volume remains constant)

(C) $0.5 \mathrm{M}, 0.5 \mathrm{M}, 1 \mathrm{M}$

(A) $0.75 \mathrm{M}, 0.75 \mathrm{M}, 1 \mathrm{M}$

(B) $0.5 \mathrm{M}, 0.5 \mathrm{M}, 0.5 \mathrm{M}$

(D) $1 \mathrm{M}, 1 \mathrm{M}, 1 \mathrm{M}$

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Redox Reactions and Electrochemistry

One mole of

Cl_{2} (g)

was passed into 2 L of cold 2 M KOH solution. After the reaction, the concentrations of

Cl^{-}, ClO^{-}

and

OH^{-}

are respectively (assume volume remains constant)

Answer: (C)

In cold dilute KOH chlorine disproportionates:

C l_{2} + 2 O H^{-} \to C l^{-} + C l O^{-} + H_{2} O

. The amount of KOH is

2 M \times 2 L = 4 mol

. One mole of

C l_{2}

consumes

2 mol

O H^{-}

and produces

1 mol

each of

C l^{-}

and

C l O^{-}

, leaving

4 - 2 = 2 mol

O H^{-}

. In the

2 L

solution the concentrations are

[C l^{-}] = 0.5 M

[C l O^{-}] = 0.5 M

and

[O H^{-}] = 1 M

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Single correct option