Test Index

Redox Reactions

Section: Chemistry

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Question : 5 of 73

Marks: +1, -0

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 ^{-}

OH ^{-}

are respectively (assume volume remains constant)

[24 Jan 2026 Shift 2]

$0.5 M , 0.5 M , 0.5 M$
$1 M , 1 M , 1 M$
$0.5 M , 0.5 M , 1 M$
$0.75 M , 0.75 M , 1 M$

Solution:

Concept:

In cold KOH, chlorine disproportionates to give chloride and hypochlorite ions.

Explanation:

The balanced reaction is:

\mathrm{Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O}

Initial moles of

\mathrm{OH^-}

in

2\text{ L}

of

2\text{ M}

KOH =

2 \times 2 = 4

mol.

Given

1

mol

\mathrm{Cl_2}

reacts.

Moles of

\mathrm{OH^-}

consumed =

2 \times 1 = 2

mol.

Moles of

\mathrm{Cl^-}

formed =

1

mol.

Moles of

\mathrm{ClO^-}

formed =

1

mol.

Moles of

\mathrm{OH^-}

remaining =

4 - 2 = 2

mol.

Volume remains constant at

2\text{ L}

.

Concentration of

\mathrm{Cl^-}

=

\frac{1}{2}=0.5\text{ M}

.

Concentration of

\mathrm{ClO^-}

=

\frac{1}{2}=0.5\text{ M}

.

Concentration of

\mathrm{OH^-}

=

\frac{2}{2}=1\text{ M}

.

Answer:

Option C:

0.5\text{ M},\ 0.5\text{ M},\ 1\text{ M}

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