Redox Reactions — JEE Main practice

38 questions

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Sample questions with solutions

Q1 · 2026

The oxidation state of chromium in the final product formed in the reaction between KI and acidified $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ solution is :

A.
+3
B.
+2
C.
+4
D.
+6

Answer: A

In acidified $\mathrm{K_2Cr_2O_7}$ , chromium starts at the $+6$ oxidation state (in $\mathrm{Cr_2O_7^{2-}}$ ).
When KI is added, $\mathrm{I^-}$ is oxidized to $\mathrm{I_2}$ , while dichromate is reduced via:

\mathrm{Cr_2O_7^{2-}+14H^++6e^-\rightarrow2Cr^{3+}+7H_2O}

So chromium in the final product is $\mathrm{Cr^{3+}}$ , i.e., oxidation state $+3$ .

Hence, the answer is Option A.

Q2 · 2026

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)

A.
$0.5 \mathrm{M}, 0.5 \mathrm{M}, 1 \mathrm{M}$
B.
$0.5 \mathrm{M}, 0.5 \mathrm{M}, 0.5 \mathrm{M}$
C.
$1 \mathrm{M}, 1 \mathrm{M}, 1 \mathrm{M}$
D.
$0.75 \mathrm{M}, 0.75 \mathrm{M}, 1 \mathrm{M}$

Answer: A

In cold KOH, chlorine undergoes disproportionation:

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

Compute the initial moles of $\mathrm{OH^-}$ in 2 L of 2 M KOH:

n(\mathrm{OH^-})=2\times2=4\text{ mol}

With 1 mol $\mathrm{Cl_2}$ reacting: $\mathrm{OH^-}$ consumed = $2\times1=2$ mol, $\mathrm{Cl^-}$ formed = 1 mol, $\mathrm{ClO^-}$ formed = 1 mol, $\mathrm{OH^-}$ remaining = $4-2=2$ mol.
Assuming the volume stays at 2 L, compute the concentrations:

\begin{aligned} [\mathrm{Cl^-}] &= \frac{1}{2} = 0.5~\mathrm{M} \\ [\mathrm{ClO^-}] &= \frac{1}{2} = 0.5~\mathrm{M} \\ [\mathrm{OH^-}] &= \frac{2}{2} = 1~\mathrm{M} \end{aligned}

Hence, the answer is Option A.

Q3 · 2026

Given below are two statements:

Statement (I): Oxidising power of halogens decreases in the order $F_2 > Cl_2 > Br_2 > I_2$ , which is the basis of "Layer test".

Statement (II): "Layer test" to identify $Br_2$ and $I_2$ in aqueous solution involves the oxidation of bromide or iodide into $Br_2$ or $I_2$ respectively with $Cl_2$ , which is a type of displacement redox reaction.

In the light of the above statements, choose the correct answer from the options given below:

A.
Both Statement I and Statement II are true
B.
Both Statement I and Statement II are false
C.
Statement I is true but Statement II is false
D.
Statement I is false but Statement II is true

Answer: A

The oxidizing power of halogens decreases as $F_2>Cl_2>Br_2>I_2$ , meaning fluorine is the strongest oxidizer and iodine the weakest. This ordering underlies the "Layer test," confirming Statement I is true.
In the Layer test, chlorine gas is passed through an aqueous solution containing bromide or iodide ions, and since $\mathrm{Cl_2}$ has a higher oxidizing power than $\mathrm{Br_2}$ or $\mathrm{I_2}$ ; it displaces and oxidizes them:

\mathrm{NaBr + Cl_2 \rightarrow Br_2 + NaCl}

\mathrm{NaI + Cl_2 \rightarrow I_2 + NaCl}

These are indeed displacement redox reactions, confirming Statement II is also true, and it correctly explains Statement I.

Hence, the answer is Option A.

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