JEE Main 2025 — Redox Reactions Question with Solution

Molarity of KI = 0.1 M

$$5I^{-}+I{O}_3^{-}+6H^{+}\to 3I_2+3H_{2}O$$ (balanced chemical equation)

$$5KI+KI{O}_3+3H_{2}S{O}_4\to 3I_2+3K_{2}S{O}_4+3H_{2}O$$

(A) Volume given (KI) = 200 mL = 0.2 L

Molarity (KI) = 0.1 M or mol L${}^{-1}$

The molarity formula is used here to calculate the number of moles of K.I.

Molarity, $$M=\frac{numberofmoles}{volumeinL}$$

So, number of moles = Molarity $\times$ Volume in L

For the given volume of KI,

number of moles = 0.1 mol L${}^{-1}$ $\times$ 0.2 L

= 0.02 mol

The stoichiometric ratio between KI and KIO${}_3$ is

$$KI:KI{O}_3$$

$$I^{-}:I{O}_3^{-}$$

$$5:1$$

$$1:\frac{1}{5}$$

For one mol $$I^{-},\frac{1}{5}$$ mol $$I{O}_3^{-}$$ is used.

So, for 0.2 mol I${}^{-}$,

Moles of $$I{O}_3^{-}=0.02\times \frac{1}{5}=0.004$$ mol

The statement (A) is correct.

(B) Volume given (KI) = 200 mL = 0.2 L

Molarity (KI) = 0.1 M

number of moles of KI (I${}^{-}$) = Molarity $\times$ Volume (L)

$$=0.1mol{L}^{-1}\times 0.2L=0.02mol$$

The stoichiometric ratio between KE and $$H_{2}H{O}_4$$ is

$$KI:H_{2}H{O}_4$$

$$5:3$$

$$1:\frac{3}{5}$$ {3 moles $$H_{2}S{O}_4$$ can give 6H${}^{+}$}

$$I^{-},\frac{3}{5}$$

For one mol $$I^{-},\frac{3}{5}$$ mol $$H_{2}S{O}_4$$ is used.

So, for 0.02 mol I${}^{-}$,

moles of $$H_{2}S{O}_4=0.02\times \frac{3}{5}=0.012$$ mol

This statement is not correct.

(C) Volume gien (KI) = 0.5 L

Molarity (KI) = 0.1 M

Number of moles of KI = Molarity $\times$ Volume

$$=0.1mol{L}^{-1}\times 0.5L=0.05mol$$

The stoichiometric ratio between I${}^{-}$ and I${}_2$ is

$$KI:I_2$$

$$I^{-}:I_2$$

$$5:3$$

$$1:\frac{3}{5}$$

For 1 mol $$I^{-},\frac{3}{5}$$ mol I${}_2$ is produced.

So, for 0.05 mol, moles of $$I_2=0.05\times \frac{3}{5}=0.03$$ mol

This statement is not correct.

(D) Equivalent weight of $$KI{O}_3=\frac{Molecularweight}{5}$$

This statement is correct.

Equivalent weight $$=\frac{Molecularweight}{Valencyfactor\to numberofelectrons}$$

Here, valency factor = 5

Each KIO${}_3$ molecule gains 5 electrons in this vacation.

KIO${}_3$ acts as the ordinating agent, gains electrons from the iodide ion in KI.

When KIO${}_3$ is redued to I${}_2$ each molecule gains 5 electrons. So, the valency factor is 5 and equivalent weight is lower than molecular weight.

This statement is correct.

Correct statements are A and D.