JEE Main 2026 — Redox Reactions Question with Solution

First, we calculate the millimoles of the reactants:
Millimoles of ${\text{MnO}}_4^{-}$ = $500\text{ mL}\times 0.2\text{ M}=100\text{ mmol}$
Millimoles of ${\text{I}}^{-}$ = $500\text{ mL}\times 1.5\text{ M}=750\text{ mmol}$

In a basic medium, ${\text{MnO}}_4^{-}$ is reduced to ${\text{MnO}}_2$. The change in oxidation state of Mn is from $+7$ to $+4$, so its n-factor is $3$.
The problem explicitly states that iodide ions are oxidized to molecular iodine (${\text{I}}_2$). The change in oxidation state for iodine is from $-1$ to $0$, so the n-factor for ${\text{I}}^{-}$ is $1$.

Equivalents of ${\text{MnO}}_4^{-}$ = $100\times 3=300\text{ meq}$
Equivalents of ${\text{I}}^{-}$ = $750\times 1=750\text{ meq}$

Since ${\text{MnO}}_4^{-}$ is the limiting reagent, the equivalents of ${\text{I}}_2$ produced will be equal to the equivalents of ${\text{MnO}}_4^{-}$ consumed.
Equivalents of ${\text{I}}_2$ formed = $300\text{ meq}$

The liberated iodine is titrated with thiosulphate (${\text{S}}_2{\text{O}}_3^{2-}$):
${\text{I}}_2+2{\text{S}}_2{\text{O}}_3^{2-}\to 2{\text{I}}^{-}+{\text{S}}_4{\text{O}}_6^{2-}$

For thiosulphate, the n-factor is $1$ (as $2$ moles of ${\text{S}}_2{\text{O}}_3^{2-}$ lose $2$ moles of electrons).
From the law of equivalence:
Equivalents of ${\text{S}}_2{\text{O}}_3^{2-}$ = Equivalents of ${\text{I}}_2$
$\text{Molarity}\times \text{Volume (in mL)}\times \text{n-factor}=300$
$x\times 300\times 1=300$
$x=1$

Answer: $1$