JEE Main 2026 — Redox Reactions Question with Solution

In acidic medium, $KMn{O}_4$ acts as an oxidising agent and gets reduced to $M{n}^{2+}$. The n-factor for $KMn{O}_4$ is $5$.

The n-factors for the given reducing agents are calculated based on the total number of moles of electrons lost per mole of the compound:

For $Fe{C}_2{O}_4$:
$F{e}^{2+}\to F{e}^{3+}+e^{-}$
$C_2{O}_4^{2-}\to 2C{O}_2+2e^{-}$
Total electrons lost $=3$, so n-factor $=3$.
Equivalents $=1\times 3=3$.

For $F{e}_2(C_2{O}_4{)}_3$:
$F{e}^{3+}$ is not oxidised.
$3C_2{O}_4^{2-}\to 6C{O}_2+6e^{-}$
Total electrons lost $=6$, so n-factor $=6$.
Equivalents $=1\times 6=6$.

For $FeS{O}_4$:
$F{e}^{2+}\to F{e}^{3+}+e^{-}$
$S{O}_4^{2-}$ is not oxidised.
Total electrons lost $=1$, so n-factor $=1$.
Equivalents $=1\times 1=1$.

For $F{e}_2(S{O}_4{)}_3$:
Neither $F{e}^{3+}$ nor $S{O}_4^{2-}$ can be oxidised further.
n-factor $=0$.
Equivalents $=0$.

Total equivalents of the reducing mixture $=3+6+1+0=10$.

Let the number of moles of $KMn{O}_4$ required be $x$.
Equivalents of $KMn{O}_4=x\times 5$.

Equating the equivalents of the oxidising and reducing agents:
$5x=10$
$x=2$

Answer: $2$