JEE Main 2026 — Redox Reactions Question with Solution

In acidic medium, $K_{2}C{r}_2{O}_7$ acts as a strong oxidizing agent where $C{r}_2{O}_7^{2-}$ is reduced to $C{r}^{3+}$.
The reduction half-reaction is: $C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}\to 2C{r}^{3+}+7H_{2}O$.
For the oxidation of $I^{-}$ to $I_2$:
The balanced oxidation half-reaction is $2I^{-}\to I_2+2e^{-}$.
To balance the electrons with the reduction half-reaction (6 electrons), we multiply the oxidation half-reaction by 3:
$6I^{-}\to 3I_2+6e^{-}$.
Thus, the number of electrons involved in the balanced redox reaction for $I^{-}$ oxidation is $X=6$.
For the oxidation of $S^{2-}$ to $S$:
The balanced oxidation half-reaction is $S^{2-}\to S+2e^{-}$.
To balance the electrons with the reduction half-reaction (6 electrons), we multiply the oxidation half-reaction by 3:
$3S^{2-}\to 3S+6e^{-}$.
Thus, the number of electrons involved in the balanced redox reaction for $S^{2-}$ oxidation is $Y=6$.
The value of $X+Y=6+6=12$.