JEE Main 2023 — Redox Reactions Question with Solution

The reaction given is a disproportionation reaction where iodine (${\mathrm{I}}_2$) undergoes both oxidation and reduction. Disproportionation reactions are a type of redox reaction where an element is simultaneously oxidized and reduced.

In the process, iodine gets oxidized from 0 oxidation state (in ${\mathrm{I}}_2$) to +5 oxidation state (in ${\mathrm{IO}}_3^{-}$), and reduced from 0 oxidation state (in ${\mathrm{I}}_2$) to -1 oxidation state (in ${\mathrm{I}}^{-}$).

The n-factor is the total change in oxidation state per molecule that undergoes the redox reaction. In this case, the n-factor for ${\mathrm{IO}}_3^{-}$ is 5 (as iodine goes from 0 to +5) and for ${\mathrm{I}}^{-}$, it's 1 (as iodine goes from 0 to -1).

Now, to balance the redox reaction, the total increase in oxidation state (total oxidation) must equal the total decrease in oxidation state (total reduction). Hence, the molar ratio of ${\mathrm{IO}}_3^{-}$ to ${\mathrm{I}}^{-}$ must be 1:5.

So, the balanced reaction would be:

${\mathrm{IO}}_3^{-}+6{\mathrm{H}}^{+}+5{\mathrm{I}}^{-}\to 3{\mathrm{I}}_2+3{\mathrm{H}}_2\mathrm{O}$

This equation yields 3 moles of ${\mathrm{I}}_2$, but the original equation needs to produce 6 moles of ${\mathrm{I}}_2$, so the entire equation is multiplied by 2:

$2{\mathrm{IO}}_3^{-}+12{\mathrm{H}}^{+}+10{\mathrm{I}}^{-}\to 6{\mathrm{I}}_2+6{\mathrm{H}}_2\mathrm{O}$

This tells us that to get 6 moles of ${\mathrm{I}}_2$, you need 10 moles of ${\mathrm{I}}^{-}$. So, $x=10$.