$$2 \mathrm{IO}_{3}^{-}+x \mathrm{I}^{-}+12 \mathrm{H}^{+} \rightarrow 6 \mathrm{I}_{2}+6 \mathrm{H}_{2} \mathrm{O}$$

What is the value of $x$ ?

<p>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.</p> <p>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 $\mathrm{I}^{-}$).</p> <p>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&#39;s 1 (as iodine goes from 0 to -1).</p> <p>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.</p> <p>So, the balanced reaction would be:</p> <p>$\mathrm{IO}_3^{-}+6 \mathrm{H}^{+}+5 \mathrm{I}^{-} \rightarrow 3 \mathrm{I}_2+3 \mathrm{H}_2 \mathrm{O}$</p> <p>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:</p> <p>$2 \mathrm{IO}_3^{-}+12 \mathrm{H}^{+}+10 \mathrm{I}^{-} \rightarrow 6 \mathrm{I}_2+6 \mathrm{H}_2 \mathrm{O}$</p> <p>This tells us that to get 6 moles of $\mathrm{I}_2$, you need 10 moles of $\mathrm{I}^{-}$. So, $x = 10$.</p>