In the direct iodination of benzene, periodic acid () can be utilized as the designated oxidizing ag — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

We must balance the redox reaction between periodic acid and hydrogen iodide. In $HIO_4$, iodine is in the $+7$ oxidation state. In $HI$, iodine is in the $-1$ oxidation state. The goal is to produce elemental iodine ($I_2$), where iodine is at an oxidation state of $0$. The reduction half-reaction for the oxidizer is: $HIO_4 + 7H^+ + 7e^- \rightarrow \frac{1}{2}I_2 + 4H_2O$. The oxidation half-reaction for $HI$ is: $HI \rightarrow \frac{1}{2}I_2 + H^+ + e^-$. To balance the $7e^-$ required by one molecule of $HIO_4$, we must multiply the $HI$ half-reaction by $7$. Therefore, exactly $7$ moles of $HI$ are consumed per mole of $HIO_4$ ($HIO_4 + 7HI \rightarrow 4I_2 + 4H_2O$).