Identify the oxidizing agent and reducing agent in the reaction \(\mathrm{IO}_{4}^{-}+7 \mathrm{I}^{-}+8 \mathrm{H}^{+} \rightarrow 4 \mathrm{I}_{2}+4 \mathrm{H}_{2} \mathrm{O}\)

The species involved in the reaction are \(\mathrm{IO}_{4}^{-}\), \(\mathrm{I}^{-}\), \(\mathrm{I}_{2}\), \(\mathrm{H}_{2}\mathrm{O}\), and \(\mathrm{H}^{+}\). We need to find the initial and final oxidation states of iodine in each of these species. We can ignore hydrogen and oxygen as they aren't involved in the redox process in this case. Initial oxidation states: - \(\mathrm{IO}_{4}^{-}\): Iodine has an oxidation state of +7 - \(\mathrm{I}^{-}\): Iodine has an oxidation state of -1 Final oxidation states: - \(\mathrm{I}_{2}\): Iodine has an oxidation state of 0 - \(\mathrm{H}_{2}\mathrm{O}\): Oxygen and hydrogen have oxidation states of -2 and +1, respectively, and there is no iodine in this compound, so we can ignore it.

Oxidation is the process of losing electrons, while reduction is the process of gaining electrons. Now, let's look at the change in oxidation states of the iodine in each species: - \(\mathrm{IO}_{4}^{-}\): Iodine's oxidation state goes from +7 to 0 in \(\mathrm{I}_{2}\). This is a decrease of 7, meaning it gains 7 electrons (reduction process). - \(\mathrm{I}^{-}\): Iodine's oxidation state goes from -1 to 0 in \(\mathrm{I}_{2}\). This is an increase of 1, meaning it loses 1 electron (oxidation process).

Based on the changes in oxidation states, we can now identify the oxidizing and reducing agents: - Oxidizing agent: The species that gets reduced (gains electrons) is the oxidizing agent. In this case, \(\mathrm{IO}_{4}^{-}\) is the oxidizing agent, as its iodine goes from an oxidation state of +7 to 0. - Reducing agent: The species that gets oxidized (loses electrons) is the reducing agent. In this case, \(\mathrm{I}^{-}\) is the reducing agent, as its iodine goes from an oxidation state of -1 to 0. So, the oxidizing agent is \(\mathrm{IO}_{4}^{-}\), and the reducing agent is \(\mathrm{I}^{-}\).