The reaction \(\mathrm{H}_{2} \mathrm{SeO}_{3}(a q)+6 \mathrm{I}^{-}(a q)+4 \mathrm{H}^{+}(a q)\) \(\longrightarrow \mathrm{Se}(s)+2 \mathrm{I}_{3}^{-}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l)\) was studied at \(0^{\circ} \mathrm{C}\), and the following data were obtained:

First, we have to identify the species that are being oxidized and reduced in the given reaction. In this case, Se (selenium) from H2SeO3 is being reduced to Se(s), and I⁻(aq) is being oxidized to I₃⁻(aq). We can now write the unbalanced half-reactions: Oxidation half-reaction: \[\mathrm{I}^-(aq) \rightarrow \mathrm{I}_3^-(aq)\] Reduction half-reaction: \[\mathrm{H}_{2} \mathrm{SeO}_{3}(aq) \rightarrow \mathrm{Se}(s)\]

Now we will balance the oxidation and reduction half-reactions: For the oxidation half-reaction: 2I⁻ → I₃⁻ + I⁻, manage charges by multiplying the iodine ion by 6. Thus, we get: \(6\mathrm{I}^{-}(aq) \rightarrow 2\mathrm{I}_{3}^{-}(aq)\) For the reduction half-reaction: To balance the oxygens, we add 3H2O(l) as a product: \(\mathrm{H}_{2} \mathrm{SeO}_{3}(aq) \rightarrow \mathrm{Se}(s) + 3\mathrm{H}_{2}\mathrm{O}(l)\) To balance hydrogen, we add 6H⁺(aq) on the left side: \(\mathrm{H}_{2} \mathrm{SeO}_{3}(aq) + 6\mathrm{H}^{+}(aq) \rightarrow \mathrm{Se}(s) + 3\mathrm{H}_{2}\mathrm{O}(l)\) Now, we subtract 2H⁺(aq) from both sides to get the balanced reduction half-reaction: \(\mathrm{H}_{2} \mathrm{SeO}_{3}(aq) + 4\mathrm{H}^{+}(aq) \rightarrow \mathrm{Se}(s) + 3\mathrm{H}_{2}\mathrm{O}(l)\)

Finally, we can combine the balanced oxidation and reduction half-reactions to get the balanced overall reaction: \(6\mathrm{I}^{-}(aq) \rightarrow 2\mathrm{I}_{3}^{-}(aq)\) (Balanced Oxidation Half-Reaction) \(\mathrm{H}_{2} \mathrm{SeO}_{3}(aq) + 4\mathrm{H}^{+}(aq) \rightarrow \mathrm{Se}(s) + 3\mathrm{H}_{2}\mathrm{O}(l)\) (Balanced Reduction Half-Reaction) \(\mathrm{H}_{2} \mathrm{SeO}_{3}(aq) + 6\mathrm{I}^{-}(aq) + 4\mathrm{H}^{+}(aq) \rightarrow \mathrm{Se}(s) + 2\mathrm{I}_{3}^{-}(aq) + 3\mathrm{H}_{2}\mathrm{O}(l)\) (Balanced Overall Reaction) We have now successfully broken down and balanced the given redox reaction.