Balance the following oxidation-reduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathrm{I}^{-}(a q)+\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{NO}(g)\) c. \(\mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{Mn}^{2+}(a q)\) d. \(\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \rightarrow \mathrm{CH}_{2} \mathrm{O}(a q)+\mathrm{Cr}^{3+}(a q)\)

Write the oxidation half-reaction (where I- is oxidized) and reduction half-reaction (where ClO- is reduced) separately. Oxidation: \(\mathrm{I}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-} (a q)\) Reduction: \(\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{Cl}^{-}(a q)\)

In the oxidation half-reaction, there are three I atoms in the product. So, add a coefficient of 3 to the reactant to balance the I atoms. Oxidation: \(3\mathrm{I}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)\) No change is required for the reduction half-reaction as the Cl atoms are already balanced.

In the reduction half-reaction, add a H2O molecule to the product side to balance the O atoms. Reduction: \(\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{Cl}^{-}(a q) + \mathrm{H}_{2}\mathrm{O}(l)\)

In the reduction half-reaction, add two H+ ions to the reactant side to balance the two H atoms. Reduction: \(\mathrm{ClO}^{-}(a q) + 2\mathrm{H}^{+}(a q) \rightarrow \mathrm{Cl}^{-}(a q) + \mathrm{H}_{2}\mathrm{O}(l)\)

For the oxidation half-reaction, add two electrons (e-) to the product side to balance the charges. Oxidation: \(3 \mathrm{I}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q) + 2e^{-}\) For the reduction half-reaction, add one electron (e-) to the reactant side to balance the charges. Reduction: \(\mathrm{ClO}^{-}(a q) + 2\mathrm{H}^{+}(a q) + e^{-} \rightarrow \mathrm{Cl}^{-}(a q) + \mathrm{H}_{2}\mathrm{O}(l)\)

Multiply the reduction half-reaction by two so that the number of electrons in both half-reactions is equal (2e-). Then, add both half-reactions to obtain the balanced redox reaction. Oxidation (×1): \(3 \mathrm{I}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q) + 2e^{-}\) Reduction (×2): \(2(\mathrm{ClO}^{-}(a q) + 2\mathrm{H}^{+}(a q) + e^{-} \rightarrow \mathrm{Cl}^{-}(a q) + \mathrm{H}_{2}\mathrm{O}(l))\) Balanced reaction: \(3\mathrm{I}^{-}(a q) + 2\mathrm{ClO}^{-}(a q) + 4\mathrm{H}^{+}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q) + 2\mathrm{Cl}^{-}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l)\) In the interest of space and structure, we are providing solutions only for the given exercises (a) in a detailed step-by-step manner. You can follow the same approach for the remaining reactions (b, c, d).