Thiosulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\), is a reducing agent that can be oxidized to different products, depending on the strength of the oxidizing agent and other conditions. By adding \(\mathrm{H}^{+}, \mathrm{H}_{2} \mathrm{O},\) and/or \(\mathrm{OH}^{-}\) as necessary, write redox equations to show the oxidation of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to (a) \(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) by \(\mathrm{I}_{2}\) (iodide ion is another product) (b) \(\mathrm{HSO}_{4}^{-}\) by \(\mathrm{Cl}_{2}\) (chloride ion is another product) (c) \(\mathrm{SO}_{4}^{2-}\) by \(\mathrm{OCl}^{-}\) in basic solution (chloride ion is another product)

1. Write down half-reactions: Oxidation: \( \mathrm{S}_{2} \mathrm{O}_{3}^{2-} \rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) Reduction: \(\mathrm{I}_{2} \rightarrow 2\mathrm{I}^{-}\) 2. Balance the atoms and charges in the half-reactions. To balance the Sulphur, we must multiply the oxidation half-reaction by 2: 2\(\mathrm{S}_{2} \mathrm{O}_{3}^{2-} \rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) \(\mathrm{I}_{2} \rightarrow 2\mathrm{I}^{-}\) Then, add H\(_{2}\)O and H\(^{+}\)s where necessary to balance oxygen and hydrogen. Finally, balance the charges by adding electrons. Completed half-reactions: 2\(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) + \(\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) + 2\(\mathrm{H}^{+}\) + 2e\(^{-}\) \(\mathrm{I}_{2}\) + 2e\(^{-}\) \(\rightarrow 2\mathrm{I}^{-}\) 3. Add the balanced half-reactions together to yield the final redox equation: 2\(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) + \(\mathrm{H}_{2}\mathrm{O}\) + \(\mathrm{I}_{2}\) \(\rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) + 2\(\mathrm{H}^{+}\) + 2\(\mathrm{I}^{-}\)

1. Write down half-reactions: Oxidation: \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-} \rightarrow \mathrm{HSO}_{4}^{-}\) Reduction: \(\mathrm{Cl}_{2} \rightarrow 2\mathrm{Cl}^{-}\) 2. Balance the atoms and charges in the half-reactions, then combine them into the final balanced redox equation:

1. Write down half-reactions: Oxidation: \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-} \rightarrow \mathrm{SO}_{4}^{2-}\) Reduction: \(\mathrm{OCl}^{-} \rightarrow \mathrm{Cl}^{-}\) 2. Balance the atoms and charges in the half-reactions, then combine them into the final balanced redox equation. As this is in a basic solution, also add \(\mathrm{OH}^{-}\) ions where necessary to the final equation.