91Ó°ÊÓ

Assign the oxidation numbers for each species in the given reactions \n\n Reaction a: \n Chlorine in \(\mathrm{Cl}_{2}(\mathrm{g})\) has an oxidation number of 0. In \(\mathrm{Cl}^{-}\), it has an oxidation number of -1 and in \(\mathrm{ClO}_{3}^{-}\), it has an oxidation number of +5. \n\n Reaction b: \n Sulfur in \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-}\) has an oxidation state of +4, in \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\), it has an oxidation number of +2 and in \(\mathrm{HSO}_{3}^{-}\) it has an oxidation number of +4.

Create and balance half reactions for both oxidation and reduction \n\n Reaction a: \n Oxidation: \(\mathrm{Cl}\to\mathrm{ClO}_{3}^{-}+6e^{-}\) \n Reduction: \(\mathrm{Cl}+e^{-}\to\mathrm{Cl}^{-}\) \n\n Reaction b: \n Oxidation: \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-} - 2e^{-}\to\mathrm{HSO}_{3}^{-}\) \n Reduction: \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-}\to\mathrm{S}_{2} \mathrm{O}_{3}^{2-}+e^{-}\)

Balance the number of electrons lost and gained in each half-reaction \n\n Reaction a: \n Multiply the reduction half-reaction by 6 and the oxidation half-reaction by 1 and then add them up: \n 6\(\mathrm{Cl} + e^{-} \to 6 \mathrm{Cl}^{-}\) \n \(\mathrm{Cl} \to \mathrm{ClO}_{3}^{-}+6e^{-}\) \n The final balanced equation for reaction a in basic solution is: \n \(\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 6 \mathrm{Cl}^{-} + \mathrm{ClO}_{3}^{-}\) \n\n Reaction b: \n Multiply the reduction half-reaction by 2 and the oxidation half-reaction by 1 and then add them up: \n 2 \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-} \to 2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}+2e^{-}\) \n \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-} - 2e^{-} \to \mathrm{HSO}_{3}^{-} \) \n The final balanced equation for reaction b in acidic solution is: \n 3 \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-} \rightarrow 2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-} + \mathrm{HSO}_{3}^{-}\)