Complete and balance the following equations, and identify the oxidizing and reducing agents: (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{IO}_{3}^{-}(a q)\) (acidic solution) (b) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow\) \(\mathrm{Mn}^{2+}(a q)+\mathrm{HCO}_{2} \mathrm{H}(a q)\) (acidic solution) (c) \(\mathrm{I}_{2}(s)+\mathrm{OCl}^{-}(a q) \longrightarrow \mathrm{IO}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) (acidic solution) (d) \(\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{N}_{2} \mathrm{O}_{3}(a q)\) (acidic solution) (e) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Br}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)+\mathrm{BrO}_{3}^{-}(a q)\) (basic solution) (f) \(\mathrm{Pb}(\mathrm{OH})_{4}^{2-}(a q)+\mathrm{ClO}^{-}(a q) \longrightarrow \mathrm{PbO}_{2}(s)+\mathrm{Cl}^{-}(a q)\) (basic solution)

Oxidation states: \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\) -> Cr(+6), I(-1) -> Cr(+3), \(\mathrm{IO}_{3}^{-}\) -> I(+5) Half-reactions: 1. Reduction: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow \mathrm{Cr}^{3+}\) 2. Oxidation: \(\mathrm{I}^{-} \longrightarrow \mathrm{IO}_{3}^{-}\)

1. Reduction: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow 2\mathrm{Cr}^{3+}\) (balance Cr) \(7\mathrm{H}_{2}\mathrm{O} + \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \longrightarrow 2\mathrm{Cr}^{3+} + 14\mathrm{H}^{+}\) (balance O with H2O and H with H+) 2. Oxidation: \(2\mathrm{I}^{-} \longrightarrow \mathrm{IO}_{3}^{-}\) + \(6\mathrm{H}_{2}\mathrm{O}\) (balance I) \(2\mathrm{I}^{-} \longrightarrow \mathrm{IO}_{3}^{-}\) + \(6\mathrm{H}_{2}\mathrm{O}+12\mathrm{H}^{+}+ 6\mathrm{e^{-}}\) (balance O with H2O, H with H+ and charge with e-)

First, make the electrons equal in both half-reactions: \(6[\) \(7\mathrm{H}_{2}\mathrm{O} + \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \longrightarrow 2\mathrm{Cr}^{3+} + 14\mathrm{H}^{+}\) \(]\) \(2[\) \(2\mathrm{I}^{-} \longrightarrow \mathrm{IO}_{3}^{-} + 6\mathrm{H}_{2}\mathrm{O}+12\mathrm{H}^{+}+ 6\mathrm{e^{-}}\) \(]\) Now add the half-reactions: \(42\mathrm{H}_{2}\mathrm{O} + 6\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 4\mathrm{I}^{-} \longrightarrow 12\mathrm{Cr}^{3+} + 84\mathrm{H}^{+} + 2\mathrm{IO}_{3}^{-} + 12\mathrm{H}_{2}\mathrm{O}+24\mathrm{H}^{+}+ 12\mathrm{e^{-}}\) Simplify the reaction: \(4\mathrm{I}^{-} + 6\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 42\mathrm{H}_{2}\mathrm{O}\longrightarrow 12\mathrm{Cr}^{3+}+\mathrm{IO}_{3}^{-}+ 60\mathrm{H}^{+}\) So the balanced reaction is: \(4\mathrm{I}^{-}(aq) + 6\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(aq) + 42\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow 12\mathrm{Cr}^{3+}(aq) + 2\mathrm{IO}_{3}^{-}(aq) + 60\mathrm{H}^{+}(aq)\) Oxidizing agent: \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\), Reducing agent: \(\mathrm{I}^{-}\)