Balance these equations for redox reactions occurring in basic solution. (a) \(\mathrm{CrO}_{4}^{2-}+\mathrm{S}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{SO}_{3}^{2-}\) (b) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}+\mathrm{N}_{2} \mathrm{H}_{4} \longrightarrow\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}+\mathrm{N}_{2}(\mathrm{g})\) (c) \(\operatorname{Fe}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{MnO}_{4}^{-} \longrightarrow\) \(\mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{MnO}_{2}(\mathrm{s})\)

First, the individual half reactions for the redox reaction are identified. The two half reactions for the equation are \(\mathrm{CrO}_{4}^{2-}\) to \(\mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s})\) and \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-}\) to \(\mathrm{SO}_{3}^{2-}\). The half reactions are balanced separately by matching oxygen atoms with water molecules and hydrogen atoms with \(\mathrm{H}^{+}\) ions, and if necessary, matching the charges with electrons. Finally, the half equations are added together, cancelling out any same term from both sides of the reaction. So, the balanced equation is \(\mathrm{CrO}_{4}^{2-} + 3\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s}) + 5\mathrm{OH}^{-}\) and \(\mathrm{S}_2\mathrm{O}_4^{2-} + 2\mathrm{OH}^{-} \rightarrow 2\mathrm{SO}_3^{2-}+\mathrm{H}_2\mathrm{O}\).

For the second equation, the half reactions are \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\) to \([\mathrm{Fe}(\mathrm{CN})_{6}]^{4-}\) and \(\mathrm{N}_2\mathrm{H}_4\) to \(\mathrm{N}_2\mathrm{(g)}\). There is no redox principle in the first half reaction equation thereby, balancing this equation with water molecules, hydrogen ions, and electrons form it into: \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-} \rightarrow [\mathrm{Fe}(\mathrm{CN})_{6}]^{4-} + e^{-}\). Balancing the second half reaction equation results in: \(\mathrm{N}_2\mathrm{H}_4 + 2 \mathrm{OH}^{-} \rightarrow \mathrm{N}_2\mathrm{(g)}+ 4\mathrm{H}_2\mathrm{O}\). And finally, the overall equation is found by combining the half reactions to give: \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}+ \mathrm{N}_2\mathrm{H}_4 +2 \mathrm{OH}^{-}\rightarrow [\mathrm{Fe}(\mathrm{CN})_{6}]^{4-} + \mathrm{N}_2\mathrm{(g)}+ 4\mathrm{H}_2\mathrm{O}\).

The half reactions for the third equation are \(\mathrm{Fe}(\mathrm{OH})_{2}(\mathrm{s})\) to \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\) and \(\mathrm{O}_2(\mathrm{g})\) to \(\mathrm{H}_2\mathrm{O}\). The final balanced equation is obtained as: \(\mathrm{Fe}(\mathrm{OH})_{2}(\mathrm{s}) + \mathrm{O}_{2}(\mathrm{g})+\mathrm{H}_2\mathrm{O}\rightarrow \mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\).

For the fourth equation, the half reactions are \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) to \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) and \(\mathrm{MnO}_{4}^{-}\) to \(\mathrm{MnO}_{2}(\mathrm{s})\). The final balanced equation is obtained as: \(3\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} + \mathrm{MnO}_{4}^{-}+\mathrm{OH}^{-} \rightarrow 3\mathrm{CH}_{3} \mathrm{COO}^{-} + \mathrm{MnO}_{2}(\mathrm{s}) + 4\mathrm{H}_2\mathrm{O}\).