91Ó°ÊÓ

For each half-reaction, begin by balancing any elements other than oxygen and hydrogen.**(a)** For \(\mathrm{UO}_{2}^{+} \rightarrow \mathrm{U}^{4+}\), uranium (U) is already balanced.**(b)** For \(\mathrm{ClO}_{3}^{-} \rightarrow \mathrm{Cl}^{-}\), chlorine (Cl) is already balanced.**(c)** For \(\mathrm{N}_{2}\mathrm{H}_{4} \rightarrow \mathrm{N}_{2}\), nitrogen (N) is already balanced.**(d)** For \(\mathrm{ClO}^{-} \rightarrow \mathrm{Cl}^{-}\), chlorine (Cl) is already balanced.

Balance the oxygen atoms by adding water molecules (\(\mathrm{H}_2\mathrm{O}\)) to the appropriate side of each equation.**(a)** \(\mathrm{UO}_{2}^{+} \rightarrow \mathrm{U}^{4+}\): Add 2 \(\mathrm{H}_2\mathrm{O}\) to the right. **(b)** \(\mathrm{ClO}_{3}^{-} \rightarrow \mathrm{Cl}^{-}\): Add 3 \(\mathrm{H}_2\mathrm{O}\) to the right.**(c)** No action needed, as there are no O atoms in reactants or products.**(d)** \(\mathrm{ClO}^{-} \rightarrow \mathrm{Cl}^{-}\): Add 1 \(\mathrm{H}_2\mathrm{O}\) to the left.

Balance the hydrogen atoms by adding \(\mathrm{H}^{+}\) ions. In basic solutions, subsequently neutralize the \(\mathrm{H}^{+}\) by adding \(\mathrm{OH}^{-}\) ions to both sides of the equation.**(a)** \(\mathrm{UO}_{2}^{+} \rightarrow \mathrm{U}^{4+} + 2\mathrm{H}_2\mathrm{O}\): Add 4 \(\mathrm{H}^{+}\) to the left.**(b)** \(\mathrm{ClO}_{3}^{-} \rightarrow \mathrm{Cl}^{-} + 3\mathrm{H}_2\mathrm{O}\): Add 6 \(\mathrm{H}^{+}\) to the left.**(c)** \(\mathrm{N}_{2}\mathrm{H}_{4} \rightarrow \mathrm{N}_{2}\): Add 4 \(\mathrm{OH}^{-}\) to the right to balance the 4 \(\mathrm{H}\), rearrange for basic: \(\mathrm{N}_{2}\mathrm{H}_{4} + 4\mathrm{OH}^{-} \rightarrow \mathrm{N}_{2} + 4\mathrm{H}_2\mathrm{O}\).**(d)** \(\mathrm{ClO}^{-} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{Cl}^{-}\): Add 2 \(\mathrm{H}^{+}\) to the right, then neutralize by adding 2 \(\mathrm{OH}^{-}\) to both sides: \(\mathrm{ClO}^{-} + \mathrm{H}_2\mathrm{O} + 2\mathrm{OH}^{-} \rightarrow \mathrm{Cl}^{-} + 2\mathrm{OH}^{-}\).

Balance the charge by adding electrons (\(\mathrm{e}^{-}\)) to the side with the greater positive charge.**(a)** \(\mathrm{UO}_{2}^{+} + 4\mathrm{H}^{+} \rightarrow \mathrm{U}^{4+} + 2\mathrm{H}_2\mathrm{O}\): Add 3 electrons to the left: \(\mathrm{UO}_{2}^{+} + 4\mathrm{H}^{+} + 3\mathrm{e}^{-} \rightarrow \mathrm{U}^{4+} + 2\mathrm{H}_2\mathrm{O}\).**(b)** \(\mathrm{ClO}_{3}^{-} + 6\mathrm{H}^{+} \rightarrow \mathrm{Cl}^{-} + 3\mathrm{H}_2\mathrm{O}\): Add 6 electrons to the right: \(\mathrm{ClO}_{3}^{-} + 6\mathrm{H}^{+} + 6\mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-} + 3\mathrm{H}_2\mathrm{O}\).**(c)** \(\mathrm{N}_2\mathrm{H}_{4} + 4\mathrm{OH}^{-} \rightarrow \mathrm{N}_2 + 4\mathrm{H}_2\mathrm{O}\): Add 4 electrons to the right: \(\mathrm{N}_2\mathrm{H}_4 + 4\mathrm{OH}^{-} \rightarrow \mathrm{N}_2 + 4\mathrm{H}_2\mathrm{O} + 4\mathrm{e}^{-}\).**(d)** \(\mathrm{ClO}^{-} + 2\mathrm{OH}^{-} \rightarrow \mathrm{Cl}^{-} + \mathrm{H}_2\mathrm{O} + 2\mathrm{OH}^{-}\): Add 2 electrons to the right: \(\mathrm{ClO}^{-} + \mathrm{H}_2\mathrm{O} + 2\mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-} + 2\mathrm{OH}^{-}\).

Ensure that the number of atoms and the charges on both sides of each half-reaction are equal.**(a)** \(\mathrm{UO}_{2}^{+} + 4\mathrm{H}^{+} + 3\mathrm{e}^{-} \rightarrow \mathrm{U}^{4+} + 2\mathrm{H}_2\mathrm{O}\): 1 U, 2 O, 4 H, total charge = +2 both sides.**(b)** \(\mathrm{ClO}_{3}^{-} + 6\mathrm{H}^{+} + 6\mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-} + 3\mathrm{H}_2\mathrm{O}\): 1 Cl, 3 O, 6 H, total charge = 0 both sides.**(c)** \(\mathrm{N}_2\mathrm{H}_4 + 4\mathrm{OH}^{-} \rightarrow \mathrm{N}_2 + 4\mathrm{H}_2\mathrm{O} + 4\mathrm{e}^{-}\): 2 N, 8 H, 4 O, total charge = -4 both sides.**(d)** \(\mathrm{ClO}^{-} + \mathrm{H}_2\mathrm{O} + 2\mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-} + 2\mathrm{OH}^{-}\): 1 Cl, 1 O, 2 H, total charge = -3 both sides.