Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxidation or a reduction. (a) \(\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Sn}^{4+}(a q)\) (acidic solution) (b) \(\mathrm{TiO}_{2}(s) \longrightarrow \mathrm{Ti}^{2+}(a q)\) (acidic solution) (c) \(\mathrm{ClO}_{3}^{-}(a q) \longrightarrow \mathrm{Cl}^{-}(a q)\) (acidic solution) (d) \(\mathrm{N}_{2}(g) \longrightarrow \mathrm{NH}_{4}^{+}(a q)\) (acidic solution) (e) \(\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{O}_{2}(g)\) (basic solution) (f \(\mathrm{SO}_{3}^{2-}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)\) (basic solution) (g) \(\mathrm{N}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g)\) (basic solution)

We can see that Sn is going from an oxidation number of +2 to +4.

Since the oxidation number is increasing, this is an oxidation half-reaction.

There are no other atoms to balance in this half-reaction.

There are no oxygen atoms in this half-reaction.

There are no hydrogen atoms in this half-reaction.

The charge on the left side is +2, and the charge on the right side is +4. To balance the charges, we must add 2 electrons (e-) to the right side: \[\mathrm{Sn}^{2+} \longrightarrow \mathrm{Sn}^{4+} + 2\mathrm{e}^-\]

Both the atoms and charge are balanced in the half-reaction: \[\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Sn}^{4+}(a q) + 2\mathrm{e}^-\] (Oxidation) (b) \(\mathrm{TiO}_{2}(s) \longrightarrow \mathrm{Ti}^{2+}(a q)\) (acidic solution)

In TiO2, Ti has an oxidation number of +4. In Ti^{2+}, the oxidation number is +2.

Since the oxidation number is decreasing, this is a reduction half-reaction.

We have 1 Ti atom on both sides.

There are 2 oxygen atoms on the left side, so we add 2 H2O molecules to the right side: \[\mathrm{TiO}_{2}(s) \longrightarrow \mathrm{Ti}^{2+}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l)\]

There are 4 hydrogen atoms on the right side. We add 4 H+ ions to the left side: \[\mathrm{TiO}_{2}(s) + 4\mathrm{H}^+ \longrightarrow \mathrm{Ti}^{2+}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l)\]

The charge on the left side is +4, and the charge on the right side is +2. To balance the charges, we must add 2 electrons (e-) to the left side: \[2\mathrm{e}^- + \mathrm{TiO}_{2}(s) + 4\mathrm{H}^+ \longrightarrow \mathrm{Ti}^{2+}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l)\]

The atoms and charge are balanced in the half-reaction: \[2\mathrm{e}^- + \mathrm{TiO}_{2}(s) + 4\mathrm{H}^+ \longrightarrow \mathrm{Ti}^{2+}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l)\] (Reduction) Complete the remaining half-reactions (c) through (g) following the same steps. _Result_ (c) \[\mathrm{ClO}_{3}^{-}(a q) + 6\mathrm{e}^- + 6\mathrm{H}^{+} \longrightarrow \mathrm{Cl}^{-}(a q) + 3\mathrm{H}_{2}\mathrm{O}(l)\] (Reduction) (d) \[3\mathrm{e}^- + \mathrm{N}_{2}(g) + 6\mathrm{H}^{+} \longrightarrow 2\mathrm{NH}_{4}^{+}(a q)\] (Reduction) (e) \[2\mathrm{OH}^{-}(aq) \longrightarrow \mathrm{O}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(l) + 4\mathrm{e}^-\] (Oxidation) (f) \[\mathrm{SO}_{3}^{2-}(a q) + \mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{SO}_{4}^{2-}(a q) + 2\mathrm{H}^{+} + 2\mathrm{e}^-\] (Oxidation) (g) \[6\mathrm{e}^- + \mathrm{N}_{2}(g) + 6\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow 2\mathrm{NH}_{3}(g) + 12\mathrm{OH}^{-}(a q)\] (Reduction)