Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxidation or a reduction. (a) \(\mathrm{Mo}^{3+}(a q) \longrightarrow \mathrm{Mo}(s)\) (acidic solution) (b) \(\mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)\) (acidic solution) (c) \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)\) (acidic solution) (d) \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (acidic solution) (e) \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (basic solution) (f) \(\mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{2}(s)\) (basic solution) (g) \(\mathrm{Cr}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q)\) (basic solution)

Step by step solution

01

Balance atoms other than H and O

In this case, the Mo atoms are already balanced.

02

Balance the O atoms

There are no O atoms in this half-reaction, so this step can be skipped.

03

Balance the H atoms

There are no H atoms in this half-reaction, so this step can be skipped.

04

Balance the charge

Add 3 electrons to the right side of the equation to balance the charge: \[\mathrm{Mo}^{3+}(a q) + 3 \mathrm{e}^{-} \longrightarrow \mathrm{Mo}(s)\]

05

Determine if it is an oxidation or reduction

Since the electrons are being gained on the right side, this is a reduction half-reaction. (b) \(\mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)\) (acidic solution)

06

Balance atoms other than H and O

The S atoms are already balanced.

07

Balance the O atoms

Add 2 \(\mathrm{H}_{2}\mathrm{O}\) molecules to the left side to balance the O atoms: \[\mathrm{H}_{2} \mathrm{SO}_{3}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)\]

08

Balance the H atoms

Add 4 \(\mathrm{H}^{+}\) ions to the right side to balance the H atoms: \[\mathrm{H}_{2} \mathrm{SO}_{3}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)+ 4\mathrm{H}^{+}(a q)\]

09

Balance the charge

Add 2 electrons to the right side of the equation to balance the charge: \[\mathrm{H}_{2} \mathrm{SO}_{3}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l) + 2 \mathrm{e}^{-} \longrightarrow \mathrm{SO}_{4}^{2-}(a q)+ 4\mathrm{H}^{+}(a q)\]

10

Determine if it is an oxidation or reduction

Since the electrons are being gained on the right side, this is a reduction half-reaction. (c) \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)\) (acidic solution)

11

Balance atoms other than H and O

The N atoms are already balanced.

12

Balance the O atoms

Add 2 \(\mathrm{H}_{2}\mathrm{O}\) molecules to the right side to balance the O atoms: \[\mathrm{NO}_{3}^{-} (a q) \longrightarrow \mathrm{NO}(g) + 2\mathrm{H}_{2}\mathrm{O}(l)\]

13

Balance the H atoms

Add 4 \(\mathrm{H}^{+}\) ions to the left side to balance the H atoms: \[4\mathrm{H}^{+}(a q)+ \mathrm{NO}_{3}^{-} (a q) \longrightarrow \mathrm{NO}(g) + 2\mathrm{H}_{2}\mathrm{O}(l)\]

14

Balance the charge

Add 3 electrons to the left side of the equation to balance the charge: \[4\mathrm{H}^{+}(a q)+ \mathrm{NO}_{3}^{-} (a q) + 3\mathrm{e}^{-} \longrightarrow \mathrm{NO}(g) + 2\mathrm{H}_{2}\mathrm{O}(l)\]

15

Determine if it is an oxidation or reduction

Since the electrons are being gained on the right side, this is a reduction half-reaction. The solutions for the remaining half-reactions (d, e, f, and g) follow similar steps to the above examples. You can practice completing and balancing them to gain a better understanding. Keep in mind that basic solutions will require the addition of \(\mathrm{OH}^{-}\) ions to balance the H and O atoms.

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