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

Identify the half-reaction type

The half-reaction is a reduction since Mo is gaining electrons (going from +3 to 0).

02

Balance Mo

Mo is already balanced.

03

Balance O using H2O

There are no oxygen atoms in this half-reaction, so nothing needs to be done.

04

Balance H using H+ (acidic solution)

There are no hydrogen atoms in this half-reaction, so nothing needs to be done.

05

Balance charge using electrons

Add 3 electrons to the left side to balance the charge: \(3e^- + Mo^{3+}(aq) \longrightarrow Mo(s)\). (b) \(\mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)\) (acidic solution)

06

Identify the half-reaction type

The half-reaction is an oxidation since sulfur is losing electrons (going from +4 to +6).

07

Balance S

S is already balanced.

08

Balance O using H2O

Add one H2O molecule to the left side: \(H_2SO_3(aq) + H_2O(l) \longrightarrow SO_4^{2-}(aq)\).

09

Balance H using H+ (acidic solution)

Add 4 H+ ions to the right side: \(H_2SO_3(aq) + H_2O(l) \longrightarrow SO_4^{2-}(aq) + 4H^+(aq)\).

10

Balance charge using electrons

Add 2 electrons to the right side to balance the charge: \(H_2SO_3(aq) + H_2O(l) \longrightarrow SO_4^{2-}(aq) + 4H^+(aq) + 2e^-\). (c) \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)\) (acidic solution)

11

Identify the half-reaction type

The half-reaction is a reduction since nitrogen is gaining electrons (going from +5 to +2).

12

Balance N

N is already balanced.

13

Balance O using H2O

Add 2 H2O molecules to the right side: \(NO_3^-(aq) \longrightarrow NO(g) + 2H_2O(l)\).

14

Balance H using H+ (acidic solution)

Add 4 H+ ions to the left side: \(NO_3^-(aq) + 4H^+(aq) \longrightarrow NO(g) + 2H_2O(l)\).

15

Balance charge using electrons

Add 3 electrons to the left side to balance the charge: \(3e^- + NO_3^-(aq) + 4H^+(aq) \longrightarrow NO(g) + 2H_2O(l)\). I will leave the rest for you to practice - follow the same steps as shown above to balance the remaining half-reactions.

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