91Ó°ÊÓ

The loss of electrons or an increase in the oxidation state of an atom, an ion, or certain atoms in a molecule is referred to as oxidation.

(a) ${\text{O}}_2\left(g\right)+NO\left(g\right)→{\text{NO}}_3^{-}\left(aq\right)$

1: Separate the half-reactions.

$$\begin{gathered} OHR:NO→N{O}_3^{-} \\ RHR:O_2→O^{2-} \end{gathered}$$

2: Balance electrons and non-O atoms.

$$\begin{gathered} OHR:NO→{\text{NO}}_3^{-}+3e^{-} \\ RHR:4e^{-}+O_2→2O^{2-} \end{gathered}$$

3: Balance reaction charges with $H^{+}$(acidic environment).

$$\begin{gathered} OHR:NO→N{O}_3^{-}+3e^{-}+4H^{+} \\ RHR:4e^{-}+O_2→2O^{2-} \end{gathered}$$

4: Balance O atoms by adding ${\text{H}}_2\text{O}$to the opposite side.

$$\begin{gathered} OHR:2H_2\text{O}+\text{NO}→{\text{NO}}_3^{-}+3e^{-}+4{\text{H}}^{+} \\ RHR:4e^{-}+O_2→2O^{2-} \end{gathered}$$

5: Multiply reactions by the least common factor and combine half-reactions.

$$\begin{gathered} 4x\text{OHR}:8{\text{H}}_2\text{O}+4\text{NO}→4{\text{NO}}_3^{-}+12e^{-}+16{\text{H}}^{+} \\ 3x\text{RHR}:12e^{-}+3{\text{O}}_2→6{\text{O}}^{2-} \\ \text{Reaction}:8{\text{H}}_2\text{O}+4\text{NO}+3{\text{O}}_2→4{\text{NO}}_3^{-}+16{\text{H}}^{+}+6{\text{O}}^{2-} \end{gathered}$$

6: Combine $2{\text{H}}^{+}and{\text{O}}^{2-}$to form ${\text{H}}_2\text{O}$and cancel common ions/ compounds on both sides accordingly.

$$\begin{gathered} \text{Reaction}:8{\text{H}}_2\text{O}+4\text{NO}+3{\text{O}}_2→4{\text{NO}}_3^{-}+4{\text{H}}^{+}+6{\text{H}}_2\text{O} \\ 2{\text{H}}_2\text{O}+4\text{NO}+3{\text{O}}_2→4{\text{NO}}_3^{-}+4{\text{H}}^{+} \\ \text{OA}:{\text{O}}_2, \\ \text{RA}:\text{NO} \end{gathered}$$

(b) ${\text{CrO}}_{4\left(aq\right)}^{2-}+C{u}_{\left(s\right)}→\text{Cr}(\text{OH}{)}_{3\left(s\right)}+\text{Cu}(\text{OH}{)}_{2\left(s\right)}$

1: Separate the half reactions.

$$\begin{gathered} OHR:Cu→Cu(OH{)}_2 \\ RHR:Cr{O}_4^{2-}→Cr(OH{)}_3 \end{gathered}$$

2: Balance electrons and non-O atoms.

$$\begin{gathered} OHR:Cu→Cu(OH{)}_2+2e^{-} \\ RHR:Cr{O}_4^{2-}+3e^{-}→Cr(OH{)}_3 \end{gathered}$$

3: Balance reaction charges with ${\text{OH}}^{-}$(basic environment).

$$\begin{gathered} OHR:2O{H}^{-}+Cu→Cu(OH{)}_2+2e^{-} \\ RHR:Cr{O}_4^{2-}+3e^{-}→Cr(OH{)}_3+5O{H}^{-} \end{gathered}$$

4: Balance O atoms by adding ${\text{H}}_2\text{O}$to the opposite side.

$$\begin{gathered} OHR:2O{H}^{-}+Cu→Cu(OH{)}_2+2e^{-} \\ RHR:4H_{2}O+Cr{O}_4^{2-}+3e^{-}→Cr(OH{)}_3+5O{H}^{-} \end{gathered}$$

5: Multiply reactions by least common factor and combine half reactions. Cancel common ions/ compounds on both sides accordingly.

$$\begin{gathered} 3xOHR:6O{H}^{-}+3Cu→3Cu(OH{)}_2+6e^{-} \\ 2x\text{RHR}:8H_2\text{O}+2{\text{CrO}}_4^{2-}+6e^{-}→2\text{Cr}(\text{OH}{)}_3+10{\text{OH}}^{-} \\ \text{Reaction}:8{\text{H}}_2\text{O}+2{\text{CrO}}_4^{2-}+3\text{Cu}→3\text{Cu}(\text{OH}{)}_2+2\text{Cr}(\text{OH}{)}_3+4O{\text{H}}^{-} \\ \text{OA}:{\text{CrO}}_4^{2-},\text{RA}:\text{Cu} \end{gathered}$$

(c) ${\text{AsO}}_{4\left(aq\right)}^{3-}+{\text{NO}}_{2\left(aq\right)}^{-}→{\text{NO}}_{3\left(aq\right)}^{-}+{\text{AsO}}_{2\left(aq\right)}^{-}$

1: Separate the half reactions.

$$\begin{gathered} OHR:{\text{NO}}_2^{-}→{\text{NO}}_3^{-} \\ \text{RHR}:{\text{AsO}}_4^{3-}→{\text{AsO}}_2^{-} \end{gathered}$$

2: Balance electrons and non-O atoms.

$$\begin{gathered} OHR:{\text{NO}}_2^{-}→{\text{NO}}_3^{-}+2e^{-} \\ RHR:2e^{-}+{\text{AsO}}_4^{3-}→{\text{AsO}}_2^{-} \end{gathered}$$

3: Balance reaction charges with$O{H}^{-}$ (basic environment).

$$\begin{gathered} \text{OHR}:2{\text{OH}}^{-}+{\text{NO}}_2^{-}→{\text{NO}}_3^{-}+2e^{-} \\ \text{RHR}:2e^{-}+{\text{AsO}}_4^{3-}→{\text{AsO}}_2^{-}+4{\text{OH}}^{-} \end{gathered}$$

4: Balance O atoms by adding${\text{H}}_2\text{O}$ to the opposite side.

$$\begin{gathered} \text{OHR}:2{\text{OH}}^{-}+{\text{NO}}_2^{-}→{\text{NO}}_3^{-}+2e^{-}+{\text{H}}_2\text{O} \\ \text{RHR}:2e^{-}+{\text{AsO}}_4^{3-}+2{\text{H}}_2\text{O}→{\text{AsO}}_2^{-}+4{\text{OH}}^{-} \end{gathered}$$

5: Combine half reactions and cancel common ions/ compounds on both sides accordingly.

$$\begin{gathered} \text{OHR}:2{\text{OH}}^{-}+{\text{NO}}_2^{-}→{\text{NO}}_3^{-}+2e^{-}+{\text{H}}_2\text{O} \\ RHR:2e^{-}+{\text{AsO}}_4^{3-}+2{\text{H}}_2\text{O}→{\text{AsO}}_2^{-}+4O{\text{H}}^{-} \\ \text{Reaction}:{\text{NO}}_2^{-}+{\text{AsO}}_4^{3-}+{\text{H}}_2\text{O}→{\text{NO}}_3^{-}+{\text{AsO}}_2^{-}+2{\text{OH}}^{-} \\ \text{OA}:{\text{NO}}_2^{-},\text{RA}:{\text{AsO}}_4^{3-} \end{gathered}$$