Q21.13P - Chemistry: Molecular Nature Of Matter And Change

Question

Balance the following skeleton reactions and identify the oxidizing and reducing agents: 

(a) ${\mathit{O}}_{\mathbf{2}}\left(g\right)\mathbf{ }\mathbf{+}\mathit{N}\mathit{O}\left(g\right)\mathbf{\to }\mathit{N}{\mathit{O}}_{\mathbf{3}}\left(aq\right)\mathbf{ }\left[acidic\right]$

(b) $\mathit{C}\mathit{r}{\mathit{O}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\left(aq\right)\mathbf{ }\mathbf{+}\mathit{C}\mathit{u}\left(s\right)\mathbf{\to }\mathit{C}\mathit{r}{\left(OH\right)}_{\mathbf{3}}\left(s\right)\mathbf{+}\mathbf{ }\mathit{C}\mathit{u}{\left(OH\right)}_{\mathbf{2}}\left(s\right)\left[basic\right]$

(c) $\mathit{A}\mathit{s}{\mathit{O}}_{\mathbf{4}}^{\mathbf{3}\mathbf{-}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{ }\mathbf{+}\mathit{N}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{\to }\mathit{A}\mathit{s}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{ }\mathbf{+}\mathit{N}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{\left[}\mathbf{b}\mathbf{a}\mathbf{s}\mathbf{i}\mathbf{c}\mathbf{\right]}$

Step-by-Step Solution

Answer

$$\begin{gathered} \left(a\right)2{\text{H}}_2{\text{O}}_{\left(l\right)}+4{\text{NO}}_{\left(g\right)}+3{\text{O}}_{2\left(g\right)}\to 4{\text{NO}}_{3\left(aq\right)}^{-}+4{\text{H}}_{\left(aq\right)}^{+};OA:{\text{O}}_2,\text{RA}:\text{NO} \\ \left(b\right)8{\text{H}}_2{\text{O}}_{\left(l\right)}+2{\text{CrO}}_{4\left(aq\right)}^{2-}+3\text{Cu}{u}_{\left(s\right)}\to 3\text{Cu}(\text{OH}{)}_{2\left(s\right)}+2\text{Cr}(\text{OH}{)}_{3\left(s\right)}+4{\text{OH}}_{\left(aq\right)}^{-};\text{OA}:{\text{CrOO}}_4^{2-},\text{RA}:\text{Cu} \\ \left(c\right){\text{NO}}_{2\left(aq\right)}^{-}+{\text{AsO}}_{4\left(aq\right)}^{3-}+{\text{H}}_2{\text{O}}_{\left(l\right)}\to {\text{NO}}_{3\left(aq\right)}^{-}+{\text{AsO}}_{2\left(aq\right)}^{-}+2{\text{OH}}_{\left(aq\right)}^{-};\text{OA}:{\text{NO}}_2^{-},\text{RA}:{\text{AsO}}_4^{3-} \end{gathered}$$

Definition of Oxidation

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.

Balancing of the given reaction in part a.

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

1: Separate the half-reactions.

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

2: Balance electrons and non-O atoms.

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

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

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

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

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

Balancing of the given reaction in part a.

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

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

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.

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

Balancing of the given reaction in part b.

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

1: Separate the half reactions.

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

2: Balance electrons and non-O atoms.

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

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

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

Balancing of the given reaction in part b.

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

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

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

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

Balancing of the given reaction in part c.

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

1: Separate the half reactions.

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

2: Balance electrons and non-O atoms.

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

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

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

Balancing of the given reaction in part c.

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

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

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

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