Q21.16P - Chemistry: Molecular Nature Of Matter And Change

Question

 

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

(a) $\mathbf{ }\mathit{S}\mathit{b}\left(s\right)\mathbf{ }\mathbf{+}\mathit{N}{\mathit{O}}_{\mathbf{3}}^{\mathbf{-}}\left(aq\right)\mathbf{\to }\mathit{S}{\mathit{b}}_{\mathbf{4}}{\mathit{O}}_{\mathbf{6}}\left(s\right)\mathbf{+}\mathbf{ }\mathit{N}\mathit{O}\left(g\right)\left[acidic\right]$

(b) $\mathit{M}{\mathit{n}}^{\mathbf{2}\mathbf{+}}\left(aq\right)\mathbf{+}\mathbf{ }\mathit{B}\mathit{i}{\mathit{O}}_{\mathbf{3}}^{\mathbf{-}}\left(aq\right)\mathbf{\to }\mathit{M}\mathit{n}{\mathit{O}}_{\mathbf{4}}^{\mathbf{-}}\left(aq\right)\mathbf{+}\mathbf{ }\mathit{B}{\mathit{i}}^{\mathbf{3}\mathbf{+}}\left(aq\right)\left[acidic\right]$

(c) $\mathit{F}\mathit{e}{\left(OH\right)}_{\mathbf{2}}\left(s\right)\mathbf{ }\mathbf{+}\mathit{P}\mathit{b}{{\left(OH\right)}_{\mathbf{3}}}^{\mathbf{-}}\left(aq\right)\mathbf{\to }\mathit{F}\mathit{e}{\left(OH\right)}_{\mathbf{3}}\left(s\right)\mathbf{+}\mathbf{ }\mathit{P}\mathit{b}\left(s\right)\left[basic\right]$

Step-by-Step Solution

Answer

$$\begin{gathered} \left(a\right)4\text{Sb}\left(\text{s}\right)+4{\text{NO}}_3^{-}\left(\text{aq}\right)+4{\text{H}}^{+}\to {\text{Sb}}_4{O}_6\left(s\right)+4\text{NO}\left(g\right)+2{\text{H}}_{2}O \\ \left(b\right)2{\text{Mn}}^{2+}\left(aq\right)+5{\text{BiO}}_3^{-}\left(\text{aq}\right)+14{\text{H}}^{+}\to 2{\text{MnO}}_4^{-}\left(\text{aq}\right)+5{\text{Bi}}^{3+}\left(\text{aq}\right)+7{\text{H}}_2\text{O} \\ \left(c\right)2\text{Fe}(\text{OH}{)}_2( \text{s})+\text{Pb}\left(\text{OH}{)}_3^{-}\right(\text{aq})\to 2\text{Fe}(\text{OH}{)}_3( \text{s})+\text{Pb}(\text{s})+{\text{OH}}^{-} \end{gathered}$$

Definition of Redox Reaction

The oxidation states of atoms are changed in a redox process. Redox reactions are defined by the real or formal movement of electrons between chemical species, with one species (the reducing agent) oxidizing (losing electrons) and another species (the oxidizing agent) reducing (gains electrons).

Balancing of the given reaction in part a.

(a)$\text{Sb}\left(\text{s}\right)+{\text{NO}}_3^{-}\left(\text{aq}\right)\to {\text{Sb}}_4{O}_6( \text{s})+\text{NO}\left(\text{g}\right)\left[acidic\right]$

To balance the equation, we will follow the following steps:

1: Separate the half-reactions.

$$\begin{gathered} {\text{NO}}_3^{-}\left(\text{aq}\right)\to \text{NO}\left(\text{g}\right) \\ \text{Sb}\left(\text{s}\right)\to {\text{Sb}}_4{O}_6( \text{s}) \end{gathered}$$

2: Balance all the elements other than Oxygen and Hydrogen.

$$\begin{gathered} {\text{NO}}_3^{-}\left(\text{aq}\right)\to \text{NO}\left(\text{g}\right) \\ 4\text{Sb}\left(\text{s}\right)\to {\text{Sb}}_4{O}_6( \text{s}) \end{gathered}$$

3: Now add ${\text{H}}_{2}O$ molecules to balance oxygen atoms.

$$\begin{gathered} {\text{NO}}_3^{-}\left(\text{aq}\right)\to \text{NO}\left(\text{g}\right)+2{\text{H}}_{2}O \\ 4\text{Sb}\left(\text{s}\right)+6{\text{H}}_2\text{O}\to {\text{Sb}}_4{O}_6( \text{s}) \end{gathered}$$

Balancing of the given reaction in part a.

4: To balance hydrogen atoms we add protons, $H^{+}$.

$$\begin{gathered} {\text{NO}}_3^{-}\left(aq\right)+4{\text{H}}^{+}\to \text{NO}\left(\text{g}\right)+2{\text{H}}_2\text{O} \\ 4\text{Sb}\left(\text{s}\right)+6{\text{H}}_2\text{O}\to {\text{Sb}}_4{O}_6\left(s\right)+12{\text{H}}^{+} \end{gathered}$$

5: Now balance the charge with electrons, $e^{-}$

$$\begin{gathered} {\text{NO}}_3^{-}\left(aq\right)+4{\text{H}}^{+}+3{\text{e}}^{-}\to \text{NO}\left(\text{g}\right)+2{\text{H}}_2\text{O} \\ 4\text{Sb}\left(\text{s}\right)+6{\text{H}}_2\text{O}\to {\text{Sb}}_4{O}_6\left(s\right)+12H^{+}+12{\text{e}}^{-} \end{gathered}$$

6: Scale the reactions to make the electron count equal.

$$\begin{gathered} 4{\text{NO}}_3^{-}\left(aq\right)+16H^{+}+12e^{-}\to 4\text{NO}\left(\text{g}\right)+8H_{2}O \\ 4Sb\left(s\right)+6H_{2}O\to S{b}_4{O}_6\left(s\right)+12H^{+}+12e^{-} \end{gathered}$$

Balancing of the given reaction in part a.

7: Add the reactions.

$4{\text{NO}}_3^{-}\left(\text{aq}\right)+16{\text{H}}^{+}+12{\text{e}}^{-}+4\text{Sb}\left(\text{s}\right)+6{\text{H}}_2\text{O}\to 4\text{NO}\left(\text{g}\right)+8{\text{H}}_2\text{O}+{\text{Sb}}_4{O}_6\left(s\right)+12H^{+}+12{\text{e}}^{-}$

8: Cancel out common terms.

$4{\text{NO}}_3^{-}\left(\text{aq}\right)+4{\text{H}}^{+}+4\text{Sb}\left(\text{s}\right)\to 4\text{NO}\left(\text{g}\right)+2{\text{H}}_2\text{O}+{\text{Sb}}_2{O}_6( \text{s})$

Hence, we get the following balanced reaction:

$4\text{Sb}\left(\text{s}\right)+4{\text{NO}}_3^{-}\left(\text{aq}\right)+4{\text{H}}^{+}\to {\text{Sb}}_4{O}_6\left(s\right)+4\text{NO}\left(g\right)+2{\text{H}}_2\text{O}$

Oxidizing agent: ${\text{NO}}_3^{-}\left(aq\right)$

Reducing agent: $\text{Sb}\left(\text{s}\right)$

Balancing of the given reaction in part b.

(b) 

${\text{Mn}}^{2+}\left(\text{aq}\right)+{\text{BiO}}_3^{-}\left(\text{aq}\right)\to {\text{MnO}}_4^{-}\left(\text{aq}\right)+{\text{Bi}}^{3+}\left(\text{aq}\right)\left[acidic\right]$

To balance the equation, we will follow the following steps:

1: Separate the half-reactions.

$$\begin{gathered} {\text{BiO}}_3^{-}\left(aq\right)\to {\text{Bi}}^{3+}\left(aq\right) \\ {\text{Mn}}^{2+}\left(aq\right)\to {\text{MnO}}_4^{-}\left(aq\right) \end{gathered}$$

2: Balance all the elements other than Oxygen and Hydrogen.

$$\begin{gathered} {\text{BiO}}_3^{-}\left(aq\right)\to {\text{Bi}}^{3+}\left(aq\right) \\ {\text{Mn}}^{2+}\left(aq\right)\to {\text{MnO}}_4^{-}\left(aq\right) \end{gathered}$$

3: Now add ${\text{H}}_2\text{O}$ molecules to balance oxygen atoms.

$$\begin{gathered} {\text{BiO}}_3^{-}\left(aq\right)\to {\text{Bi}}^{3+}\left(\text{aq}\right)+3{\text{H}}_2\text{O} \\ {\text{Mn}}^{2+}\left(\text{aq}\right)+4{\text{H}}_2\text{O}\to {\text{MnO}}_4^{-}\left(aq\right) \end{gathered}$$

Balancing of the given reaction in part b.

4: To balance hydrogen atoms we add protons,

$$\begin{gathered} {\text{BiO}}_3^{-}\left(\text{aq}\right)+6{\text{H}}^{+}\to {\text{Bi}}^{3+}\left(\text{aq}\right)+3{\text{H}}_2\text{O} \\ {\text{Mn}}^{2+}\left(\text{aq}\right)+4{\text{H}}_2\text{O}\to {\text{MnO}}_4^{-}\left(\text{aq}\right)+8{\text{H}}^{+} \end{gathered}$$

5: Now balance the charge with electrons, $e^{-}$.

$$\begin{gathered} {\text{BiO}}_3^{-}\left(\text{aq}\right)+6{\text{H}}^{+}+2{\text{e}}^{-}\to {\text{Bi}}^{3+}\left(\text{aq}\right)+3{\text{H}}_2\text{O} \\ {\text{Mn}}^{2+}\left(\text{aq}\right)+4{\text{H}}_2\text{O}\to {\text{MnO}}_4^{-}\left(\text{aq}\right)+8{\text{H}}^{+}+5{\text{e}}^{-} \end{gathered}$$

6: Scale the reactions to make the electron count equal.

$$\begin{gathered} 5{\text{BiO}}_3^{-}\left(\text{aq}\right)+30H^{+}+10e^{-}\to 5{\text{Bi}}^{3+}\left(\text{aq}\right)+15H_{2}O \\ 2M{n}^{2+}\left(aq\right)+8H_{2}O\to 2{\text{MnO}}_4^{-}\left(\text{aq}\right)+16H^{+}+10e^{-} \end{gathered}$$.

Balancing of the given reaction in part b.

7: Add the reactions.

$5{\text{BiO}}_3^{-}\left(\text{aq}\right)+30{\text{H}}^{+}+10{\text{e}}^{-}+2{\text{Mn}}^{2+}\left(\text{aq}\right)+8{\text{H}}_2\text{O}\to 5{\text{Bi}}^{3+}\left(\text{aq}\right)+15{\text{H}}_2\text{O}+2{\text{MnO}}_4^{-}\left(\text{aq}\right)+16{\text{H}}^{+}+10{\text{e}}^{-}$

8: Cancel out common terms.

$5{\text{BiO}}_3^{-}\left(\text{aq}\right)+14{\text{H}}^{+}+2{\text{Mn}}^{2+}\left(\text{aq}\right)\to 5{\text{Bi}}^{3+}\left(\text{aq}\right)+7{\text{H}}_2\text{O}+2{\text{MnO}}_4^{-}\left(\text{aq}\right)$

Hence, we get the following balanced reaction:

$2{\text{Mn}}^{2+}\left(\text{aq}\right)+5{\text{BiO}}_3^{-}\left(\text{aq}\right)+14{\text{H}}^{+}\to 2{\text{MnO}}_4^{-}\left(\text{aq}\right)+5{\text{Bi}}^{3+}\left(\text{aq}\right)+7{\text{H}}_2\text{O}$

Oxidizing agent: ${\text{BiO}}_3^{-}\left(\text{aq}\right)$

Reducing agent: ${\text{Mn}}^{2+}\left(aq\right)$

Balancing of the given reaction in part c.

(c) $\text{Fe}(\text{OH}{)}_2( \text{s})+\text{Pb}\left(\text{OH}{)}_3^{-}\right(\text{aq})\to \text{Fe}(\text{OH}{)}_3( \text{s})+\text{Pb}(\text{s}\left)\right[\text{basic}]$

To balance the equation, we will follow following steps:

1: Separate the half-reactions.

$$\begin{gathered} \text{Pb}\left(\text{OH}{)}_3^{-}\right(aq)\to \text{Pb} \\ \text{Fe}(\text{OH}{)}_2\to \text{Fe}(\text{OH}{)}_3( \text{s}) \end{gathered}$$

2: Balance all the elements other than Oxygen and Hydrogen.

$$\begin{gathered} \text{Pb}\left(\text{OH}{)}_3^{-}\right(aq)\to \text{Pb}(s) \\ \text{Fe}(\text{OH}{)}_2\to \text{Fe}(\text{OH}{)}_3( \text{s}) \end{gathered}$$

3: Now add ${\text{H}}_{2}O$ molecules to balance oxygen atoms.

$$\begin{gathered} \text{Pb}\left(\text{OH}{)}_3^{-}\right(aq)\to \text{Pb}(\text{s})+3{\text{H}}_2\text{O} \\ \text{Fe}(\text{OH}{)}_2+{\text{H}}_2\text{O}\to \text{Fe}(\text{OH}{)}_3( \text{s}) \end{gathered}$$

Balancing of the given reaction in part c.

4: To balance hydrogen atoms we add protons, $H^{+}$.

$$\begin{gathered} \text{Pb}\left(\text{OH}{)}_3^{-}\right(\text{aq})+3{\text{H}}^{+}\to \text{Pb}(\text{s})+3{\text{H}}_2\text{O} \\ \text{Fe}(\text{OH}{)}_2+{\text{H}}_2\text{O}\to \text{Fe}(\text{OH}{)}_3( \text{s})+{\text{H}}^{+} \end{gathered}$$

5: Now balance the charge with electrons, $e^{-}$.

$$\begin{gathered} \text{Pb}\left(\text{OH}{)}_3^{-}\right(\text{aq})+3{\text{H}}^{+}+2{\text{e}}^{-}\to \text{Pb}(\text{s})+3{\text{H}}_2\text{O} \\ \text{Fe}(\text{OH}{)}_2+{\text{H}}_2\text{O}\to \text{Fe}(\text{OH}{)}_3( \text{s})+{\text{H}}^{+}+{\text{e}}^{-} \end{gathered}$$

6: Scale the reactions to make the electron count equal.

$$\begin{gathered} \text{Pb}\left(\text{OH}{)}_3^{-}\right(\text{aq})+3{\text{H}}^{+}+2{\text{e}}^{-}\to \text{Pb}(\text{s})+3{\text{H}}_2\text{O} \\ 2\text{Fe}(\text{OH}{)}_2+2{\text{H}}_2\text{O}\to 2\text{Fe}(\text{OH}{)}_3( \text{s})+2{\text{H}}^{+}+2{\text{e}}^{-} \end{gathered}$$

7: Add the reactions.

$\text{Pb}\left(\text{OH}{)}_3^{-}\right(\text{aq})+3{\text{H}}^{+}+2{\text{e}}^{-}+2\text{Fe}(\text{OH}{)}_2+2{\text{H}}_2\text{O}\to \text{Pb}(\text{s})+3{\text{H}}_2\text{O}+2\text{Fe}(\text{OH}{)}_3( \text{s})+2{\text{H}}^{+}+2{\text{e}}^{-}$

Balancing of the given reaction in part c.

8: Cancel out common terms.

$\text{Pb}\left(\text{OH}{)}_3^{-}\right(\text{aq})+{\text{H}}^{+}+2\text{Fe}(\text{OH}{)}_2\to \text{Pb}(\text{s})+{\text{H}}_2\text{O}+2\text{Fe}(\text{OH}{)}_3( \text{s})$

9: Since, the reaction is in basic medium we add $O{H}^{-}$ to balance the $H^{+}$ ions.

$\text{Pb}\left(\text{OH}{)}_3^{-}\right(\text{aq})+{\text{H}}^{+}+{\text{OH}}^{-}+2\text{Fe}(\text{OH}{)}_2\to \text{Pb}(\text{s})+{\text{H}}_2\text{O}+2\text{Fe}(\text{OH}{)}_3( \text{s})$

10: Combine $O{H}^{-}$ ions and $H^{+}$ ions present on the same side to form water molecule.

$\text{Pb}\left(\text{OH}{)}_3^{-}\right(\text{aq})+{\text{H}}_2\text{O}+2\text{Fe}(\text{OH}{)}_2\to \text{Pb}(\text{s})+{\text{H}}_2\text{O}+2\text{Fe}(\text{OH}{)}_3( \text{s})+{\text{OH}}^{-}$

11: Cancel out common terms.

$\text{Pb}\left(\text{OH}{)}_3^{-}\right(\text{aq})+2\text{Fe}(\text{OH}{)}_2\to \text{Pb}(\text{s})+2\text{Fe}(\text{OH}{)}_3( \text{s})+{\text{OH}}^{-}$

Hence, we get the following balanced reaction:

$2\text{Fe}(\text{OH}{)}_2( \text{s})+\text{Pb}\left(\text{OH}{)}_3^{-}\right(\text{aq})\to 2\text{Fe}(\text{OH}{)}_3( \text{s})+\text{Pb}(\text{s})+{\text{OH}}^{-}$

Oxidizing agent: $\text{Pb}\left(\text{OH}{)}_3^{-}\right(aq)$.

Reducing agent: $\text{Fe}(\text{OH}{)}_2$.