Chapter 14, Problem 79QP

(a)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced and also, the oxidizing agent, reducing agent, the elements being oxidized and reduced are to be identified.

Explanation of Solution

The half-cell reactions for the given chemical equations can be obtained by writing the oxidation numbers of each species present in the given reaction.

$\begin{array}{l}+2-1+1-1+3-1-1+1\\ {\text{CoCl}}_2\left(s\right)+{\text{Na}}_{\text{2}}{\text{O}}_2\left(aq\right)\to \text{Co}{\left(\text{OH}\right)}_3\left(s\right)+{\text{Cl}}^{-}\left(aq\right)+{\text{Na}}^{+}\left(aq\right)\end{array}$

As the compound, ${\text{Na}}_{\text{2}}{\text{O}}_2$ undergoes removal of oxygen atoms and ${\text{CoCl}}_2$ undergoes the addition of oxygen atoms, ${\text{Na}}_{\text{2}}{\text{O}}_2$ acts as an oxidizing agent and ${\text{CoCl}}_2$ acts as a reducing agent in the given reaction.

On the other hand, the oxidation number of oxygen atom changes from $-1$ to $-2$ and the oxidation number of cobalt atom changes from $+2$ to $+3$ . Therefore, oxygen is being reduced and cobalt is being oxidized.

On the basis of this, the oxidation (anode) and reduction half-cell reactions (cathode) can be written as,

$\begin{array}{c}\text{Anode}:{\text{CoCl}}_2\left(s\right)\to \text{Co}{\left(\text{OH}\right)}_3\left(s\right)+{\text{Cl}}^{-}\left(aq\right)\\ \text{Cathode:}{\text{Na}}_{\text{2}}{\text{O}}_2\left(aq\right)\to {\text{OH}}^{-}\left(aq\right)+{\text{Na}}^{+}\left(aq\right)\end{array}$

Now, balance the reaction occurring at anode by using the stepwise approach as,

$\begin{array}{c}{\text{CoCl}}_2\left(s\right)\to \text{Co}{\left(\text{OH}\right)}_3\left(s\right)+2{\text{Cl}}^{-}\left(aq\right)\\ {\text{CoCl}}_2\left(s\right)\to \text{Co}{\left(\text{OH}\right)}_3\left(s\right)+2{\text{Cl}}^{-}\left(aq\right)+{\text{e}}^{-}\\ {\text{3OH}}^{-}\left(aq\right)+{\text{CoCl}}_2\left(s\right)\to \text{Co}{\left(\text{OH}\right)}_3\left(s\right)+2{\text{Cl}}^{-}\left(aq\right)+{\text{e}}^{-}\end{array}$

Now, balance the reaction occurring at cathode by using the stepwise approach as,

$\begin{array}{c}{\text{Na}}_{\text{2}}{\text{O}}_2\left(aq\right)\to 2{\text{OH}}^{-}\left(aq\right)+2{\text{Na}}^{+}\left(aq\right)\\ {\text{Na}}_{\text{2}}{\text{O}}_2\left(aq\right)+2{\text{e}}^{-}\to 2{\text{OH}}^{-}\left(aq\right)+2{\text{Na}}^{+}\left(aq\right)\\ {\text{Na}}_{\text{2}}{\text{O}}_2\left(aq\right)+2{\text{e}}^{-}\to 2{\text{OH}}^{-}\left(aq\right)+2{\text{Na}}^{+}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)\\ {\text{Na}}_{\text{2}}{\text{O}}_2\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)+2{\text{e}}^{-}\to 4{\text{OH}}^{-}\left(aq\right)+2{\text{Na}}^{+}\left(aq\right)\end{array}$

Multiply the reaction occurring at anode with $2$ and add it with the cathode-half reaction in order to obtain the overall reaction.

$\begin{array}{c}2\left({\text{3OH}}^{-}\left(aq\right)+{\text{CoCl}}_2\left(s\right)\to \text{Co}{\left(\text{OH}\right)}_3\left(s\right)+2{\text{Cl}}^{-}\left(aq\right)+{\text{e}}^{-}\right)\\ {\text{6OH}}^{-}\left(aq\right)+2{\text{CoCl}}_2\left(s\right)\to 2\text{Co}{\left(\text{OH}\right)}_3\left(s\right)+4{\text{Cl}}^{-}\left(aq\right)+\cancel{2{\text{e}}^{-}}\\ {\text{Na}}_{\text{2}}{\text{O}}_2\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)+\cancel{2{\text{e}}^{-}}\to 4{\text{OH}}^{-}\left(aq\right)+2{\text{Na}}^{+}\left(aq\right)\\ \underset{¯}{\overline{{\text{6OH}}^{-}\left(aq\right)+2{\text{CoCl}}_2\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)+{\text{Na}}_{\text{2}}{\text{O}}_2\left(aq\right)\to 2\text{Co}{\left(\text{OH}\right)}_3\left(s\right)+4{\text{Cl}}^{-}\left(aq\right)+4{\text{OH}}^{-}\left(aq\right)+2{\text{Na}}^{+}\left(aq\right)}}\end{array}$

The overall reaction can be obtained by balancing the hydroxide ions in the final equation as,

${\text{2OH}}^{-}\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)+2{\text{CoCl}}_2\left(s\right)+{\text{Na}}_{\text{2}}{\text{O}}_2\left(aq\right)\to 2\text{Co}{\left(\text{OH}\right)}_3\left(s\right)+4{\text{Cl}}^{-}\left(aq\right)+2{\text{Na}}^{+}\left(aq\right)$

(b)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced and also, the oxidizing agent, reducing agent, the elements being oxidized and reduced are to be identified.

Explanation of Solution

The half-cell reactions for the given chemical equations can be obtained by writing the oxidation numbers of each species present in the given reaction.

$\begin{array}{l}+3-2+1-1+5-2-1\\ {\text{Bi}}_2{\text{O}}_3\left(s\right)+{\text{ClO}}^{-}\left(aq\right)\to {\text{BiO}}_3^{-}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)\end{array}$

As the compound ${\text{ClO}}^{-}$ undergoes removal of oxygen atoms and ${\text{Bi}}_2{\text{O}}_3$ involves the gain of electrons, ${\text{ClO}}^{-}$ acts as an oxidizing agent and ${\text{Bi}}_2{\text{O}}_3$ acts as reducing agent in the given reaction. On the other hand, the oxidation number of chlorine atom changes from $+1$ to $-1$ and the oxidation number of bismuth atom changes from $+3$ to $+5$ , therefore, chlorine is being reduced and bismuth is being oxidized.

On the basis of this, the oxidation (anode) and reduction half-cell reactions (cathode) can be written as,

$\begin{array}{c}\text{Anode}:{\text{Bi}}_2{\text{O}}_3\left(s\right)\to {\text{BiO}}_3^{-}\left(aq\right)\\ \text{Cathode:}{\text{ClO}}^{-}\left(aq\right)\to {\text{Cl}}^{-}\left(aq\right)\end{array}$

Now, balance the reaction occurring at anode by using the stepwise approach as,

$\begin{array}{c}{\text{Bi}}_2{\text{O}}_3\left(s\right)\to 2{\text{BiO}}_3^{-}\left(aq\right)\\ {\text{Bi}}_2{\text{O}}_3\left(s\right)\to 2{\text{BiO}}_3^{-}\left(aq\right)+4{\text{e}}^{-}\\ {\text{6OH}}^{-}\left(aq\right)+{\text{Bi}}_2{\text{O}}_3\left(s\right)\to 2{\text{BiO}}_3^{-}\left(aq\right)+4{\text{e}}^{-}\\ {\text{6OH}}^{-}\left(aq\right)+{\text{Bi}}_2{\text{O}}_3\left(s\right)\to 2{\text{BiO}}_3^{-}\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)+4{\text{e}}^{-}\end{array}$

Now, balance the reaction occurring at cathode by using the stepwise approach as,

$\begin{array}{c}{\text{ClO}}^{-}\left(aq\right)+2{\text{e}}^{-}\to {\text{Cl}}^{-}\left(aq\right)\\ {\text{ClO}}^{-}\left(aq\right)+2{\text{e}}^{-}\to 2{\text{OH}}^{-}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)\\ {\text{ClO}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)+2{\text{e}}^{-}\to 2{\text{OH}}^{-}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)\end{array}$

Multiply the reaction occurring at cathode with $2$ and add it with the cathode-half reaction in order to obtain the overall reaction.

$\begin{array}{c}2\left({\text{ClO}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)+2{\text{e}}^{-}\to 2{\text{OH}}^{-}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)\right)\\ {\text{2ClO}}^{-}\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)+\cancel{4{\text{e}}^{-}}\to 4{\text{OH}}^{-}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)\\ {\text{6OH}}^{-}\left(aq\right)+{\text{Bi}}_2{\text{O}}_3\left(s\right)\to 2{\text{BiO}}_3^{-}\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)+\cancel{4{\text{e}}^{-}}\\ \underset{¯}{\overline{{\text{2ClO}}^{-}\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)+{\text{6OH}}^{-}\left(aq\right)+{\text{Bi}}_2{\text{O}}_3\left(s\right)\to 4{\text{OH}}^{-}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)+2{\text{BiO}}_3^{-}\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)}}\end{array}$

The overall reaction can be obtained by balancing the hydroxide ions and water molecules in the final equation as,

${\text{2ClO}}^{-}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)+{\text{Bi}}_2{\text{O}}_3\left(s\right)\to 2{\text{Cl}}^{-}\left(aq\right)+2{\text{BiO}}_3^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$