Chapter 4, Problem 4.124AP

(a)

Interpretation Introduction

Interpretation: The oxidation numbers to the elements are to be assigned and the redox reaction is to be balanced. The net ionic equation describing the reaction

For formation of chlorine is to be stated. The equation for the conversion of ${\text{Cl}}_{\text{2}}$ to $\text{HCl}$ and $\text{HOCl}$ is to be balanced.

Concept introduction: The reaction in which reduction and oxidation process occurs simultaneously is termed as redox reaction. Redox reaction involves all the chemical reactions in which atoms will change their oxidation states.

In oxidation there is loss of electrons occur which increases the oxidation state of an atom whereas in reduction process there is gaining of electrons occur which decreases the oxidation state of an atom.

Oxidation number is defined as the charge present on ions. When an atom forms a chemical bond with another atom, the number of electrons that an atom gain or loses is the oxidation number. It is also known as oxidation state.

To determine: The oxidation number of the elements in each compound of the given reaction and balanced redox reaction in acid solution.

Answer to Problem 4.124AP

Solution

The oxidation numbers of the elements in each compound have been rightfully stated.

The balanced redox reaction in acid solution is as follows:

$\begin{array}{l}4\text{NaCl}\left(aq\right)+2{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)+{\text{MnO}}_{\text{2}}\left(aq\right)\to \\ 2{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)+{\text{MnCl}}_{\text{2}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{Cl}}_{\text{2}}\left(g\right)\end{array}$

Explanation of Solution

Explanation

The given element is ${\text{MnO}}_{\text{2}}$ in the redox reaction.

The oxidation state of oxygen is usually $-2$ as it requires two more electrons in order to achieve stable electronic configuration.

The oxidation number of $\text{Mn}$ is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on MnO}}_2=\left[\begin{array}{c}\left(\text{Number of Mn}\text{atoms}\times \text{Oxidation number of Mn}\right)\\ +\left(\text{Number of O atoms}\times \text{Oxidation number of O}\right)\end{array}\right]$

Substitute the number of atoms and their oxidation number in the above formula.

$\begin{array}{c}{\text{Charge on MnO}}_2=\left[\left(1\times x\right)+\left(2\times \left(-2\right)\right)\right]\\ 0=x-4\\ x=+4\end{array}$

Hence, the oxidation number of $\text{Mn}$ in ${\text{MnO}}_{\text{2}}$ is $\underset{\_}{+4}$.

The oxidation numbers of the elements in ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ is as follows:

The given element is ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ in the redox reaction.

The oxidation state of oxygen is usually $-2$ as it requires two more electrons in order to achieve stable electronic configuration and ${\text{Na}}^{\text{+}}$ has oxidation state of $+1$ ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$.

The oxidation number of $\text{S}$ in ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on Na}}_2{\text{SO}}_{\text{4}}=\left[\begin{array}{c}\left(\text{Number of Na}\text{atoms}\times \text{oxidation number of Na}\right)\\ +\left(\text{Number of}\text{S}\text{atoms}\times \text{oxidation number of S}\right)\\ +\left(\text{Number of O atoms}\times \text{oxidation number of O}\right)\end{array}\right]$

Substitute the number of atoms and their oxidation number in the above formula.

$\begin{array}{c}{\text{Charge on Na}}_{\text{2}}{\text{SO}}_{\text{4}}=\left[\left(2\times \left(+1\right)\right)+\left(x\times 1\right)+\left(4\times \left(-2\right)\right)\right]\\ 0=\left[\left(2\right)+\left(x\right)+\left(-8\right)\right]\\ x=2-8\\ x=+6\end{array}$

Hence, the oxidation number of $\text{S}$ in ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\underset{\_}{+6}$.

The oxidation numbers of the elements in ${\text{MnCl}}_{\text{2}}$ is as follows

The given element is ${\text{MnCl}}_{\text{2}}$ in the redox reaction. The oxidation state of chlorine is usually $-1$ as it requires one more electrons in order to achieve stable electronic configuration.

The oxidation number of $\text{Mn}$ is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on MnCl}}_2=\left[\begin{array}{c}\left(\text{Number of Mn}\text{atoms}\times \text{oxidation number of Mn}\right)\\ +\left(\text{Number of Cl atoms}\times \text{oxidation number of Cl}\right)\end{array}\right]$

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of manganese.

$\begin{array}{c}{\text{Charge on MnCl}}_2=\left[\left(1\times x\right)+\left(2\times \left(-1\right)\right)\right]\\ 0=x-2\\ x=+2\end{array}$

Hence, the oxidation number of $\text{Mn}$ in ${\text{MnCl}}_{\text{2}}$ is $\underset{\_}{+2}$.

The given unbalanced redox reaction is,

$\text{NaCl}\left(aq\right)+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)+{\text{MnO}}_{\text{2}}\left(aq\right){\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)+{\text{MnCl}}_{\text{2}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{Cl}}_{\text{2}}\left(g\right)$

To balance the above redox reaction the half reactions must be shown.

The reduction half reaction in which $\text{Mn}$ is reduced is,

${\text{MnO}}_{\text{2}}\left(aq\right)+2{\text{e}}^{-}\to {\text{MnCl}}_{\text{2}}\left(g\right)$

The oxidation half reaction is in which $\text{Cl}$ is oxidized is,

$\text{NaCl}\left(aq\right)+1{\text{e}}^{-}\to {\text{Cl}}_2\left(g\right)$

Now balance the hydrogen atoms on both the sides by adding ${\text{4H}}^{\text{+}}$ ions and ${\text{2H}}_{\text{2}}\text{O}$ molecule to the reduction half reaction,

${\text{MnO}}_{\text{2}}\left(aq\right)+2{\text{e}}^{-}+4{\text{H}}^{+}\left(aq\right)\to {\text{MnCl}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The oxidation half reaction is multiplied by $2$ to balance the charge .

$\text{2NaCl}\left(aq\right)\to 2{\text{Cl}}_2\left(g\right)+2{\text{e}}^{-}$

To balance $\text{Cl}$ atoms add coefficient of $2$ at the reactant side of oxidation half reaction,

$\text{4NaCl}\left(aq\right)\to 2{\text{Cl}}_2\left(g\right)+2{\text{e}}^{-}$

Now, add both the two half reactions which is shown as follows;

$\begin{array}{c}\text{4NaCl}\left(aq\right)\to 2{\text{Cl}}_2\left(g\right)+2{\text{e}}^{-}\\ {\text{MnO}}_{\text{2}}\left(aq\right)+2{\text{e}}^{-}+4{\text{H}}^{+}\left(aq\right)\to {\text{MnCl}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\\ {\text{MnO}}_{\text{2}}\left(aq\right)+4\text{NaCl}\left(aq\right)+4{\text{H}}^{+}\left(aq\right)\to {\text{MnCl}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)+2{\text{Cl}}_2\left(g\right)\end{array}$

As given the reaction is taking place in acid solution, the redox reaction will be balanced by replacing $4{\text{H}}^{+}$ ions with ${\text{2H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)$ because $4{\text{H}}^{+}$ ions is equals to ${\text{2H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)$.

${\text{MnO}}_{\text{2}}\left(aq\right)+4\text{NaCl}\left(aq\right)+2{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{MnCl}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)+2{\text{Cl}}_2\left(g\right)$

Now balance the chlorine atom by removing one chlorine atom from the product side of the reaction then the reaction becomes as,

${\text{MnO}}_{\text{2}}\left(aq\right)+4\text{NaCl}\left(aq\right)+2{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{MnCl}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{Cl}}_2\left(g\right)$

In the above redox reaction, there are four sodium atoms and four sulfate atoms are placed on the left hand side  so we need to add four sodium atoms and four sulfate atoms on the right hand side of the equation in the form of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ which will be shown as follows,

${\text{MnO}}_{\text{2}}\left(aq\right)+4\text{NaCl}\left(aq\right)+2{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to 2{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right){\text{+MnCl}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{Cl}}_2\left(g\right)$

Thus, the complete and balanced redox reaction is,

${\text{MnO}}_{\text{2}}\left(aq\right)+4\text{NaCl}\left(aq\right)+2{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to 2{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right){\text{+MnCl}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{Cl}}_2\left(g\right)$

(b)

Interpretation Introduction

To determine: A net ionic equation describing the formation of chlorine.

Answer to Problem 4.124AP

Solution

A net ionic equation describing the formation of chlorine is,

${\text{Mn}}^{4+}\left(aq\right)+4{\text{Cl}}^{-}\left(aq\right)\to {\text{Mn}}^{2+}\left(g\right)+2{\text{Cl}}_{\text{2}}\left(g\right)$

Explanation of Solution

Explanation

The given unbalanced redox reaction is,

$\text{NaCl}\left(aq\right)+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)+{\text{MnO}}_{\text{2}}\left(aq\right){\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)+{\text{MnCl}}_{\text{2}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{Cl}}_{\text{2}}\left(g\right)$

In the above reaction the formation of chlorine atom occurs in the reduction half reaction which is,

${\text{MnO}}_{\text{2}}\left(aq\right)+2{\text{e}}^{-}\to {\text{MnCl}}_{\text{2}}\left(g\right)$

The oxidation state of $\text{Mn}$ in ${\text{MnO}}_{\text{2}}$ is $\underset{\_}{+4}$ whereas the oxidation state of $\text{Mn}$ in ${\text{MnCl}}_{\text{2}}$ is $\underset{\_}{+2}$. The reaction will remove the ions that will be the same that is oxygen atoms will be vanished from the reactant side and will add appropriate amount of chlorine atoms to equalize the charge on manganese ions to both the sides of the reaction so the reaction will be shown as

${\text{Mn}}^{4+}\left(aq\right)+4{\text{Cl}}^{-}\left(aq\right)\to {\text{Mn}}^{2+}\left(g\right)+2{\text{Cl}}_{\text{2}}\left(g\right)$

Therefore, a net ionic equation describing the formation of chlorine is,

${\text{Mn}}^{4+}\left(aq\right)+4{\text{Cl}}^{-}\left(aq\right)\to {\text{Mn}}^{2+}\left(g\right)+2{\text{Cl}}_{\text{2}}\left(g\right)$

(c)

Interpretation Introduction

To determine: The balanced equation for the conversion of ${\text{Cl}}_{\text{2}}$ to $\text{HCl}$ and $\text{HOCl}$

Answer to Problem 4.124AP

Solution

The balanced equation for the conversion of ${\text{Cl}}_{\text{2}}$ to $\text{HCl}$ and $\text{HOCl}$ is,

$3{\text{Cl}}_2\left(g\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 5\text{HCl}\left(aq\right)+{\text{HClO}}_3\left(aq\right)$

Explanation of Solution

Explanation

The reaction of water with ${\text{Cl}}_{\text{2}}\left(g\right)$ will give $\text{HCl}\left(aq\right)$ and $\text{HOCl}\left(aq\right)$  the equation as ,

${\text{Cl}}_2\left(g\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{HCl}\left(aq\right)+{\text{HClO}}_3\left(aq\right)$

To balance the equation the amount of chlorine add coefficient of $2$ in front of ${\text{Cl}}_{\text{2}}$ and add coefficient $5$ in front of $\text{HCl}$ on the product side, therefore the balanced reaction will be,

${\text{2Cl}}_2\left(g\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 5\text{HCl}\left(aq\right)+{\text{HClO}}_3\left(aq\right)$

Therefore The balanced equation for the conversion of ${\text{Cl}}_{\text{2}}$ to $\text{HCl}$ and $\text{HOCl}$ is

${\text{2Cl}}_2\left(g\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 5\text{HCl}\left(aq\right)+{\text{HClO}}_3\left(aq\right)$

Conclusion

We conclude that

1. a) The oxidation numbers of the elements in each compound of the given reaction are ,

   ${\text{Na}}^{+}=\underset{\_}{+1},\text{Cl}=\underset{\_}{-1,}\text{H}=\underset{\_}{+1},\text{O}=\underset{\_}{-2},\text{Mn}\text{in}{\text{MnO}}_{\text{2}}=\underset{\_}{+4},\text{S}\text{in}{\text{Na}}_{\text{2}}{\text{SO}}_4=\underset{\_}{+6}\text{and}\text{Mn}\text{in}{\text{MnCl}}_{\text{2}}=\underset{\_}{+2}$ The balanced redox reaction in acid solution is as follows:

   $4\text{NaCl}\left(aq\right)+2{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)+{\text{MnO}}_{\text{2}}\left(aq\right)\to 2{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)+{\text{MnCl}}_{\text{2}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{Cl}}_{\text{2}}\left(g\right)$

2. b) A net ionic equation describing the formation of chlorine is,

   ${\text{Mn}}^{4+}\left(aq\right)+4{\text{Cl}}^{-}\left(aq\right)\to {\text{Mn}}^{2+}\left(g\right)+2{\text{Cl}}_{\text{2}}\left(g\right)$

3. c) The balanced equation for the conversion of ${\text{Cl}}_{\text{2}}$ to $\text{HCl}$ and $\text{HOCl}$ is, $3{\text{Cl}}_2\left(g\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 5\text{HCl}\left(aq\right)+{\text{HClO}}_3\left(aq\right)$