Chapter 4, Problem 4.103QP

(a)

Interpretation Introduction

Interpretation: The net ionic equations are to be balanced and the elements that are oxidized and reduced in the reaction are to be identified.

Concept introduction: Oxidation is a process in which there is a loss of electron during a reaction.

Reduction is a process in which there is a gain of electron during a reaction

In ionic equation, the individual ions of the constituent compound are shown.

The reaction is balanced when the number of atoms and charge are balanced on both the sides of the reaction.

To determine: The balanced reaction of the given net ionic equation and identification of the elements that are oxidized and reduced in the reaction.

Answer to Problem 4.103QP

Solution

The balanced net equation is,

${\text{MnO}}_2\left(s\right)+2\text{HCl}\left(aq\right)+2{\text{H}}^{+}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+2{\text{Cl}}_2\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Manganese is reduced and chlorine is oxidized in the reaction.

Explanation of Solution

Explanation

The given net ionic equation is,

${\text{MnO}}_2\left(s\right)+\text{HCl}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+{\text{Cl}}_2\left(g\right)$

The above equation is not balanced, to balance the equation first number of hydrogen and oxygen atoms are to be balanced. Oxygen is balanced by adding water molecules with the coefficient of $2$ in the right hand side of the reaction and is written as,

${\text{MnO}}_2\left(s\right)+\text{HCl}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+{\text{Cl}}_2\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Now, the $+2$ charge on the right hand side of the reaction is to be balanced by adding the ${\text{H}}^{+}$ ions with the coefficient of $2$ in the left hand side of the reaction and to balance the hydrogen atoms, the coefficient $2$ is added to $\text{HCl}$ in the left hand side of the reaction and is written as,

${\text{MnO}}_2\left(s\right)+2\text{HCl}\left(aq\right)+2{\text{H}}^{+}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+2{\text{Cl}}_2\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The above equation is balanced net ionic equation.

The oxidation number of oxygen is $\left(-2\right)$ in common.

The oxidation number of manganese in ${\text{MnO}}_2$ is calculated by the formula,

$\left(\begin{array}{l}\text{Sum}\text{of}\text{oxidation}\text{number}\text{of}\text{all}\\ \text{element}\text{in}{\text{MnO}}_2\end{array}\right)=\text{Net}\text{charge}\text{on}\text{whole}\text{molecule}$

The net charge in ${\text{MnO}}_2$ is zero.

The oxidation number of manganese in ${\text{MnO}}_2$ is assumed to be $x$.

Substitute the value of oxidation number in the above expression.

$\begin{array}{c}\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{manganese}\end{array}\right)+2\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{oxygen}\end{array}\right)=0\\ x+2\left(-2\right)=0\\ x+\left(-4\right)=0\\ x=+4\end{array}$

The oxidation number of manganese in ${\text{MnO}}_2$ is $+4$.

The oxidation number of manganese in ${\text{Mn}}^{2+}$ is $+2$.

The oxidation number of manganese decreases from $+4$ in ${\text{MnO}}_2$ to $+2$ in ${\text{Mn}}^{2+}$. Thus, manganese is reduced.

Chlorine has an oxidation number zero in ${\text{Cl}}_{\text{2}}$ and chloride ion has an oxidation number $-1$.

The oxidation number of chlorine has increased from $-1$ in ${\text{Cl}}^{-}$ to $0$ in ${\text{Cl}}_{\text{2}}$. Hence, chlorine is oxidized.

(b)

Interpretation Introduction

To determine: The balanced reaction of the given net ionic equation and identification of the elements that are oxidized and reduced in the reaction.

Answer to Problem 4.103QP

Solution

The balanced net ionic equation is,

${\text{I}}_2\left(s\right)+2{\text{S}}_{\text{2}}{\text{O}}_3^{2-}\left(aq\right)\to {\text{S}}_4{\text{O}}_6^{2-}\left(aq\right)+2{\text{I}}^{-}\left(aq\right)$

Sulfur is oxidized and iodine is reduced in the reaction.

Explanation of Solution

Explanation

The given net ionic equation is,

${\text{I}}_2\left(s\right)+{\text{S}}_{\text{2}}{\text{O}}_3^{2-}\left(aq\right)\to {\text{S}}_4{\text{O}}_6^{2-}\left(aq\right)+{\text{I}}^{-}\left(aq\right)$

To balance the above equation for equal number of atoms on both the sides of the reaction, first iodine is balanced by adding coefficient $2$ to ${\text{I}}^{-}$ in the right hand side of the reaction and is written as,

${\text{I}}_2\left(s\right)+{\text{S}}_{\text{2}}{\text{O}}_3^{2-}\left(aq\right)\to {\text{S}}_4{\text{O}}_6^{2-}\left(aq\right)+2{\text{I}}^{-}\left(aq\right)$

Now, the sulfur and oxygen atoms are to be balanced by adding coefficient $2$ to ${\text{S}}_{\text{2}}{\text{O}}_3^{2-}$ in the left hand side of the reaction and is written as,

${\text{I}}_2\left(s\right)+2{\text{S}}_{\text{2}}{\text{O}}_3^{2-}\left(aq\right)\to {\text{S}}_4{\text{O}}_6^{2-}\left(aq\right)+2{\text{I}}^{-}\left(aq\right)$

The above equation is balanced net ionic equation.

The oxidation number of oxygen is $\left(-2\right)$ in common.

The oxidation number of sulfur in ${\text{S}}_{\text{2}}{\text{O}}_3^{2-}$ is calculated by the formula,

$\left(\begin{array}{l}\text{Sum}\text{of}\text{oxidation}\text{number}\text{of}\text{all}\\ \text{element}\text{in}{\text{S}}_{\text{2}}{\text{O}}_3^{2-}\end{array}\right)=\text{Net}\text{charge}\text{on}\text{whole}\text{molecule}$

The net charge on ${\text{S}}_{\text{2}}{\text{O}}_3^{2-}$ is $-2$.

The oxidation number of sulfur in ${\text{S}}_{\text{2}}{\text{O}}_3^{2-}$ is assumed to be $x$.

Substitute the value of oxidation number in the above expression.

$\begin{array}{c}2\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{sulfur}\end{array}\right)+3\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{oxygen}\end{array}\right)=-2\\ 2x+3\left(-2\right)=-2\\ 2x+\left(-6\right)=-2\\ x=+2\end{array}$

The oxidation number of sulfur in ${\text{S}}_{\text{2}}{\text{O}}_3^{2-}$ is $+2$.

The oxidation number of sulfur in ${\text{S}}_4{\text{O}}_6^{2-}$ is calculated by the formula,

$\left(\begin{array}{l}\text{Sum}\text{of}\text{oxidation}\text{number}\text{of}\text{all}\\ \text{element}\text{in}{\text{S}}_4{\text{O}}_6^{2-}\end{array}\right)=\text{net}\text{charge}\text{on}\text{whole}\text{molecule}$

The net charge on ${\text{S}}_4{\text{O}}_6^{2-}$ is $-2$.

The oxidation number of sulfur in ${\text{S}}_4{\text{O}}_6^{2-}$ is assumed to be $x$.

Substitute the value of oxidation number in the above expression.

$\begin{array}{c}4\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{sulfur}\end{array}\right)+6\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{oxygen}\end{array}\right)=-2\\ 4x+6\left(-2\right)=-2\\ 4x+\left(-12\right)=-2\\ x=+2.5\end{array}$

Thus, the average oxidation number of sulfur in ${\text{S}}_4{\text{O}}_6^{2-}$ is $+2.5$.

The oxidation number of sulfur has increased from $-2$ to $+2.5$ so, sulfur is oxidized.

The oxidation number of iodine has increased from $0$ in ${\text{I}}_{\text{2}}$ to $-1$ in ${\text{I}}^{-}$ to. Hence, iodine is reduced.

(c)

Interpretation Introduction

To determine: The balanced reaction of the given net ionic equation and identification of the elements that are oxidized and reduced in the reaction.

Answer to Problem 4.103QP

Solution

The balanced net ionic equation is,

${\text{MnO}}_4^{-}\left(aq\right)+5{\text{Fe}}^{2+}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+5{\text{Fe}}^{3+}\left(aq\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Iron is oxidized and manganese is reduced in the reaction.

Explanation of Solution

Explanation

The given net ionic equation is,

${\text{MnO}}_4^{-}\left(aq\right)+{\text{Fe}}^{2+}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+{\text{Fe}}^{3+}\left(aq\right)$

The above equation is not balanced, to balance the equation, first number of hydrogen and oxygen atoms are to be balanced. Oxygen is balanced by adding water molecules with the coefficient of $4$ in the right hand side of the reaction and is written as,

${\text{MnO}}_4^{-}\left(aq\right)+{\text{Fe}}^{2+}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+{\text{Fe}}^{3+}\left(aq\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The number of hydrogen atoms in the right hand side of the reaction is balanced by adding ${\text{H}}^{+}$ ions with the coefficient of $8$ in the left hand side of the reaction because reaction occurs in an acidic medium and is written as,

${\text{MnO}}_4^{-}\left(aq\right)+{\text{Fe}}^{2+}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+{\text{Fe}}^{3+}\left(aq\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Then, the charge is balanced on both sides of the reaction by adding coefficient of five to give a total charge of $+17$ and is written as,

${\text{MnO}}_4^{-}\left(aq\right)+5{\text{Fe}}^{2+}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+5{\text{Fe}}^{3+}\left(aq\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The above equation is balanced net ionic equation.

The oxidation number of oxygen is $\left(-2\right)$ in common.

The net charge on ${\text{MnO}}_4^{-}$ is $-1$.

The oxidation number of manganese in ${\text{MnO}}_4^{-}$ is calculated by the formula,

$\left(\begin{array}{l}\text{Sum}\text{of}\text{oxidation}\text{number}\text{of}\text{all}\\ \text{element}\text{in}{\text{MnO}}_4^{-}\end{array}\right)=\text{Net}\text{charge}\text{on}\text{whole}\text{molecule}$

The oxidation number of manganese in ${\text{MnO}}_4^{-}$ is assumed to be $x$.

Substitute the value of oxidation number in the above expression.

$\begin{array}{c}\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{manganese}\end{array}\right)+4\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{oxygen}\end{array}\right)=-1\\ x+4\left(-2\right)=-1\\ x+\left(-8\right)=-1\\ x=+7\end{array}$

The oxidation number of manganese in ${\text{MnO}}_4^{-}$ is $+7$.

The oxidation number of manganese in ${\text{Mn}}^{2+}$ is $+2$.

The oxidation number of manganese decreases from $+7$ in ${\text{MnO}}_4^{-}$ to $+2$ in ${\text{Mn}}^{2+}$. Thus, manganese is reduced.

The oxidation number of iron in ${\text{Fe}}^{{}^{2+}}$ is $+2$ and in ${\text{Fe}}^{{}^{3+}}$ it is $+3$. So, the oxidation number of iron has increased from $+2$ to $+3$.

Therefore, iron has oxidized.

Conclusion

1. a) The balanced net equation is,

   ${\text{MnO}}_2\left(s\right)+2\text{HCl}\left(aq\right)+2{\text{H}}^{+}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+2{\text{Cl}}_2\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$ Manganese is reduced and chlorine is oxidized in the reaction.

2. b) The balanced net ionic equation is,

   ${\text{I}}_2\left(s\right)+2{\text{S}}_{\text{2}}{\text{O}}_3^{2-}\left(aq\right)\to {\text{S}}_4{\text{O}}_6^{2-}\left(aq\right)+2{\text{I}}^{-}\left(aq\right)$

Sulfur is oxidized and iodine is reduced in the reaction.

1. c) The balanced net ionic equation is,

${\text{MnO}}_4^{-}\left(aq\right)+5{\text{Fe}}^{2+}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+5{\text{Fe}}^{3+}\left(aq\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Iron is oxidized and manganese is reduced in the reaction