Chapter 4, Problem 4.104QP

 (a)

Interpretation Introduction

Interpretation: The given net ionic equations are to be balanced and the elements that are oxidized and reduced 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 the both sides of the reaction.

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

Answer to Problem 4.104QP

Solution

The balanced net ionic equation is,

${\text{MnO}}_4^{-}\left(aq\right)+{\text{S}}^{2-}\left(aq\right)+4{\text{H}}^{+}\left(aq\right)\to {\text{MnO}}_2\left(s\right)+\text{S}\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Sulfur is oxidized and manganese is reduced in the reaction.

Explanation of Solution

Explanation

The given net ionic reaction is,

${\text{MnO}}_4^{-}\left(aq\right)+{\text{S}}^{2-}\left(aq\right)\to {\text{MnO}}_4\left(s\right)+\text{S}\left(s\right)$

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

To balance the above equation, first number of atoms will be balanced.

The number of oxygen atoms is balanced by adding the coefficient $2$ to water molecules.

${\text{MnO}}_4^{-}\left(aq\right)+{\text{S}}^{2-}\left(aq\right)\to {\text{MnO}}_4\left(s\right)+\text{S}\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Four hydrogen atoms are present on the right hand side of the reaction. To balance the hydrogen atoms, coefficient $4$ is added to ${\text{H}}^{+}$ in the left hand side of the reaction as the above reaction occurs in presence of the acidic solution.

The balanced net ionic equation is,

${\text{MnO}}_4^{-}\left(aq\right)+{\text{S}}^{2-}\left(aq\right)+4{\text{H}}^{+}\left(aq\right)\to {\text{MnO}}_2\left(s\right)+\text{S}\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

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

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 ${\text{MnO}}_2$ 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 on ${\text{MnO}}_2$ 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 has decreased from $+7$ in ${\text{MnO}}_4^{-}$ to $+4$ in ${\text{MnO}}_2$.

Thus, manganese is reduced.

The oxidation number of sulfur increases from $\left(-2\right)$ in ${\text{S}}^{2-}$ to $0$ in elemental sulfur $\left(\text{S}\right)$ Thus, sulfur is oxidized.

(b)

Interpretation Introduction

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

Answer to Problem 4.104QP

Solution

The balanced net ionic equation is,

${\text{IO}}_3^{-}\left(aq\right)+5{\text{I}}^{-}\left(aq\right)+6{\text{H}}^{+}\left(aq\right)\to 3{\text{I}}_2\left(s\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Iodine is both oxidized and reduced in the reaction.

Explanation of Solution

Explanation

The given net ionic reaction is,

${\text{IO}}_3^{-}\left(aq\right)+{\text{I}}^{-}\left(aq\right)\to {\text{I}}_2\left(s\right)$

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

To balance the above equation, first oxygen atoms are balanced by adding the coefficient $3$  to water molecules.

${\text{IO}}_3^{-}\left(aq\right)+{\text{I}}^{-}\left(aq\right)\to 3{\text{I}}_2\left(s\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Now, the hydrogen atoms are balanced by adding ${\text{H}}^{+}$ ions in the left hand side of the reaction with the coefficient $6$. Then the charge is balanced by adding the coefficient $5$ to iodide ion $\left({\text{I}}^{-}\right)$.

The balanced net ionic equation is,

${\text{IO}}_3^{-}\left(aq\right)+{\text{I}}^{-}\left(aq\right)+6{\text{H}}^{+}\left(aq\right)\to 3{\text{I}}_2\left(s\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

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

The net charge present on ${\text{IO}}_3^{-}$ is $\left(-1\right)$.

The oxidation number of iodine atom in ${\text{IO}}_3^{-}$ is calculated by the formula,

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

The oxidation number of iodine atom in ${\text{IO}}_3^{-}$ is assumed to be $x$.

Substitute the value of oxidation number in the above expression.

$\begin{array}{c}\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{iodine}\end{array}\right)+3\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{oxygen}\end{array}\right)=-1\\ x+3\left(-2\right)=-1\\ x+\left(-6\right)=-1\\ x=+5\end{array}$

The oxidation number of iodine atom in ${\text{IO}}_3^{-}$ is $+5$.

The oxidation number of iodine atom in ${\text{I}}_{\text{2}}$ is $0$ while it is $\left(-1\right)$ in ${\text{I}}^{-}$.

Therefore, iodine in iodate ion ${\text{IO}}_3^{-}$ is reduced to ${\text{I}}_{\text{2}}$ and ${\text{I}}^{-}$ is oxidized to ${\text{I}}_{\text{2}}$.

(c)

Interpretation Introduction

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

Answer to Problem 4.104QP

Solution

The balanced net ionic equation is,

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

Manganese is oxidized and bismuth is reduced in the reaction.

Explanation of Solution

Explanation

The given equation is,

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

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

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 $+2$.

So, the oxidation number of manganese has increased from $+2$ in ${\text{Mn}}^{2+}$ to $+7$ in ${\text{MnO}}_4^{-}$ thus manganese is oxidized.

The oxidation number of bismuth is calculated by the formula,

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

The oxidation number of bismuth is assumed to be $x$.

The net charge in ${\text{BiO}}_3^{-}$ is $\left(-1\right)$.

Substitute the value of oxidation number in the above expression.

$\begin{array}{c}\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{bismuth}\end{array}\right)+3\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{oxygen}\end{array}\right)=-1\\ x+3\left(-2\right)=-1\\ x+\left(-6\right)=-1\\ x=+5\end{array}$

The oxidation number of bismuth in ${\text{BiO}}_3^{-}$ is $+5$.

The oxidation number of bismuth in ${\text{Bi}}^{3+}$ is $+3$.

So, the oxidation number is decreasing from $+5$ in ${\text{BiO}}_3^{-}$ to $+3$ in ${\text{Bi}}^{3+}$. Thus, bismuth is reduced.

The above equation is not balanced.

To balance the equation, coefficient $2$ is added to manganese ions and the coefficient $5$ is added to bismuth ions in the reaction.

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

First oxygen atoms are balanced by adding the coefficient $7$ to water molecules in the left hand side of the reaction.

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

Then, hydrogen atoms are balanced by adding the coefficient $14$ to ${\text{H}}^{+}$ ions in the left hand side of the reaction.

The balanced net ionic equation is,

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

Therefore, manganese is oxidized and bismuth is reduced in the reaction.