Chapter 4, Problem 4.105QP

(a)

Interpretation Introduction

Interpretation: The oxidation numbers of all the elements in the reactants and products of the given equations are to be determined and the elements that are oxidized and reduced in the given reactions are to be identified.

Concept introduction: The oxidation state or number in a compound is the charge present on the individual atom if the all bonds are assumed to be ionic. The charge present on the ionic compound is equal to the some of the charges of all the atoms present in that ionic compound.

The oxidation number is assigned by the following set of rules,

• The oxidation number present on a neutral compound is always zero.
• The oxidation number of atoms present in a element in the pure form is zero.
• In the compounds, if halogens are present the oxidation number of halogens is $-1$ but when oxygen is present with them they have different oxidation number.
• The common oxidation number of oxygen in most compounds is $-2$ and oxidation number of hydrogen is $+1$.

To determine: The oxidation numbers of all the elements in the reactants and products of the given equation and the identification of the elements that are oxidized and reduced in the given reaction.

Answer to Problem 4.105QP

Solution

The oxidation number of all the reactants and products in the given reaction are,

In ${\text{SiO}}_2$, silicon has oxidation number $\underset{\_}{+4}$ and oxygen has oxidation number $\underset{\_}{-2}$.

In ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}$, oxidation number of $\text{Fe}$ is $+\underset{\_}{\frac{8}{3}}$ and oxygen has oxidation number $\underset{\_}{-2}$.

In ${\text{Fe}}_{\text{2}}{\text{SiO}}_4$, oxidation number of $\text{Fe}$ is $\underset{\_}{+2}$, oxygen has oxidation number $\underset{\_}{-2}$ and silicon has oxidation number $\underset{\_}{+4}$.

In ${\text{O}}_{\text{2}}$, oxidation number of oxygen is $\underset{\_}{0}$.

Iron has reduced and oxygen has oxidized in the reaction.

Explanation of Solution

Explanation

The given reaction is,

$3{\text{SiO}}_2\left(s\right)+2{\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}\left(s\right)\to 3{\text{Fe}}_{\text{2}}{\text{SiO}}_4\left(s\right){\text{+O}}_{\text{2}}\left(g\right)$

The reactants in the reaction are ${\text{SiO}}_2$ and ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}\left(s\right)$.

The reactant, ${\text{SiO}}_2$ contain silicon and oxygen elements.

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

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

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

The oxidation number of oxygen is $-2$.

Net charge on ${\text{SiO}}_2$ is zero.

Substitute the value of oxidation number in the above expression.

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

The oxidation number of silicon in the given reactant ${\text{SiO}}_2$ is $+4$.

The reactant, ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}$ contain iron $\left(\text{Fe}\right)$ and oxygen.

The oxidation number of oxygen is $-2$.

The oxidation number of $\text{Fe}$ in ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}$ is calculated by the formula,

$\left(\begin{array}{l}\text{Sum}\text{of}\text{oxidation}\text{number}\text{of}\text{all}\\ \text{element}\text{in}{\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}\end{array}\right)=\text{Net}\text{charge}\text{on}\text{whole}\text{molecule}$

The oxidation number of $\text{Fe}$ in ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}$ is assumed to be $x$.

Net charge on ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}$ is zero.

Substitute the value of oxidation number in the above expression.

$\begin{array}{c}3\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{iron}\end{array}\right)+4\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{oxygen}\end{array}\right)=0\\ 3x+\left[4\times \left(-2\right)\right]=0\\ 3x-8=0\\ x=+\frac{8}{3}\end{array}$

The oxidation number of $\text{Fe}$ in the given reactant is $+\frac{8}{3}$.

The products formed in the above reaction are ${\text{Fe}}_{\text{2}}{\text{SiO}}_4$ and ${\text{O}}_{\text{2}}$.

The product, ${\text{Fe}}_{\text{2}}{\text{SiO}}_4$ contain iron, silicon and oxygen.

The oxidation number of oxygen is $-2$ and the oxidation number of silicon in ${\text{SiO}}_2$ is $+4$ as calculated above.

The oxidation number of $\text{Fe}$ in ${\text{Fe}}_{\text{2}}{\text{SiO}}_4$ is calculated by the formula,

$\left(\begin{array}{l}\text{Sum}\text{of}\text{oxidation}\text{number}\text{of}\text{all}\\ \text{element}\text{in}{\text{Fe}}_{\text{2}}{\text{SiO}}_4\end{array}\right)=\text{Net}\text{charge}\text{on}\text{whole}\text{molecule}$

The oxidation number of $\text{Fe}$ in ${\text{Fe}}_{\text{2}}{\text{SiO}}_4$ is assumed to be $x$.

Net charge on ${\text{Fe}}_{\text{2}}{\text{SiO}}_4$ is zero.

Substitute the value of oxidation number in the above expression.

$\begin{array}{c}2\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{iron}\end{array}\right)+\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{silicon}\end{array}\right)+4\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{oxygen}\end{array}\right)=0\\ 2x+4+\left(4\left(-2\right)\right)=0\\ 2x+4-8=0\\ x=+2\end{array}$

The oxidation number of $\text{Fe}$ in ${\text{Fe}}_{\text{2}}{\text{SiO}}_4$ is $+2$.

The oxidation number of oxygen is zero in ${\text{O}}_{\text{2}}$ because it is in pure form.

Therefore, the oxidation number of the elements that are reactants and products in the reaction are as follows.

In ${\text{SiO}}_2$, silicon has oxidation number $\underset{\_}{+4}$ and oxygen has oxidation number $\underset{\_}{-2}$.

In ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}$, oxidation number of $\text{Fe}$ is $\underset{\_}{+\frac{8}{3}}$ and oxygen has oxidation number $\underset{\_}{-2}$.

In ${\text{Fe}}_{\text{2}}{\text{SiO}}_4$, oxidation number of $\text{Fe}$ is $\underset{\_}{+2}$, oxygen has oxidation number $\underset{\_}{-2}$ and silicon has oxidation number $\underset{\_}{+4}$.

In ${\text{O}}_{\text{2}}$, oxidation number of oxygen is $\underset{\_}{0}$.

Therefore, iron has reduced from $+\frac{8}{3}$ to $+2$ oxidation state and oxygen has oxidized from $-2$ to $0$ oxidation state.

(b)

Interpretation Introduction

To determine: The oxidation numbers of all the elements in the reactants and products of the given equation and the identification of the elements that are oxidized and reduced in the given reaction.

Answer to Problem 4.105QP

Solution

The oxidation number of all the reactants and products in the given reaction are,

In ${\text{SiO}}_2$, silicon has oxidation number $\underset{\_}{+4}$ and oxygen has oxidation number $\underset{\_}{-2}$.

In $\text{Fe}$, the oxidation number of $\text{Fe}$ is $\underset{\_}{0}$.

In ${\text{O}}_{\text{2}}$, the oxidation number of oxygen is $\underset{\_}{0}$.

In ${\text{Fe}}_{\text{2}}{\text{SiO}}_4$, The oxidation number of $\text{Fe}$ is $\underset{\_}{+2}$, silicon has oxidation number $\underset{\_}{+4}$ and oxygen has oxidation number $-2$.

Iron has oxidized and oxygen has reduced in the reaction.

Explanation of Solution

Explanation

The given reaction is,

${\text{SiO}}_2\left(s\right)+2\text{Fe}\left(s\right)+{\text{O}}_2\left(g\right)\to {\text{Fe}}_{\text{2}}{\text{SiO}}_4\left(s\right)$

The reactant in the reaction are ${\text{SiO}}_2$, $\text{Fe}$ and ${\text{O}}_{\text{2}}$.

The reactant, ${\text{SiO}}_2$ contain silicon and oxygen elements.

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

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

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

The oxidation number of oxygen is $-2$.

Net charge on ${\text{SiO}}_2$ is zero.

Substitute the value of oxidation number in the above expression.

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

The oxidation number of silicon in the given reactant ${\text{SiO}}_2$ is $+4$.

The oxidation number of $\text{Fe}$ is $0$ because it is present in its elemental form.

The oxidation number of oxygen is $0$ in ${\text{O}}_{\text{2}}$ because it is present in the pure form.

The product formed in the above reaction is ${\text{Fe}}_{\text{2}}{\text{SiO}}_4$.

The product, ${\text{Fe}}_{\text{2}}{\text{SiO}}_4$ contain iron, silicon and oxygen.

The oxidation number of oxygen is $-2$ and the oxidation number of silicon in ${\text{SiO}}_2$ is $+4$ as calculated above.

The oxidation number of $\text{Fe}$ in ${\text{Fe}}_{\text{2}}{\text{SiO}}_4$ is calculated by the formula,

$\left(\begin{array}{l}\text{Sum}\text{of}\text{oxidation}\text{number}\text{of}\text{all}\\ \text{element}\text{in}{\text{Fe}}_{\text{2}}{\text{SiO}}_4\end{array}\right)=\text{Net}\text{charge}\text{on}\text{whole}\text{molecule}$

The oxidation number of $\text{Fe}$ in ${\text{Fe}}_{\text{2}}{\text{SiO}}_4$ is assumed to be $x$.

Net charge on ${\text{Fe}}_{\text{2}}{\text{SiO}}_4$ is zero.

Substitute the value of oxidation number in the above expression.

$\begin{array}{c}2\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{iron}\end{array}\right)+\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{silicon}\end{array}\right)+4\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{oxygen}\end{array}\right)=0\\ 2x+4+\left(4\left(-2\right)\right)=0\\ 2x+4-8=0\\ x=+2\end{array}$

Therefore, the oxidation number of $\text{Fe}$ in ${\text{Fe}}_{\text{2}}{\text{SiO}}_4$ is $+2$.

Therefore, the oxidation number of the elements that are reactants and products in the reaction are as follows.

In ${\text{SiO}}_2$, silicon has oxidation number $\underset{\_}{+4}$ and oxygen has oxidation number $-2$.

In $\text{Fe}$, the oxidation number of $\text{Fe}$ is $\underset{\_}{0}$.

In ${\text{O}}_{\text{2}}$, the oxidation number of oxygen is $\underset{\_}{0}$.

In ${\text{Fe}}_{\text{2}}{\text{SiO}}_4$, The oxidation number of $\text{Fe}$ is $\underset{\_}{+2}$, silicon has oxidation number $\underset{\_}{+4}$ and oxygen has oxidation number $-2$.

Therefore, iron has oxidized from $0$ to $+2$ oxidation state and oxygen has reduced to $-2$ from $0$ oxidation state.

(c)

Interpretation Introduction

To determine: The oxidation numbers of all the elements in the reactants and products of the given equation and the identification of the elements that are oxidized and reduced in the given reaction.

Answer to Problem 4.105QP

Solution

The oxidation number of all the reactants and products in the given reaction are,

In $\text{FeO}$, the oxidation number of $\text{Fe}$ is $\underset{\_}{+2}$ and oxidation number of oxygen is $\underset{\_}{-2}$.

In ${\text{O}}_{\text{2}}$, the oxidation number of oxygen is $\underset{\_}{0}$.

In ${\text{H}}_{\text{2}}\text{O}$, the oxidation number of oxygen is $\underset{\_}{-2}$ and oxidation number of hydrogen is $\underset{\_}{+1}$.

In $\text{Fe}{\left(\text{OH}\right)}_3$, the oxidation number of $\text{Fe}$ is $\underset{\_}{+3}$, oxidation number of hydrogen is $\underset{\_}{+1}$ and oxidation number of oxygen is $\underset{\_}{-2}$.

Iron has oxidized and oxygen has reduced in the reaction.

Explanation of Solution

Explanation

The given reaction is,

$\text{4FeO}\left(s\right)+{\text{O}}_2\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\to 4\text{Fe}{\left(\text{OH}\right)}_3\left(s\right)$

The reactants are $\text{FeO}$, ${\text{O}}_2$ and ${\text{H}}_{\text{2}}\text{O}$.

The reactant, $\text{FeO}$ contain iron and oxygen.

The oxidation number of oxygen is $-2$.

Net charge on $\text{FeO}$ is zero.

The oxidation number of $\text{Fe}$ in $\text{FeO}$ is calculated by the formula,

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

The oxidation number of $\text{Fe}$ in $\text{FeO}$ is assumed to be $x$.

Substitute the value of oxidation number in the above expression.

$\begin{array}{c}\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{iron}\end{array}\right)+\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{oxygen}\end{array}\right)=0\\ x+\_2=0\\ x=+2\\ \end{array}$

The oxidation number of $\text{Fe}$ in the given reactant is $+2$.

The oxidation number of oxygen is $0$ in ${\text{O}}_{\text{2}}$ because it is present in the pure form.

In ${\text{H}}_{\text{2}}\text{O}$, the oxidation number of oxygen is $-2$ and oxidation number of hydrogen is $+1$.

The product form in the above reaction is $\text{Fe}{\left(\text{OH}\right)}_3$.

The product $\text{Fe}{\left(\text{OH}\right)}_3$ contains iron, hydrogen and oxygen.

The oxidation number of hydrogen is $+1$ and oxidation number of oxygen is $-2$.

The oxidation number of $\text{Fe}$ in $\text{Fe}{\left(\text{OH}\right)}_3$ is calculated by the formula,

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

The oxidation number of $\text{Fe}$ in $\text{Fe}{\left(\text{OH}\right)}_3$ is assumed to be $x$.

Substitute the value of oxidation number in the above expression.

$\begin{array}{c}\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{iron}\end{array}\right)+3\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{oxygen}\end{array}\right)+3\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{hydrogen}\end{array}\right)=0\\ x+3\left(-2\right)+3\left(+1\right)=0\\ x=+3\\ \end{array}$

The oxidation number of $\text{Fe}$ in the given reactant is $+3$.

Therefore, the oxidation number of the elements that are reactants and products in the reaction are as follows.

In $\text{FeO}$, the oxidation number of $\text{Fe}$ is $\underset{\_}{+2}$ and oxidation number of oxygen is $\underset{\_}{-2}$.

In ${\text{O}}_{\text{2}}$, the oxidation number of oxygen is $\underset{\_}{0}$.

In ${\text{H}}_{\text{2}}\text{O}$, the oxidation number of oxygen is $\underset{\_}{-2}$ and oxidation number of hydrogen is $\underset{\_}{+1}$.

In $\text{Fe}{\left(\text{OH}\right)}_3$, the oxidation number of $\text{Fe}$ is $\underset{\_}{+3}$, oxidation number of hydrogen is $\underset{\_}{+1}$ and oxidation number of oxygen is $\underset{\_}{-2}$.

Therefore, Iron has oxidized from $+2$ to $+3$ oxidation state and oxygen has reduced from $0$ to $-2$ in the reaction.

Conclusion

1. a) The oxidation number of all the reactants and products in the given reaction has been rightfully stated. Iron has reduced and oxygen has oxidized in the reaction.
2. b) The oxidation number of all the reactants and products in the given reaction has been rightfully stated. Iron has oxidized and oxygen has reduced in the reaction.
3. c) The oxidation number of all the reactants and products in the given reaction has been rightfully stated. Iron has oxidized and oxygen has reduced in the reaction