Chapter 4, Problem 4.104QP

(a)

Interpretation Introduction

Interpretation: The oxidation numbers of each of the elements in the given reactions are to be stated and the elements that are oxidized or reduced in the reaction 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,

1. 1. The oxidation number present on a neutral compound is always zero.
2. 2. The oxidation number of atoms present in a element in the pure form is zero.
3. 3. 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.
4. 4. The common oxidation number of oxygen in most compounds is $-2$ and oxidation number of hydrogen is $+1$.

To determine: The oxidation number of each of the elements in the given reaction and identification of the elements that are oxidized or reduced in the reaction.

Answer to Problem 4.104QP

Solution

The oxidation number of the elements that are reactants and products in the reaction are,

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

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

In ${\text{H}}_4{\text{SiO}}_4$, oxidation number of hydrogen is $\underset{\_}{+1}$, silicon has oxidation number $\underset{\_}{+4}$ and oxygen has oxidation number $\underset{\_}{-2}$.

No elements is oxidized or reduced in the reaction.

Explanation of Solution

Explanation

The given reaction is,

${\text{SiO}}_2\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{H}}_4{\text{SiO}}_4\left(aq\right)$

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

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 $+4$.

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

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

The product ${\text{H}}_4{\text{SiO}}_4$ contains silicon, hydrogen and oxygen.

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

The oxidation number of silicon in ${\text{H}}_4{\text{SiO}}_4$ is calculated by the formula,

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

The oxidation number of silicon in ${\text{H}}_4{\text{SiO}}_4$ is assumed to be $x$.

Net charge on ${\text{H}}_4{\text{SiO}}_4$ is zero.

Substitute the value of oxidation number in the above expression.

$\begin{array}{c}4\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{hydrogen}\end{array}\right)+\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{silicon}\end{array}\right)+4\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{oxygen}\end{array}\right)=0\\ 4\left(+1\right)+x+\left[4\times \left(-2\right)\right]=0\\ x=8-4\\ x=+4\end{array}$

The oxidation number of silicon in ${\text{H}}_4{\text{SiO}}_4$ is $+4$.

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

In ${\text{H}}_4{\text{SiO}}_4$, oxidation number of hydrogen is $\underset{\_}{+1}$, silicon has oxidation number $\underset{\_}{+4}$ and oxygen has oxidation number $\underset{\_}{-2}$.

No elements is oxidized or reduced in the reaction because oxidation state of all the elements remains the same.

(b)

Interpretation Introduction

To determine: The oxidation number of each of the elements in the given reaction and identification of the elements that are oxidized or reduced in the reaction.

Answer to Problem 4.104QP

Solution

The oxidation number of the elements that are reactants and products in the reaction are,

In ${\text{MnCO}}_3$, manganese has oxidation number $\underset{\_}{+2}$, carbon has oxidation number $\underset{\_}{+4}$ and oxygen has oxidation number $\underset{\_}{-2}$.

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

In ${\text{MnO}}_{\text{2}}$ manganese has oxidation number $\underset{\_}{+4}$ and oxygen has oxidation number $\underset{\_}{-2}$.

In ${\text{CO}}_{\text{2}}$ the oxidation number of oxygen is $\underset{\_}{-2}$ and carbon has oxidation number $\underset{\_}{+4}$.

Manganese is oxidized and oxygen is reduced in the reaction.

Explanation of Solution

Explanation

The given reaction is,

${\text{2MnCO}}_3\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)\to 2{\text{MnO}}_{\text{2}}\left(s\right){\text{+2CO}}_{\text{2}}\left(g\right)$

The reactants in the above reaction are ${\text{MnCO}}_3$ and ${\text{O}}_{\text{2}}$.

The reactant ${\text{MnCO}}_3$ contain manganese $\left(\text{Mn}\right)$, carbon and oxygen elements.

The oxidation number of carbon is $+4$ and oxidation number of oxygen is $-2$.

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

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

The oxidation number of manganese in ${\text{MnCO}}_3$ is assumed to be $x$.

Net charge on ${\text{MnCO}}_3$ is zero.

Substitute the value of oxidation number in the above expression.

$\begin{array}{c}4\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{manganese}\end{array}\right)+\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{carbon}\end{array}\right)+3\left(\begin{array}{l}\text{Oxidation}\text{number}\\ \text{of}\text{oxygen}\end{array}\right)=0\\ x+4+\left[3\times \left(-2\right)\right]=0\\ x+4-6=0\\ x=+2\end{array}$

The oxidation number of manganese in ${\text{MnCO}}_3$ is $+2$.

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

The products in the above reaction are ${\text{MnO}}_{\text{2}}$ and ${\text{CO}}_{\text{2}}$.

The product, ${\text{MnO}}_{\text{2}}$ contain manganese $\left(\text{Mn}\right)$ and oxygen elements.

The oxidation number of oxygen is $-2$.

The oxidation number of manganese in ${\text{MnO}}_{\text{2}}$ is calculated by the formula,

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

The oxidation number of manganese in ${\text{MnO}}_{\text{2}}$ is assumed to be $x$.

Net charge on ${\text{MnO}}_{\text{2}}$ is zero.

Substitute the value of oxidation number in the above expression.

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

The oxidation number of manganese in ${\text{MnO}}_{\text{2}}$ is $+4$.

The product, ${\text{CO}}_{\text{2}}$ contain carbon and oxygen.

The oxidation number of oxygen is $-2$.

The oxidation number of carbon in ${\text{CO}}_{\text{2}}$ is calculated by the formula,

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

The oxidation number of carbon in ${\text{CO}}_{\text{2}}$ is assumed to be $x$.

Net charge on ${\text{CO}}_{\text{2}}$ is zero.

Substitute the value of oxidation number in the above expression.

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

The oxidation number of carbon in ${\text{CO}}_{\text{2}}$ is $+4$.

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

In ${\text{MnCO}}_3$, manganese has oxidation number $\underset{\_}{+2}$, carbon has oxidation number $\underset{\_}{+4}$ and oxygen has oxidation number $\underset{\_}{-2}$.

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

In ${\text{MnO}}_{\text{2}}$ manganese has oxidation number $\underset{\_}{+4}$ and oxygen has oxidation number $\underset{\_}{-2}$.

In ${\text{CO}}_{\text{2}}$ the oxidation number of oxygen is $\underset{\_}{-2}$ and carbon has oxidation number $\underset{\_}{+4}$.

Therefore, manganese is oxidized from $+2$ to $+4$ oxidation state. and oxygen is reduced from $0$ to $-2$ in the reaction.

(c)

Interpretation Introduction

To determine: The oxidation number of each of the elements in the given reaction and identification of the elements that are oxidized or reduced in the reaction.

Answer to Problem 4.104QP

Solution

The oxidation number of the elements that are reactants and products in the reaction are,

In ${\text{NO}}_2$, the oxidation number of nitrogen is $\underset{\_}{+4}$ and oxygen has oxidation number $\underset{\_}{-2}$.

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

In ${\text{NO}}_3^{-}$, the oxidation number of nitrogen is $\underset{\_}{+5}$ and oxygen has oxidation number $\underset{\_}{-2}$.

In $\text{NO}$, the oxidation number of nitrogen is $\underset{\_}{+2}$ and oxygen has oxidation number $\underset{\_}{-2}$.

In ${\text{H}}^{+}$, oxidation number of hydrogen is $\underset{\_}{+1}$.

Nitrogen is oxidized as well as reduced in the reaction.

Explanation of Solution

Explanation

The given reaction is,

${\text{3NO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2{\text{NO}}_3^{-}\left(aq\right)\text{+NO}\left(g\right)+2{\text{H}}^{+}\left(aq\right)$

The reactants in the above reaction are ${\text{NO}}_2$ and ${\text{H}}_{\text{2}}\text{O}$.

The reactant, ${\text{NO}}_2$ contain nitrogen $\left(\text{N}\right)$ and oxygen elements.

The oxidation number of oxygen is $-2$.

The oxidation number of nitrogen in ${\text{NO}}_2$ is calculated by the formula,

$\left(\begin{array}{l}\text{Sum}\text{of}\text{oxidation}\text{number}\text{of}\text{all}\\ \text{element}\text{in}{\text{NO}}_2\end{array}\right)=\text{Net}\text{charge}\text{on}\text{whole}\text{molecule}$

The oxidation number of nitrogen in ${\text{NO}}_2$ is assumed to be $x$.

Net charge on ${\text{NO}}_2$ is zero.

Substitute the value of oxidation number in the above expression.

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

The oxidation number of nitrogen in ${\text{NO}}_2$ is $+4$.

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

The products in the above reaction are ${\text{NO}}_3^{-}$, $\text{NO}$ and ${\text{H}}^{+}$.

In the product, ${\text{H}}^{+}$ oxidation number of hydrogen is $+1$.

The product, ${\text{NO}}_3^{-}$ contain nitrogen $\left(\text{N}\right)$ and oxygen elements.

The oxidation number of oxygen is $-2$.

The oxidation number of nitrogen in ${\text{NO}}_3^{-}$ is calculated by the formula,

$\left(\begin{array}{l}\text{Sum}\text{of}\text{oxidation}\text{number}\text{of}\text{all}\\ \text{element}\text{in}{\text{NO}}_3^{-}\end{array}\right)=\text{Net}\text{charge}\text{on}\text{whole}\text{molecule}$

The oxidation number of nitrogen in ${\text{NO}}_3^{-}$ is assumed to be $x$.

Net charge on ${\text{NO}}_3^{-}$ is $-1$.

Substitute the value of oxidation number in the above expression.

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

The oxidation number of nitrogen in ${\text{NO}}_3^{-}$ is $+5$.

The product, $\text{NO}$ contain nitrogen $\left(\text{N}\right)$ and oxygen elements.

The oxidation number of oxygen is $-2$.

The oxidation number of nitrogen in $\text{NO}$ is calculated by the formula,

$\left(\begin{array}{l}\text{Sum}\text{of}\text{oxidation}\text{number}\text{of}\text{all}\\ \text{element}\text{in}\text{NO}\end{array}\right)=\text{Net}\text{charge}\text{on}\text{whole}\text{molecule}$

The oxidation number of nitrogen in $\text{NO}$ is assumed to be $x$.

Net charge on $\text{NO}$ is zero.

Substitute the value of oxidation number in the above expression.

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

The oxidation number of nitrogen in $\text{NO}$ is $+2$.

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

In ${\text{NO}}_2$, the oxidation number of nitrogen is $\underset{\_}{+4}$ and oxygen has oxidation number $\underset{\_}{-2}$.

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

In ${\text{NO}}_3^{-}$, the oxidation number of nitrogen is $\underset{\_}{+5}$ and oxygen has oxidation number $\underset{\_}{-2}$.

In $\text{NO}$, the oxidation number of nitrogen is $\underset{\_}{+2}$ and oxygen has oxidation number $\underset{\_}{-2}$.

In ${\text{H}}^{+}$, oxidation number of hydrogen is $\underset{\_}{+1}$.

Therefore, nitrogen is oxidized from $+4$ to $+5$ oxidation state and nitrogen is also reduced from $+4$ to $+2$ oxidation state in the reaction.

Conclusion

1. a) The oxidation number of each element in the given reaction has been rightfully stated. No elements is oxidized or reduced in the reaction.
2. b) The oxidation number of each element in the given reaction has been rightfully stated. Manganese is oxidized and oxygen is reduced in the reaction.
3. c) The oxidation number of each element in the given reaction has been rightfully stated. Nitrogen is oxidized as well as reduced in the reaction