Chapter 4, Problem 4.140AP

(a)

Interpretation Introduction

Interpretation: The questions corresponding to the given reaction are to be answered.

Concept introduction: 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.

The reactants and the products in a. given ionic equation must be balanced for the given number of ions or atoms of each element.

To determine: The oxidation numbers of nitrogen in the reactants and products of each reaction.

Answer to Problem 4.140AP

Solution

The oxidation number of molecular ${\text{N}}_{\text{2}}$ is always be $\underset{\_}{0}$ and the oxidation number of $\text{N}$ in ${\text{NH}}_3$ is $\underset{\_}{-3}$ in the first reaction of nitrogen fixation.

The oxidation number of $\text{N}$ in ${\text{NH}}_4{}^{+}$ is $\underset{\_}{-3}$ and the oxidation number of $\text{N}$ in ${\text{NO}}_2{}^{-}$ is $\underset{\_}{+3}$ in the nitrification reaction.

Explanation of Solution

Explanation

The first given reaction of nitrogen fixation is,

${\text{N}}_{\text{2}}\left(g\right)+{\text{H}}^{\text{+}}\left(aq\right)+{\text{M}}^{\text{2+}}\left(aq\right)\to {\text{NH}}_{\text{3}}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)+{\text{M}}^{\text{3+}}\left(aq\right)$

Where,

• $\text{M}$ is the transition element such as iron.

The oxidation state of molecular ${\text{N}}_{\text{2}}$ is always be $\underset{\_}{0}$.

The oxidation number nitrogen in the given compound ${\text{NH}}_3$ is calculated as follows.

The oxidation state hydrogen is usually $+1$. The oxidation number of $\text{N}$ in ${\text{NH}}_3$ is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on NH}}_3=\left[\begin{array}{c}\left(\text{Number of}\text{N}\text{atoms}\times \text{oxidation number of N}\right)\\ +\left(\text{Number of H atoms}\times \text{oxidation number of H}\right)\end{array}\right]$

Since, the charge on ${\text{NH}}_3$ is $0$.

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

$\begin{array}{c}{\text{Charge on NH}}_3=\left[\left(x\right)+\left(3\times \left(1\right)\right)\right]\\ 0=\left[x+3\right]\\ x=\underset{\_}{-3}\end{array}$

Thus, the oxidation number of $\text{N}$ in ${\text{NH}}_3$ is $\underset{\_}{-3}$.

Hence, the oxidation number of molecular ${\text{N}}_{\text{2}}$ is always be $\underset{\_}{0}$ and the oxidation number of $\text{N}$ in ${\text{NH}}_3$ is $\underset{\_}{-3}$ in the first reaction of nitrogen fixation.

The second nitrification reaction is,

${\text{NH}}_4{}^{+}\left(aq\right)+{\text{M}}^{\text{3+}}\left(aq\right)\to {\text{NO}}_2{}^{-}\left(aq\right)+{\text{M}}^{\text{2+}}\left(aq\right)$

Where,

• $\text{M}$ is the transition element such as iron.

The oxidation number nitrogen in the given compound ${\text{NH}}_4{}^{+}$ is calculated as follows.

The oxidation state hydrogen is usually $+1$.

The oxidation number of $\text{N}$ in ${\text{NH}}_4{}^{+}$ is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on NH}}_4{}^{+}=\left[\begin{array}{c}\left(\text{Number of}\text{N}\text{atoms}\times \text{oxidation number of N}\right)\\ +\left(\text{Number of H atoms}\times \text{oxidation number of H}\right)\end{array}\right]$

Since, the charge on ${\text{NH}}_4{}^{+}$ is $+1$ .

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

$\begin{array}{c}{\text{Charge on NH}}_4{}^{+}=\left[\left(x\right)+\left(4\times \left(1\right)\right)\right]\\ +1=\left[x+4\right]\\ x=\underset{\_}{-3}\end{array}$

Thus, the oxidation number of $\text{N}$ in ${\text{NH}}_4{}^{+}$ is $\underset{\_}{-3}$.

The oxidation number nitrogen in the given compound ${\text{NO}}_2{}^{-}$ is calculated as follows.

The common oxidation state oxygen is usually $-2$. The oxidation number of $\text{N}$ in ${\text{NO}}_2{}^{-}$ is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on NO}}_2{}^{-}=\left[\begin{array}{c}\left(\text{Number of}\text{N}\text{atoms}\times \text{oxidation number of N}\right)\\ +\left(\text{Number of O atoms}\times \text{oxidation number of O}\right)\end{array}\right]$

Since, the charge on ${\text{NO}}_2{}^{-}$ is $-1$.

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

$\begin{array}{c}{\text{Charge on NO}}_2{}^{-}=\left[\left(x\right)+\left(2\times \left(-2\right)\right)\right]\\ -1=\left[x-4\right]\\ x=\underset{\_}{+3}\end{array}$

Thus, the oxidation number of $\text{N}$ in ${\text{NO}}_2{}^{-}$ is $\underset{\_}{+3}$.

Hence, the oxidation number of $\text{N}$ in ${\text{NH}}_4{}^{+}$ is $\underset{\_}{-3}$ and the oxidation number of $\text{N}$ in ${\text{NO}}_2{}^{-}$ is $\underset{\_}{+3}$ in the second reaction of nitrogen fixation.

Therefore, The oxidation number of molecular ${\text{N}}_{\text{2}}$ is always be $\underset{\_}{0}$ and the oxidation number of $\text{N}$ in ${\text{NH}}_3$ is $\underset{\_}{-3}$ in the first reaction of nitrogen fixation. The oxidation number of $\text{N}$ in ${\text{NH}}_4{}^{+}$ is $\underset{\_}{-3}$ and the oxidation number of $\text{N}$ in ${\text{NO}}_2{}^{-}$ is $\underset{\_}{+3}$ in the second nitrification reaction.

(b)

Interpretation Introduction

To determine: The compounds or ions that are being reduced in each reaction.

Answer to Problem 4.140AP

Solution

In the first reaction of nitrogen fixation the molecule ${\text{N}}_{\text{2}}\left(g\right)$ reduced to ${\text{NH}}_3\left(aq\right)$ and ${\text{H}}^{+}\left(aq\right)$ ion reduced to ${\text{H}}_{\text{2}}\left(g\right)$ molecule.

In the nitrification reaction ${\text{M}}^{\text{3+}}$ ion reduced to ${\text{M}}^{2+}$ ion.

Explanation of Solution

Explanation

The first given reaction of nitrogen fixation is,

${\text{N}}_{\text{2}}\left(g\right)+{\text{H}}^{\text{+}}\left(aq\right)+{\text{M}}^{\text{2+}}\left(aq\right)\to {\text{NH}}_{\text{3}}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)+{\text{M}}^{\text{3+}}\left(aq\right)$

Where,

• $\text{M}$ is the transition element such as iron.

In the above reaction the oxidation number of molecular ${\text{N}}_{\text{2}}$ is $0$ and the oxidation number of $\text{N}$ in ${\text{NH}}_3$ is $-3$ as shown above.

Thus, there is a decrease in the oxidation state. Hence, ${\text{N}}_{\text{2}}\left(g\right)$ reduced to ${\text{NH}}_3\left(aq\right)$.

The oxidation state of ${\text{H}}^{\text{+}}$ ion is $+1$ and the oxidation state of molecular ${\text{H}}_{\text{2}}$ is always $0$.

Thus, there is a decrease in the oxidation state. Hence, ${\text{H}}^{+}\left(aq\right)$ ion reduced to ${\text{H}}_{\text{2}}\left(g\right)$ molecule.

The second given nitrification reaction is,

${\text{NH}}_4{}^{+}\left(aq\right)+{\text{M}}^{\text{3+}}\left(aq\right)\to {\text{NO}}_2{}^{-}\left(aq\right)+{\text{M}}^{\text{2+}}\left(aq\right)$

Where,

• $\text{M}$ is the transition element such as iron.

Here, the oxidation number of $\text{N}$ in ${\text{NH}}_4{}^{+}$ is $-3$ and the oxidation number of $\text{N}$ in ${\text{NO}}_2{}^{-}$ is $+3$ in the second reaction of nitrogen fixation.

Thus, there is increase in the oxidation state. Hence, oxidation process is occurring.

The oxidation state of ${\text{M}}^{\text{3+}}$ is $+3$ and oxidation state of ${\text{M}}^{2+}$ is $+2$. Thus, there is decrease in the oxidation state of $\text{M}$. Hence, ${\text{M}}^{\text{3+}}$ ion reduced to ${\text{M}}^{2+}$ ion.

Therefore, in the first reaction of nitrogen fixation the molecule ${\text{N}}_{\text{2}}\left(g\right)$ reduced to ${\text{NH}}_3\left(aq\right)$ and ${\text{H}}^{+}\left(aq\right)$ ion reduced to ${\text{H}}_{\text{2}}\left(g\right)$ molecule. In the nitrification reaction ${\text{M}}^{\text{3+}}$ ion reduced to ${\text{M}}^{2+}$ ion.

(c)

Interpretation Introduction

To determine: The balanced equations of nitrogen fixation.

Answer to Problem 4.140AP

Solution

The balanced equation of nitrogen fixation is,

${\text{N}}_{\text{2}}\left(g\right)+8{\text{H}}^{\text{+}}\left(aq\right)+{\text{M}}^{\text{2+}}\left(aq\right)+7{\text{e}}^{-}\to 2{\text{NH}}_{\text{3}}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)+{\text{M}}^{\text{3+}}\left(aq\right)$

The balanced equation of nitrification reaction is,

${\text{NH}}_4{}^{+}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(aq\right){\text{+M}}^{\text{3+}}\left(aq\right)+3{\text{e}}^{-}\to {\text{NO}}_2{}^{-}\left(aq\right)+{\text{M}}^{\text{2+}}\left(aq\right)+4{\text{H}}_{\text{2}}\left(g\right)$

Explanation of Solution

Explanation

The first given unbalanced reaction of nitrogen fixation is,

${\text{N}}_{\text{2}}\left(g\right)+{\text{H}}^{\text{+}}\left(aq\right)+{\text{M}}^{\text{2+}}\left(aq\right)\to {\text{NH}}_{\text{3}}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)+{\text{M}}^{\text{3+}}\left(aq\right)$

Where,

• $\text{M}$ is the transition element such as iron.

The number of nitrogen atom is balanced by placing coefficient of $2$ before ${\text{NH}}_3\left(aq\right)$ to the product side of the equation. The equation becomes as,

${\text{N}}_{\text{2}}\left(g\right)+{\text{H}}^{\text{+}}\left(aq\right)+{\text{M}}^{\text{2+}}\left(aq\right)\to 2{\text{NH}}_{\text{3}}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)+{\text{M}}^{\text{3+}}\left(aq\right)$

Now, to balance the hydrogen atom to both sides place coefficient of $8$ in front of ${\text{H}}^{+}$ ions to the reactant side of the equation as,

${\text{N}}_{\text{2}}\left(g\right)+8{\text{H}}^{\text{+}}\left(aq\right)+{\text{M}}^{\text{2+}}\left(aq\right)\to 2{\text{NH}}_{\text{3}}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)+{\text{M}}^{\text{3+}}\left(aq\right)$

There is unequal net charge on the equation. There is total $+10$ charge on the reactant side from eight ${\text{H}}^{+}$ ions and two from ${\text{M}}^{2+}$ ion and on the product side there is total of only $+3$ charge from ${\text{M}}^{3+}$ ion. Therefore, to balance the charge add $7$ electrons to the reactant side of the equation as,

${\text{N}}_{\text{2}}\left(g\right)+8{\text{H}}^{\text{+}}\left(aq\right)+{\text{M}}^{\text{2+}}\left(aq\right)+7{\text{e}}^{-}\to 2{\text{NH}}_{\text{3}}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)+{\text{M}}^{\text{3+}}\left(aq\right)$

Hence, the balanced equation of nitrogen fixation is,

${\text{N}}_{\text{2}}\left(g\right)+8{\text{H}}^{\text{+}}\left(aq\right)+{\text{M}}^{\text{2+}}\left(aq\right)+7{\text{e}}^{-}\to 2{\text{NH}}_{\text{3}}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)+{\text{M}}^{\text{3+}}\left(aq\right)$

The second given nitrification reaction is,

${\text{NH}}_4{}^{+}\left(aq\right)+{\text{M}}^{\text{3+}}\left(aq\right)\to {\text{NO}}_2{}^{-}\left(aq\right)+{\text{M}}^{\text{2+}}\left(aq\right)$

Where,

• $\text{M}$ is the transition element such as iron.

The numbers of oxygen atoms are balanced by adding water molecule to the reactant side as,

${\text{NH}}_4{}^{+}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(aq\right)+{\text{M}}^{3+}\left(aq\right)\to {\text{NO}}_2{}^{-}\left(aq\right)+{\text{M}}^{2+}\left(aq\right)$

The numbers of oxygen atoms are balanced by placing coefficient of $2$ in front of water molecule to the reactant side as,

${\text{NH}}_4{}^{+}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(aq\right)+{\text{M}}^{3+}\left(aq\right)\to {\text{NO}}_2{}^{-}\left(aq\right)+{\text{M}}^{2+}\left(aq\right)$

To balance hydrogen ions add four molecules of ${\text{H}}_2$ to the product side of the equation as,

${\text{NH}}_4{}^{+}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(aq\right)+{\text{M}}^{3+}\left(aq\right)\to {\text{NO}}_2{}^{-}\left(aq\right)+{\text{M}}^{2+}\left(aq\right)+4{\text{H}}_{\text{2}}\left(g\right)$

The numbers of all the atoms are balanced now. There is unequal distribution of charge both the side as there is total of $+4$  charge on the reactant side and there is only $+1$ charge on the product side. Thus, to balance the charge on both the sides add $3$ electrons to the reactant side of the equation as,

${\text{NH}}_4{}^{+}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(aq\right){\text{+M}}^{\text{3+}}\left(aq\right)+3{\text{e}}^{-}\to {\text{NO}}_2{}^{-}\left(aq\right)+{\text{M}}^{\text{2+}}\left(aq\right)+4{\text{H}}_{\text{2}}\left(g\right)$.

Therefore, the balanced equation of nitrification reaction is,

${\text{NH}}_4{}^{+}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(aq\right){\text{+M}}^{\text{3+}}\left(aq\right)+3{\text{e}}^{-}\to {\text{NO}}_2{}^{-}\left(aq\right)+{\text{M}}^{\text{2+}}\left(aq\right)+4{\text{H}}_{\text{2}}\left(g\right)$

Conclusion

1. a) The oxidation number of molecular ${\text{N}}_{\text{2}}$ is always be $\underset{\_}{0}$ and the oxidation number of $\text{N}$ in ${\text{NH}}_3$ is $\underset{\_}{-3}$ in the first reaction of nitrogen fixation. The oxidation number of $\text{N}$ in ${\text{NH}}_4{}^{+}$ is $\underset{\_}{-3}$ and the oxidation number of $\text{N}$ in ${\text{NO}}_2{}^{-}$ is $\underset{\_}{+3}$ in the second reaction of nitrogen fixation.
2. b) In the first reaction of nitrogen fixation the molecule ${\text{N}}_{\text{2}}\left(g\right)$ reduced to ${\text{NH}}_3\left(aq\right)$ and ${\text{H}}^{+}\left(aq\right)$ ion reduced to ${\text{H}}_{\text{2}}\left(g\right)$ molecule. In the second reaction of nitrogen fixation ${\text{M}}^{\text{3+}}$ ion reduced to ${\text{M}}^{2+}$ ion.
3. c) The balanced equation of nitrogen fixation is,

   ${\text{N}}_{\text{2}}\left(g\right)+8{\text{H}}^{\text{+}}\left(aq\right)+{\text{M}}^{\text{2+}}\left(aq\right)+7{\text{e}}^{-}\to 2{\text{NH}}_{\text{3}}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)+{\text{M}}^{\text{3+}}\left(aq\right)$.

   The balanced equation of nitrification reaction is,

   ${\text{NH}}_4{}^{+}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(aq\right){\text{+M}}^{\text{3+}}\left(aq\right)+3{\text{e}}^{-}\to {\text{NO}}_2{}^{-}\left(aq\right)+{\text{M}}^{\text{2+}}\left(aq\right)+4{\text{H}}_{\text{2}}\left(g\right)$