Chapter 13, Problem 1DE

Which element is oxidized, and which is reduced in the following reactions?

1. N₂(g) + 3 H2(g) → 2 NH₃(g)
3Fe(NO3)2(aq) + 2Al(s) →

2. 3 Fe(s) + 2 Al(NO₃)₃(aq)
3. Cl₂(aq) + 2 NaI(aq) →I₂(aq) + 2 NaC1(a9)
4. PbS(s) + 4H₂O₂(aq) → PbSO₄(s) + 4H₂O(l)

Interpretation Introduction

To determine: The elements that are getting reduced and oxidized.

Answer to Problem 1DE

Solution: In the given reaction, nitrogen $\left(\text{N}\right)$ is getting reduced and hydrogen $\left(\text{H}\right)$ is getting oxidized.

Explanation of Solution

The given reaction is,

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

The loss of the electrons by the substance is known as oxidation and the gain of the electrons by the substance is known as reduction. The substance which donates the electrons is said to be oxidized whereas the substance which accepts the electrons is said to be reduced.

Oxidation number represents the oxidation state of the ions. An increase in oxidation number occurs by the loss of electrons while the oxidation number decreases by the gaining of electrons.

The rules for assigning the oxidation number are,

• The oxidation number of a free element is always taken as zero.
• The oxidation number of $\text{H}$ is $+1$ , but it becomes $-1$ when combined with less electronegative atom.
• The oxidation number of $\text{O}$ is $-2$ but in peroxides is $-1$ .
• The oxidation number of group 1 element is taken as $+1$ .
• The oxidation number of group 2 element is taken as $+2$ .
• The oxidation number of group 17 element is taken as $-1$ .
• In a neutral compound, the sum of the oxidation number of all the atoms is taken as $0$ .
• For a polyatomic ion, the sum of the oxidation numbers is equal to the charge on the ion.

The oxidation number of a free element is always taken as zero. Therefore, the oxidation state of ${\text{N}}_{\text{2}}\left(g\right)$ and ${\text{H}}_{\text{2}}\left(g\right)$ is $0$ .

${\text{NH}}_{\text{3}}$ is a neutral compound. In a neutral compound, the sum of the oxidation number of all the atoms is taken as $0$ .

Therefore,

  $O_{\text{N}}+3O_{\text{H}}=0$

Here,

• $O_{\text{N}}$ is the oxidation state of $\text{N}$ .
• $O_{\text{H}}$ is the oxidation state of $\text{H}$ .

Substitute the value $+1$ for the oxidation number of $\text{H}$ .

 $\begin{array}{c}{O}_{\text{N}}+3\left(1\right)=0\\ O_{\text{N}}=-3\end{array}$

Hence, the oxidation number of $\text{N}$ in ${\text{NH}}_{\text{3}}$ is $-3$ and the oxidation number of $\text{H}$ in ${\text{NH}}_{\text{3}}$ is $+1$ .

In this reaction, the oxidation number of hydrogen $\left(\text{H}\right)$ is increased from $0$ to $+1$ . Hence, hydrogen atom $\left(\text{H}\right)$ is getting oxidized in this reaction.

The oxidation number of nitrogen $\left(\text{N}\right)$ changes from $0$ to $-3$ . Therefore, a decrease in oxidation number indicates that the nitrogen $\left(\text{N}\right)$ is getting reduced in this reaction.

Conclusion

The oxidation number of hydrogen $\left(\text{H}\right)$ is increased from $0$ to $+1$ . Hence, hydrogen atom $\left(\text{H}\right)$ is getting oxidized in this reaction.

The oxidation number of nitrogen $\left(\text{N}\right)$ is decreased from $0$ to $-3$ . Therefore, nitrogen $\left(\text{N}\right)$ is getting reduced in this reaction.

Interpretation Introduction

To determine: The elements that are getting reduced and oxidized.

Answer to Problem 1DE

Solution: In the given reaction, iron $\left(\text{Fe}\right)$ is getting reduced and aluminum $\left(\text{Al}\right)$ is getting oxidized.

Explanation of Solution

The given reaction is,

$\text{3Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(aq\right)+2\text{Al}\left(s\right)\stackrel{}{\to }3\text{Fe}\left(s\right)+2\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\left(aq\right)$

The loss of the electrons by the substance is known as oxidation and the gain of the electrons by the substance is known as reduction. The substance which donates the electrons is said to be oxidized whereas the substance which accepts the electrons is said to be reduced.

Oxidation number represents the oxidation state of the ions. An increase in oxidation number occurs by the loss of electrons while the oxidation number decreases by the gaining of electrons.

The oxidation number of a free element is always taken as zero. Therefore, the oxidation state of $\text{Fe}\left(s\right)$ and $\text{Al}\left(s\right)$ is $0$ .

$\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ and $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ are the neutral compounds. In a neutral compound, the sum of the oxidation number of all the atoms is taken as $0$ .

The oxidation number of $\text{Fe}$ in $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is calculated by the formula,

Therefore,

$2O_{{\text{NO}}_3}+O_{\text{Fe}}=0$

Here,

• $O_{{\text{NO}}_3}$ is the oxidation state of ${\text{NO}}_3$ .
• $O_{\text{Fe}}$ is the oxidation state of $\text{Fe}$ .

Substitute the value $-1$ for the oxidation number of nitrate ion $\left({\text{NO}}_{\text{3}}\right)$ .

 $\begin{array}{c}2\left(-1\right)+O_{\text{Fe}}=0\\ O_{\text{Fe}}=+2\end{array}$

Hence, the oxidation number of $\text{Fe}$ in $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $+2$ .

The oxidation number of $\text{Al}$ in $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_3$ is calculated by the formula,

Therefore,

  $3O_{{\text{NO}}_3}+O_{\text{Al}}=0$

Here,

• $O_{{\text{NO}}_3}$ is the oxidation state of ${\text{NO}}_3$ .
• $O_{\text{Al}}$ is the oxidation state of $\text{Al}$ .

Substitute the value $-1$ for the oxidation number of nitrate ion $\left({\text{NO}}_{\text{3}}\right)$ .

$\begin{array}{c}3\left(-1\right)+O_{\text{Al}}=0\\ O_{\text{Al}}=+3\end{array}$

Hence, the oxidation number of $\text{Al}$ in $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_3$ is $+3$ .

In this reaction, the oxidation number of aluminum $\left(\text{Al}\right)$ is increased from $0$ to $+3$ . Hence, aluminum $\left(\text{Al}\right)$ is getting oxidized in this reaction.

The oxidation number of iron $\left(\text{Fe}\right)$ changes from $+2$ to $0$ . Therefore, a decrease in oxidation number indicates that iron $\left(\text{Fe}\right)$ is getting reduced in this reaction.

Conclusion

The oxidation number of aluminum $\left(\text{Al}\right)$ is increased from $0$ to $+3$ . Hence, aluminum $\left(\text{Al}\right)$ is getting oxidized in this reaction.

The oxidation number of iron $\left(\text{Fe}\right)$ is decreased from $+2$ to $0$ . Therefore, iron $\left(\text{Fe}\right)$ is getting reduced in this reaction.

Interpretation Introduction

To determine: The elements that are getting reduced and oxidized.

Answer to Problem 1DE

Solution: In the given reaction, chlorine $\left(\text{Cl}\right)$ is getting reduced and iodine $\left(\text{I}\right)$ is getting oxidized.

Explanation of Solution

The given reaction is,

${\text{Cl}}_2\left(aq\right)+2\text{NaI}\left(aq\right)\stackrel{}{\to }{\text{I}}_{\text{2}}\left(aq\right)+2\text{NaCl}\left(aq\right)$

The loss of the electrons by the substance is known as oxidation and the gain of the electrons by the substance is known as reduction. The substance which donates the electrons is said to be oxidized whereas the substance which accepts the electrons is said to be reduced.

Oxidation number represents the oxidation state of the ions. An increase in oxidation number occurs by the loss of electrons while the oxidation number decreases by the gaining of electrons.

The oxidation number of a free element is always taken as zero. Therefore, the oxidation state of ${\text{Cl}}_2\left(aq\right)$ and ${\text{I}}_2\left(aq\right)$ is $0$ .

$\text{NaI}$ and $\text{NaCl}$ are the neutral compounds. In a neutral compound, the sum of the oxidation number of all the atoms is taken as $0$ .

The oxidation number of a neutral atom $\text{NaCl}$ is calculated by the formula,

Therefore,

$O_{\text{Na}}+O_{\text{Cl}}=0$

Here,

• $O_{\text{Na}}$ is the oxidation state of $\text{Na}$ .
• $O_{\text{Cl}}$ is the oxidation state of $\text{Cl}$ .

Substitute the value $+1$ for the oxidation number of $\text{Na}$ .

$\begin{array}{c}\left(+1\right)+O_{\text{Cl}}=0\\ O_{\text{Cl}}=-1\end{array}$

Hence, the oxidation number of $\text{Cl}$ in $\text{NaCl}$ is $-1$ .

The oxidation number of a neutral atom $\text{NaI}$ is calculated by the formula,

Therefore,

$O_{\text{Na}}+O_{\text{I}}=0$

Here,

• $O_{\text{Na}}$ is the oxidation state of $\text{Na}$ .
• $O_{\text{I}}$ is the oxidation state of $\text{I}$ .

Substitute the value $+1$ for the oxidation number of $\text{Na}$ .

$\begin{array}{c}\left(+1\right)+O_{\text{I}}=0\\ O_{\text{I}}=-1\end{array}$

Hence, the oxidation number of $\text{I}$ in $\text{NaI}$ is $-1$ .

In this reaction, the oxidation number of iodine $\left(\text{I}\right)$ is increased from $-1$ to $0$ . Hence, iodine $\left(\text{I}\right)$ is getting oxidized in this reaction.

The oxidation number of chlorine $\left(\text{Cl}\right)$ changes from $0$ to $-1$ . Therefore, a decrease in oxidation number indicates that chlorine $\left(\text{Cl}\right)$ is getting reduced in this reaction.

Conclusion

The oxidation number of iodine $\left(\text{I}\right)$ is increased from $-1$ to $0$ . Hence, iodine $\left(\text{I}\right)$ is getting oxidized in this reaction.

The oxidation number of chlorine $\left(\text{Cl}\right)$ is decreased from $0$ to $-1$ . Therefore, chlorine $\left(\text{Cl}\right)$ is getting reduced in this reaction.

Interpretation Introduction

To determine: The elements that are getting reduced and oxidized.

Answer to Problem 1DE

Solution: In the given reaction, oxygen $\left(\text{O}\right)$ is getting reduced and sulfur $\left(\text{S}\right)$ is getting oxidized.

Explanation of Solution

The given reaction is,

$\text{PbS}\left(s\right)+4{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)\stackrel{}{\to }{\text{PbSO}}_4\left(s\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The loss of the electrons by the substance is known as oxidation and the gain of the electrons by the substance is known as reduction. The substance which donates the electrons is said to be oxidized whereas the substance which accepts the electrons is said to be reduced.

Oxidation number represents the oxidation state of the ions. An increase in oxidation number occurs by the loss of electrons while the oxidation number decreases by the gaining of electrons.

$\text{PbS}$ , ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ , ${\text{PbSO}}_{\text{4}}$ and ${\text{H}}_{\text{2}}\text{O}$ are the neutral compounds. In a neutral compound, the sum of the oxidation number of all the atoms is taken as $0$ .

The oxidation number of a neutral atom $\text{PbS}$ is calculated by the formula,

Therefore,

$O_{\text{Pb}}+O_{\text{S}}=0$

Here,

• $O_{\text{Pb}}$ is the oxidation state of $\text{Pb}$ .
• $O_{\text{S}}$ is the oxidation state of $\text{S}$ .

Substitute the value $+2$ for the oxidation number of $\text{Pb}$ .

$\begin{array}{c}\left(+2\right)+O_{\text{S}}=0\\ O_{\text{S}}=-2\end{array}$

Hence, the oxidation number of $\text{S}$ in $\text{PbS}$ is $-2$ .

The oxidation number of $\text{S}$ in ${\text{PbSO}}_4$ is calculated by the formula,

Therefore,

$O_{\text{Pb}}+O_{\text{S}}+4O_{\text{O}}=0$

Here,

• $O_{\text{Pb}}$ is the oxidation state of $\text{Pb}$ .
• $O_{\text{S}}$ is the oxidation state of $\text{S}$ .
• $O_{\text{O}}$ is the oxidation state of $\text{O}$ .

Substitute the value $+2$ for the oxidation number of $\text{Pb}$ and $-2$ for the oxidation number of $\text{O}$ .

$\begin{array}{c}\left(+2\right)+O_{\text{S}}+4\left(-2\right)=0\\ O_{\text{S}}=+6\end{array}$

Hence, the oxidation number of $\text{S}$ in ${\text{PbSO}}_4$ is $+6$ .

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

Therefore,

$2O_{\text{H}}+O_{\text{O}}=0$

Here,

• $O_{\text{H}}$ is the oxidation state of $\text{H}$ .
• $O_{\text{O}}$ is the oxidation state of $\text{O}$ .

Substitute the value $+1$ for the oxidation number of $\text{H}$ .

$\begin{array}{c}2\left(+1\right)+O_{\text{O}}=0\\ O_{\text{O}}=-2\end{array}$

Hence, the oxidation number of $\text{O}$ in ${\text{H}}_{\text{2}}\text{O}$ is $-2$ .

The oxidation number of $\text{O}$ in ${\text{H}}_{\text{2}}{\text{O}}_2$ is calculated by the formula,

Therefore,

$2O_{\text{H}}+2O_{\text{O}}=0$

Here,

• $O_{\text{H}}$ is the oxidation state of $\text{H}$ .
• $O_{\text{O}}$ is the oxidation state of $\text{O}$ .

Substitute the value $+1$ for the oxidation number of $\text{H}$ .

$\begin{array}{c}2\left(+1\right)+2O_{\text{O}}=0\\ O_{\text{O}}=-1\end{array}$

Hence, the oxidation number of $\text{O}$ in ${\text{H}}_{\text{2}}{\text{O}}_2$ is $-1$ .

In this reaction, the oxidation number of sulfur $\left(\text{S}\right)$ is increased from $-2$ to $+6$ . Hence sulfur $\left(\text{S}\right)$ is getting oxidized in this reaction.

The oxidation number of oxygen $\left(\text{O}\right)$ changes from $-1$ to $-2$ . Therefore, a decrease in oxidation number indicates that oxygen $\left(\text{O}\right)$ is getting reduced in this reaction.

Conclusion

The oxidation number of sulfur $\left(\text{S}\right)$ is increased from $-2$ to $+6$ . Hence sulfur $\left(\text{S}\right)$ is getting oxidized in this reaction.

The oxidation number of oxygen $\left(\text{O}\right)$ is decreased from $-1$ to $-2$ . Therefore, oxygen $\left(\text{O}\right)$ is getting reduced in this reaction.