Chapter 17, Problem 7QRT

(a)

Interpretation Introduction

Interpretation:

The oxidation number has to be assigned to each atom of the given reaction.  The substance that is oxidized, reduced, oxidizing agent and reducing agent have to be stated.

Concept Introduction:

Redox reactions are the reactions in which oxidation and reduction occur simultaneously.  Loss of electrons or gain of oxygen is termed as oxidation.  On the other hand, addition of electrons or hydrogen is known as reduction.  Increase in oxidation number signifies oxidation and decrease in oxidation number signifies reduction.

Answer to Problem 7QRT

The oxidation number of $\text{Fe}$ changes from $\underset{\_}{0}$ to $\underset{\_}{+2}$.  So, $\text{Fe}\left(\text{s}\right)$ gets oxidized and acts as a reducing agent.

The oxidation number of $\text{Br}$ changes from $\underset{\_}{0}$ to $\underset{\_}{-1}$.  So, $\text{Br}\left(l\right)$ gets reduced and acts as an oxidizing agent

Explanation of Solution

The given reaction is shown below.

    $\text{Fe}\left(\text{s}\right)+\text{Br}\left(l\right)\to {\text{FeBr}}_2\left(\text{s}\right)$

The oxidation number of any element in its native form is zero.  So, the oxidation number of $\text{Fe}\left(\text{s}\right)$ and $\text{Br}\left(l\right)$ is zero.

In almost all compounds oxidation number of bromine is $-1$.  Assume the oxidation number of iron to be x.

Apply charge balance formula in ${\text{FeBr}}_2$ as shown below.

  $\begin{array}{c}\left(\text{2}\times \text{Oxidation}\text{state}\text{of}\text{Br}\right)+\text{Oxidation}\text{state}\text{of}\text{Fe}=\text{Overall}\text{charge}\text{on}{\text{FeBr}}_2\\ 2\left(-1\right)+x=0\\ x=+2\end{array}$.

Therefore, the oxidation number of iron in ${\text{FeBr}}_2$ is $\underset{\_}{+2}$.

The element that gets oxidized reduces the other one and is known as reducing agent.  The element that gets reduced oxidizes the other one and is known as oxidizing agent.

In the given reaction, the oxidation number of $\text{Fe}$ changes from $0$ to $+2$.  So, $\text{Fe}\left(\text{s}\right)$ gets oxidized and acts as a reducing agent.

The oxidation number of $\text{Br}$ changes from $0$ to $-1$.  So, $\text{Br}\left(l\right)$ gets reduced and acts as an oxidizing agent.

(b)

Interpretation Introduction

Interpretation:

The oxidation number has to be assigned to each atom of the given reaction.  The substance that is oxidized, reduced, oxidizing agent and reducing agent have to be stated.

Concept Introduction:

Refer to part (a).

Answer to Problem 7QRT

The oxidation number of $\text{Al}$ changes from $\underset{\_}{0}$ to $\underset{\_}{+3}$.  So, $\text{Al}\left(\text{s}\right)$ gets oxidized and acts as a reducing agent.

The oxidation number of $\text{Cl}$ changes from $\underset{\_}{0}$ to $\underset{\_}{-1}$.  So, ${\text{Cl}}_2\left(\text{g}\right)$ gets reduced and acts as an oxidizing agent

Explanation of Solution

The given reaction is shown below.

    $\text{2Al}\left(\text{s}\right)+{\text{3Cl}}_2\left(\text{g}\right)\to 2{\text{AlCl}}_3\left(\text{s}\right)$

The oxidation number of any element in its native form is zero.  So, the oxidation number of $\text{Al}\left(\text{s}\right)$ and ${\text{Cl}}_2\left(\text{g}\right)$ is zero.

In almost all compounds oxidation number of chlorine is $-1$.  Assume the oxidation number of aluminium to be x.

Apply charge balance formula in ${\text{AlCl}}_3$ as shown below.

  $\begin{array}{c}\left(\text{3}\times \text{Oxidation}\text{state}\text{of}\text{Cl}\right)+\text{Oxidation}\text{state}\text{of}\text{Al}=\text{Overall}\text{charge}\text{on}{\text{AlCl}}_3\\ 3\left(-1\right)+x=0\\ x=+3\end{array}$.

Therefore, the oxidation number of aluminium in ${\text{AlCl}}_3$ is $\underset{\_}{+3}$.

The element that gets oxidized reduces the other one and is known as reducing agent.  The element that gets reduced oxidizes the other one and is known as oxidizing agent.

In the given reaction, the oxidation number of $\text{Al}$ changes from $0$ to $+3$.  So, $\text{Al}\left(\text{s}\right)$ gets oxidized and acts as a reducing agent.

The oxidation number of $\text{Cl}$ changes from $0$ to $-1$.  So, ${\text{Cl}}_2\left(\text{g}\right)$ gets reduced and acts as an oxidizing agent.

(c)

Interpretation Introduction

Interpretation:

The oxidation number has to be assigned to each atom of the given reaction.  The substance that is oxidized, reduced, oxidizing agent and reducing agent have to be stated.

Concept Introduction:

Refer to part (a).

Answer to Problem 7QRT

The oxidation number of sulfur changes from $\underset{\_}{+6}$ to $\underset{\_}{-2}$.  So, ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ gets reduced and acts as an oxidizing agent.

The oxidation number of $\text{I}$ changes from $\underset{\_}{-1}$ to $\underset{\_}{0}$.  So, $\text{HI}$ gets oxidized and acts as a reducing agent.

Explanation of Solution

The given reaction is shown below.

    $\text{8HI}\left(\text{aq}\right)+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)\to {\text{H}}_{\text{2}}\text{S}\left(\text{aq}\right)+4{\text{I}}_{\text{2}}\left(\text{s}\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The oxidation number of any element in its native form is zero.  So, the oxidation number of ${\text{I}}_{\text{2}}\left(\text{s}\right)$ is zero.

In almost all compounds oxidation number of iodine is $-1$, hydrogen is $+1$ and oxygen is $-2$.  Assume the oxidation number of sulfur to be x.

Apply charge balance formula in ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ as shown below.

  $\begin{array}{c}\left(\begin{array}{c}\left(\text{2}\times \text{Oxidation}\text{state}\text{of}\text{H}\right)+4\times \text{Oxidation}\text{state}\text{of}\text{O}\\ +\text{Oxidation}\text{state}\text{of}\text{S}\end{array}\right)=\text{Overall}\text{charge}\text{on}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\\ 2\left(+1\right)+4\left(-2\right)+x=0\\ x=8-2\\ =6\end{array}$.

Therefore, the oxidation number of sulfur in ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\underset{\_}{+6}$.

Apply charge balance formula in ${\text{H}}_{\text{2}}\text{S}$ as shown below.

  $\begin{array}{c}\left(\left(\text{2}\times \text{Oxidation}\text{state}\text{of}\text{H}\right)+\text{Oxidation}\text{state}\text{of}\text{S}\right)=\text{Overall}\text{charge}\text{on}{\text{H}}_{\text{2}}\text{S}\\ 2\left(+1\right)+x=0\\ x=-2\end{array}$.

Therefore, the oxidation number of sulfur in ${\text{H}}_{\text{2}}\text{S}$ is $\underset{\_}{-2}$.

The element that gets oxidized reduces the other one and is known as reducing agent.  The element that gets reduced oxidizes the other one and is known as oxidizing agent.

In the given reaction, the oxidation number of sulfur changes from $+6$ to $-2$.  So, ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ gets reduced and acts as an oxidizing agent.

The oxidation number of $\text{I}$ changes from $-1$ to $0$.  So, $\text{HI}$ gets oxidized and acts as a reducing agent.

(d)

Interpretation Introduction

Interpretation:

The oxidation number has to be assigned to each atom of the given reaction.  The substance that is oxidized, reduced, oxidizing agent and reducing agent have to be stated.

Concept Introduction:

Refer to part (a).

Answer to Problem 7QRT

The oxidation number of ${\text{Fe}}^{\text{2+}}$ changes from $\underset{\_}{+2}$ to $\underset{\_}{+3}$.  So, ${\text{Fe}}^{\text{2+}}$ gets oxidized and acts as a reducing agent.

The oxidation number of $\text{O}$ changes from $\underset{\_}{-1}$ to $\underset{\_}{-2}$.  So, ${\text{H}}_2{\text{O}}_{\text{2}}$ gets reduced and acts as an oxidizing agent.

Explanation of Solution

The given reaction is shown below.

    ${\text{H}}_2{\text{O}}_{\text{2}}\left(\text{aq}\right)+{\text{2Fe}}^{2+}\left(\text{aq}\right)+{\text{2H}}^{+}\left(\text{aq}\right)\to {\text{2Fe}}^{3+}\left(\text{aq}\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The oxidation number of any element carrying charge is equal to that charge.  So, the oxidation number of ${\text{H}}^{\text{+}}$, ${\text{Fe}}^{\text{2+}}$ and ${\text{Fe}}^{\text{3+}}$ is $\underset{\_}{+1},\underset{\_}{+2}$ and $\underset{\_}{+3}$ respectively.

In almost all compounds oxidation number of hydrogen is $+1$.  Assume the oxidation number of oxygen to be x.

Apply charge balance formula in ${\text{H}}_{\text{2}}{\text{O}}_2$ as shown below.

  $\begin{array}{c}\left(\left(\text{2}\times \text{Oxidation}\text{state}\text{of}\text{H}\right)+2\times \text{Oxidation}\text{state}\text{of}\text{O}\right)=\text{Overall}\text{charge}\text{on}{\text{H}}_{\text{2}}{\text{O}}_2\\ 2\left(+1\right)+2x=0\\ x=-\frac{2}{2}\\ =-1\end{array}$.

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

Apply charge balance formula in ${\text{H}}_{\text{2}}\text{O}$ as shown below.

  $\begin{array}{c}\left(\left(\text{2}\times \text{Oxidation}\text{state}\text{of}\text{H}\right)+\text{Oxidation}\text{state}\text{of}\text{O}\right)=\text{Overall}\text{charge}\text{on}{\text{H}}_{\text{2}}\text{O}\\ 2\left(+1\right)+x=0\\ x=-2\end{array}$.

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

The element that gets oxidized reduces the other one and is known as reducing agent.  The element that gets reduced oxidizes the other one and is known as oxidizing agent.

In the given reaction, the oxidation number of ${\text{Fe}}^{\text{2+}}$ changes from $+2$ to $+3$.  So, ${\text{Fe}}^{\text{2+}}$ gets oxidized and acts as a reducing agent.

The oxidation number of $\text{O}$ changes from $-1$ to $-2$.  So, ${\text{H}}_2{\text{O}}_{\text{2}}$ gets reduced and acts as an oxidizing agent.

(e)

Interpretation Introduction

Interpretation:

The oxidation number has to be assigned to each atom of the given reaction.  The substance that is oxidized, reduced, oxidizing agent and reducing agent have to be stated.

Concept Introduction:

Refer to part (a).

Answer to Problem 7QRT

The oxidation number of ${\text{Fe}}^{\text{2+}}$ changes from $\underset{\_}{+2}$ to $\underset{\_}{+3}$.  So, $\text{FeS}$ gets oxidized and acts as a reducing agent.

In the given reaction, the oxidation number of $\text{S}$ changes from $\underset{\_}{-2}$ to $\underset{\_}{+6}$.  So, $\text{FeS}$ gets oxidized and acts as a reducing agent.

The oxidation number of $\text{N}$ changes from $\underset{\_}{+3}$ to $\underset{\_}{+2}$.  So, ${\text{NO}}_3^{-}$ gets reduced and acts as an oxidizing agent.

Explanation of Solution

The given reaction is shown below.

    $\text{FeS}\left(\text{s}\right)+3{\text{NO}}_3^{-}\left(\text{aq}\right)+4{\text{H}}^{+}\left(\text{aq}\right)\to 3\text{NO}\left(\text{g}\right)+{\text{SO}}_4^{2-}\left(\text{aq}\right)+{\text{Fe}}^{\text{3+}}\left(\text{aq}\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)$

The oxidation number of any element carrying charge is equal to that charge.  So, the oxidation number of ${\text{H}}^{\text{+}}$ and ${\text{Fe}}^{\text{3+}}$ is $\underset{\_}{+1}$ and $\underset{\_}{+3}$ respectively.

In almost all compounds oxidation number of hydrogen is $+1$ and oxygen is $-2$.  Assume the oxidation number of nitrogen to be x.

Apply charge balance formula in ${\text{NO}}_3^{-}$ as shown below.

  $\begin{array}{c}\left(\text{Oxidation}\text{state}\text{of}\text{N}+3\times \text{Oxidation}\text{state}\text{of}\text{O}\right)=\text{Overall}\text{charge}\text{on}{\text{NO}}_3^{-}\\ x+2\left(-2\right)=-1\\ x=-1+4\\ =3\end{array}$.

Therefore, the oxidation number of nitrogen in ${\text{NO}}_3^{-}$ is $\underset{\_}{+3}$.

Apply charge balance formula in $\text{NO}$ as shown below.

  $\begin{array}{c}\left(\left(\text{Oxidation}\text{state}\text{of}\text{N}\right)+\text{Oxidation}\text{state}\text{of}\text{O}\right)=\text{Overall}\text{charge}\text{on}\text{NO}\\ x+\left(-2\right)=0\\ x=2\end{array}$.

Therefore, the oxidation number of nitrogen in $\text{NO}$ is $\underset{\_}{+2}$.

Apply charge balance formula in ${\text{SO}}_4^{2-}$ as shown below.

  $\begin{array}{c}\left(\text{Oxidation}\text{state}\text{of}\text{S}+4\times \text{Oxidation}\text{state}\text{of}\text{O}\right)=\text{Overall}\text{charge}\text{on}{\text{SO}}_4^{2-}\\ x+4\left(-2\right)=-2\\ x=-2+8\\ =6\end{array}$.

Therefore, the oxidation number of nitrogen in ${\text{SO}}_4^{2-}$ is $\underset{\_}{+6}$.

Apply charge balance formula in $\text{FeS}$ as shown below.

  $\begin{array}{c}\left(\left(\text{Oxidation}\text{state}\text{of}\text{Fe}\right)+\text{Oxidation}\text{state}\text{of}\text{S}\right)=\text{Overall}\text{charge}\text{on}\text{FeS}\\ x+\left(-2\right)=0\\ x=2\end{array}$.

Therefore, the oxidation number of iron in $\text{FeS}$ is $\underset{\_}{+2}$ and that of sulfur is $\underset{\_}{-2}$.

The element that gets oxidized reduces the other one and is known as reducing agent.  The element that gets reduced oxidizes the other one and is known as oxidizing agent.

In the given reaction, the oxidation number of ${\text{Fe}}^{\text{2+}}$ changes from $+2$ to $+3$.  So, $\text{FeS}$ gets oxidized and acts as a reducing agent.

In the given reaction, the oxidation number of $\text{S}$ changes from $-2$ to $+6$.  So, $\text{FeS}$ gets oxidized and acts as a reducing agent.

The oxidation number of $\text{N}$ changes from $+3$ to $+2$.  So, ${\text{NO}}_3^{-}$ gets reduced and acts as an oxidizing agent.