Chapter 17, Problem 13QRT

(a)

Interpretation Introduction

Interpretation:

The balanced half reactions for the given reaction 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.  The galvanic cell, voltaic cell electrolytic cell are the types of electrochemical cells.  In each cell, the oxidation half-reaction takes place at the anode while half-reaction reaction takes place at the cathode.

Answer to Problem 13QRT

The half reaction that represents oxidation is as follows.

    $\text{Fe}\left(\text{s}\right)\to {\text{Fe}}^{\text{2+}}\left(\text{aq}\right)+2e^{-}$

The half reaction that represents reduction is as follows.

    $\frac{1}{2}{\text{Br}}_2\left(l\right)+e^{-}\to {\text{Br}}^{-}\left(\text{aq}\right)$

Explanation of Solution

The given reaction is shown below.

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

The oxidation number of any element in its native form is zero.  So, the oxidation number of $\text{Fe}\left(\text{s}\right)$ and ${\text{Br}}_2\left(l\right)$ 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 $+2$.

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

The oxidation number of $\text{Br}$ changes from $0$ to $-1$.  So, ${\text{Br}}_2\left(l\right)$ gets reduced.

The half reaction that represents oxidation is as follows.

    $\text{Fe}\left(\text{s}\right)\to {\text{Fe}}^{\text{2+}}\left(\text{aq}\right)+2e^{-}$        (1)

The half reaction that represents reduction is as follows.

    $\frac{1}{2}{\text{Br}}_2\left(l\right)+e^{-}\to {\text{Br}}^{-}\left(\text{aq}\right)$        (2)

(b)

Interpretation Introduction

Interpretation:

The balanced half reactions for the given reaction have to be stated.

Concept Introduction:

Refer to part (a).

Answer to Problem 13QRT

The half reaction that represents oxidation is as follows.

    $\text{Al}\left(\text{s}\right)\to {\text{Al}}^{\text{3+}}\left(\text{aq}\right)+3e^{-}$

The half reaction that represents reduction is as follows.

    $\frac{1}{2}{\text{Cl}}_2\left(\text{g}\right)+e^{-}\to {\text{Cl}}^{-}\left(\text{aq}\right)$

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 $+3$.

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

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

The half reaction that represents oxidation is as follows.

    $\text{Al}\left(\text{s}\right)\to {\text{Al}}^{\text{3+}}\left(\text{aq}\right)+3e^{-}$        (3)

The half reaction that represents reduction is as follows.

    $\frac{1}{2}{\text{Cl}}_2\left(\text{g}\right)+e^{-}\to {\text{Cl}}^{-}\left(\text{aq}\right)$        (4)

(c)

Interpretation Introduction

Interpretation:

The balanced half reactions for the given reaction have to be stated.

Concept Introduction:

Refer to part (a).

Answer to Problem 13QRT

The half reaction that represents oxidation is as follows.

    ${\text{I}}^{-}\left(\text{aq}\right)\to \frac{1}{2}{\text{I}}_2\left(\text{s}\right)+e^{-}$

The balanced half cell reaction represents reduction is shown below.

    ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)+8{\text{H}}^{\text{+}}\left(\text{aq}\right)+8e^{-}\to {\text{H}}_{\text{2}}\text{S}\left(\text{aq}\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)$

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 $+6$.

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 $-2$.

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.

The oxidation number of $\text{I}$ changes from $-1$ to $0$.  So, $\text{HI}$ gets oxidized.

The half reaction that represents oxidation is as follows.

    ${\text{I}}^{-}\left(\text{aq}\right)\to \frac{1}{2}{\text{I}}_2\left(\text{s}\right)+e^{-}$        (5)

The half reaction that represents reduction is as follows.

    ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)\to {\text{H}}_{\text{2}}\text{S}\left(\text{aq}\right)$

The number of oxygen atoms is balanced by adding four molecules of water on product side as shown below.

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

In acidic medium, the number of hydrogen atoms is balanced by adding ${\text{H}}^{\text{+}}$ ions on reactant side as shown below.

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

The charge on reactant side is $+8$ and on product side is $0$.  Therefore, in order to balance the charge, eight electrons are added on reactant side as follows.

    ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)+8{\text{H}}^{\text{+}}\left(\text{aq}\right)+8e^{-}\to {\text{H}}_{\text{2}}\text{S}\left(\text{aq}\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)$

The balanced half cell reaction represents reduction is shown below.

    ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)+8{\text{H}}^{\text{+}}\left(\text{aq}\right)+8e^{-}\to {\text{H}}_{\text{2}}\text{S}\left(\text{aq}\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)$        (6)

(d)

Interpretation Introduction

Interpretation:

The balanced half reactions for the given reaction have to be stated.

Concept Introduction:

Refer to part (a).

Answer to Problem 13QRT

The half reaction that represents oxidation is as follows.

    ${\text{Fe}}^{\text{2+}}\left(\text{aq}\right)\to {\text{Fe}}^{\text{3+}}\left(\text{aq}\right)+e^{-}$

The balanced half cell reaction represents reduction is shown below.

    ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)+2e^{-}\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

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 $+1,+2$ and $+3$ 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 $-1$.

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 $-2$.

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

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

The half reaction that represents oxidation is as follows.

    ${\text{Fe}}^{\text{2+}}\left(\text{aq}\right)\to {\text{Fe}}^{\text{3+}}\left(\text{aq}\right)+e^{-}$        (7)

The reaction for the conversion of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ to ${\text{H}}_{\text{2}}\text{O}$ is shown below.

    ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)\to {\text{H}}_{\text{2}}\text{O}\left(l\right)$

The number of oxygen atoms is balanced by adding one molecule of water on product side as shown below.

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

In acidic medium, the number of hydrogen atoms is balanced by adding ${\text{H}}^{\text{+}}$ ions on reactant side as shown below.

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

The charge on reactant side is $+2$ and on product side is $0$.  Therefore, in order to balance the charge, two electrons are added on reactant side as follows.

    ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)+2e^{-}\to {\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The balanced half cell reaction represents reduction is shown below.

    ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\left(\text{aq}\right)+2e^{-}\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)$        (8)

(e)

Interpretation Introduction

Interpretation:

The balanced half reactions for the given reaction have to be stated.

Concept Introduction:

Refer to part (a).

Answer to Problem 13QRT

The half cell reaction for oxidation of $\text{FeS}$ is shown below.

    $\text{FeS}\left(\text{s}\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{Fe}}^{3+}\left(\text{aq}\right)+{\text{SO}}_4^{2-}\left(\text{aq}\right)+8{\text{H}}^{\text{+}}+9e^{-}$

The half cell reaction for reduction of ${\text{NO}}_3^{-}$ ion is shown below.

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

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 $+1$ and $+3$ 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 $+3$.

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 $+2$.

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 $+6$.

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 $+2$ and that of sulfur is $-2$.

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

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

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

The reaction for the oxidation of $\text{FeS}$ is shown below.

    $\text{FeS}\left(\text{s}\right)\to {\text{Fe}}^{3+}\left(\text{aq}\right)+{\text{SO}}_4^{2-}\left(\text{aq}\right)$

The number of oxygen atoms is balanced by adding four molecules of water on reactant side as shown below.

    $\text{FeS}\left(\text{s}\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{Fe}}^{3+}\left(\text{aq}\right)+{\text{SO}}_4^{2-}\left(\text{aq}\right)$

In acidic medium, the number of hydrogen atoms is balanced by adding ${\text{H}}^{\text{+}}$ ions on product side as shown below.

    $\text{FeS}\left(\text{s}\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{Fe}}^{3+}\left(\text{aq}\right)+{\text{SO}}_4^{2-}\left(\text{aq}\right)+8{\text{H}}^{\text{+}}$

The charge on reactant side is zero and the charge on product side is $+9$.  Therefore, in order to balance the charge, nine electrons are added on product side as follows.

    $\text{FeS}\left(\text{s}\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{Fe}}^{3+}\left(\text{aq}\right)+{\text{SO}}_4^{2-}\left(\text{aq}\right)+8{\text{H}}^{\text{+}}+9e^{-}$

The half cell reaction for oxidation of $\text{FeS}$ is shown below.

    $\text{FeS}\left(\text{s}\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{Fe}}^{3+}\left(\text{aq}\right)+{\text{SO}}_4^{2-}\left(\text{aq}\right)+8{\text{H}}^{\text{+}}+9e^{-}$        (9)

The reaction for the reduction of ${\text{NO}}_3^{-}$ ion is shown below.

    ${\text{NO}}_3^{-}\left(\text{aq}\right)\to \text{NO}\left(\text{g}\right)$

The number of oxygen atoms is balanced by adding two molecules of water on product side as shown below.

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

In acidic medium, the number of hydrogen atoms is balanced by adding ${\text{H}}^{\text{+}}$ ions on product side as shown below.

    ${\text{NO}}_3^{-}\left(\text{aq}\right)+4{\text{H}}^{\text{+}}\left(\text{aq}\right)\to \text{NO}\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The charge on reactant side is $+3$ and on product side is zero.  Therefore, in order to balance the charge, three electrons are added on reactant side as follows.

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

The half cell reaction for reduction of ${\text{NO}}_3^{-}$ ion is shown below.

    ${\text{NO}}_3^{-}\left(\text{aq}\right)+4{\text{H}}^{\text{+}}\left(\text{aq}\right)+3e^{-}\to \text{NO}\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$        (10)