Chapter 18, Problem 47A

(a)

Interpretation Introduction

Interpretation: The following oxidation-reduction reaction needs to be balanced an acidic medium.

${\text{MnO}}_4^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)\to {\text{Mn}}^{\text{2+}}\left(\text{aq}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)$

Concept Introduction: In oxidation-reduction reaction, transfer of electron/s takes place. The transfer of electron, results in the formation of ions. In oxidation, loss of electron/s takes place and in reduction, gain of electron/s takes place.

Answer to Problem 47A

The balance oxidation-reduction reaction is-

${\text{2MnO}}_4^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)+6{\text{H}}^{+}\to 2{\text{Mn}}^{\text{2+}}\left(\text{aq}\right)+8{\text{H}}_{\text{2}}\text{O}+5{\text{O}}_{\text{2}}\left(\text{g}\right)$

Explanation of Solution

The oxidation state of each element in this reaction is calculated as follows:

${\text{MnO}}_4^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)\to {\text{Mn}}^{\text{2+}}\left(\text{aq}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)$

Let oxidation state of Mn is x. Then, in ${\text{MnO}}_4^{-}$ its oxidation state is-

$\begin{array}{c}x+4\left(-2\right)=-1\\ x=8-1\\ =7\end{array}$

Oxidation state of O is $-1$ in ${\text{H}}_2{\text{O}}_{\text{2}}$ . ${\text{O}}_{\text{2}}$ is in a free state so, its oxidation state is zero. In product side, oxidation state of Mn is +2.

Oxidation state of Mn decreases from $+7$ to $+2$ . Hence, it is reduced. The reduction half-reaction is as follows:

${\text{MnO}}_4^{-}\left(\text{aq}\right)\to {\text{Mn}}^{\text{2+}}\left(\text{aq}\right)$

Another element except O is balanced. Balance O by using ${\text{H}}_{\text{2}}\text{O}$ and H by adding ${\text{H}}^{+}$ . Now, add electrons to balance charge. To balance charge on both sides add five electrons to the left. Hence, the balanced equation is-

${\text{MnO}}_4^{-}\left(\text{aq}\right)+8{\text{H}}^{+}+5{\text{e}}^{-}\to {\text{Mn}}^{\text{2+}}\left(\text{aq}\right)+4{\text{H}}_{\text{2}}\text{O}$ …… (1)

Oxidation state of O increases from $0$ to $+2$ . Hence, it is oxidized. The oxidation half-reaction is as follows:

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

Balance H by adding ${\text{H}}^{+}$ . Now, add electrons to balance charge. To balance charge on both sides, add two electrons to the left. Hence, the balanced equation is-

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

Multiply equation (1) by 2 and equation (2) by 5. After that, add both the equation to get the net balance equation.

  $\begin{array}{c}{\text{2MnO}}_4^{-}\left(\text{aq}\right)+16{\text{H}}^{+}+10{\text{e}}^{-}\to 2{\text{Mn}}^{\text{2+}}\left(\text{aq}\right)+8{\text{H}}_{\text{2}}\text{O}\\ {\text{5H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)\to 5{\text{O}}_{\text{2}}\left(\text{g}\right)+10{\text{H}}^{+}+10{\text{e}}^{-}\\ {\text{2MnO}}_4^{-}\left(\text{aq}\right)+{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)+6{\text{H}}^{+}\to 2{\text{Mn}}^{\text{2+}}\left(\text{aq}\right)+8{\text{H}}_{\text{2}}\text{O}+5{\text{O}}_{\text{2}}\left(\text{g}\right)\end{array}$

In the overall reaction, charge and elements are balanced on both sides.

(b)

Interpretation Introduction

Interpretation: The following oxidation-reduction reaction needs to be balanced in acidic medium.

${\text{BrO}}_3^{-}\left(\text{aq}\right)+{\text{Cu}}^{\text{+}}\left(\text{aq}\right)\to {\text{Br}}^{-}\left(\text{aq}\right)+{\text{Cu}}^{\text{2+}}\left(\text{aq}\right)$

Concept Introduction: In oxidation-reduction reaction, transfer of electron/s takes place. The transfer of electron, results in the formation of ions. In oxidation, loss of electron/s takes place and in reduction, gain of electron/s takes place.

Answer to Problem 47A

The balance reduction- oxidation reaction is- ${\text{BrO}}_3^{-}\left(\text{aq}\right)+6{\text{H}}^{\text{+}}+{\text{6Cu}}^{\text{+}}\left(\text{aq}\right)\to {\text{Br}}^{-}\left(\text{aq}\right)+3{\text{H}}_{\text{2}}\text{O+}6{\text{Cu}}^{\text{2+}}$ .

Explanation of Solution

The oxidation state of the each element in this reaction is given below:

${\text{BrO}}_3^{-}\left(\text{aq}\right)+{\text{Cu}}^{\text{+}}\left(\text{aq}\right)\to {\text{Br}}^{-}\left(\text{aq}\right)+{\text{Cu}}^{\text{2+}}\left(\text{aq}\right)$

Oxidation state of Br is $+5$ in ${\text{BrO}}_3^{-}$ and O is $-2$ . Oxidation state of ${\text{Cu}}^{\text{+}}$ is +1. Oxidation state of ${\text{Br}}^{-}$ and ${\text{Cu}}^{\text{2+}}$ is $-1$ and $+2$ respectively.

Oxidation state of Br decreases from $+5$ to $-1$ . Hence, it is reduced. The reduction half-reaction is as follows:

${\text{BrO}}_3^{-}\left(\text{aq}\right)\to {\text{Br}}^{-}\left(\text{aq}\right)$

Other element except O is balanced. Balance O by using ${\text{H}}_{\text{2}}\text{O}$ and H by adding ${\text{H}}^{+}$ . Now, add electrons to balance charge. To balance charge on both side add six electrons to the left. Hence, the balanced equation is-

${\text{BrO}}_3^{-}\left(\text{aq}\right)+6{\text{H}}^{\text{+}}+6{\text{e}}^{-}\to {\text{Br}}^{-}\left(\text{aq}\right)+3{\text{H}}_{\text{2}}\text{O}$ …… (1)

Oxidation state of Cu increases from $-1$ to $+2$ . Hence, it is oxidized. The oxidation half-reaction is as follows:

${\text{Cu}}^{\text{+}}\left(\text{aq}\right)\to {\text{Cu}}^{\text{2+}}$

Both elements are balanced. Now, add electrons to balance charge. To balance charge on both side add one electron to the right. Hence, the balanced equation is-

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

Multiply the equation (2) by 6. Add equation (1) and (2) to get the net balance equation.

$\begin{array}{c}{\text{BrO}}_3^{-}\left(\text{aq}\right)+6{\text{H}}^{\text{+}}+6{\text{e}}^{-}\to {\text{Br}}^{-}\left(\text{aq}\right)+3{\text{H}}_{\text{2}}\text{O}\\ {\text{6Cu}}^{\text{+}}\left(\text{aq}\right)\to 6{\text{Cu}}^{\text{2+}}+6{\text{e}}^{-}\\ {\text{BrO}}_3^{-}\left(\text{aq}\right)+6{\text{H}}^{\text{+}}+{\text{6Cu}}^{\text{+}}\left(\text{aq}\right)\to {\text{Br}}^{-}\left(\text{aq}\right)+3{\text{H}}_{\text{2}}\text{O+}6{\text{Cu}}^{\text{2+}}\end{array}$

In the overall reaction, charge and elements are balanced in both sides.

(c)

Interpretation Introduction

Interpretation: The following oxidation-reduction reaction needs to be balanced as an acidic medium.

${\text{HNO}}_{\text{2}}\left(\text{aq}\right)+{\text{I}}^{-}\left(\text{aq}\right)\to \text{NO}\left(\text{g}\right)+{\text{I}}_{\text{2}}\left(\text{aq}\right)$

Concept Introduction: In oxidation-reduction reaction, transfer of electron/s takes place. The transfer of electron, results in the formation of ions. In oxidation, loss of electron/s takes place and in reduction, gain of electron/s takes place.

Answer to Problem 47A

The balance oxidation-reduction reaction is-

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

Explanation of Solution

The oxidation state of the each element in this reaction is given below:

${\text{HNO}}_{\text{2}}\left(\text{aq}\right)+{\text{I}}^{-}\left(\text{aq}\right)\to \text{NO}\left(\text{g}\right)+{\text{I}}_{\text{2}}\left(\text{aq}\right)$

${\text{I}}_{\text{2}}$ is in a free state so, its oxidation state is zero. Oxidation state of N in ${\text{HNO}}_{\text{2}}$ is $+3$ , O is $-2$ and H is +1. In product side, oxidation state of N in $\text{NO}$ is +2 and O is $-2$ .

Oxidation state of N decreases from $+3$ to $+2$ . Hence, it is reduced. The reduction half-reaction is as follows:

${\text{HNO}}_{\text{2}}\left(\text{aq}\right)\to \text{NO}\left(\text{g}\right)$

Other element except O is balanced. Balance O by using ${\text{H}}_{\text{2}}\text{O}$ and H by adding ${\text{H}}^{+}$ . Now, add electrons to balance charge. To balance charge on both sides add one electron to the left. Hence, the balanced equation is-

${\text{HNO}}_{\text{2}}\left(\text{aq}\right)+{\text{H}}^{\text{+}}+{\text{e}}^{-}\to \text{NO}\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}$ …… (1)

Oxidation state of ${\text{I}}^{-}$ is increases from $-1$ to $0$ . Hence, it is oxidized. The oxidation half-reaction is as follows:

${\text{I}}^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{aq}\right)$

To balance I on both side, multiply reactant side I by to 2.

${\text{2I}}^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{aq}\right)$

To balance charge on both side add $2{\text{e}}^{-}$ to the right. Hence, the balanced equation is-

${\text{2I}}^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{aq}\right)+2{\text{e}}^{-}$ …… (2)

Multiply equation (1) by 2. After that, add both the equation to get the net balance equation.

$\begin{array}{l}{\text{2HNO}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}+2{\text{e}}^{-}\to 2\text{NO}\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\\ {\text{2I}}^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{aq}\right)+2{\text{e}}^{-}\\ {\text{2HNO}}_{\text{2}}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}+{\text{2I}}^{-}\left(\text{aq}\right)\to 2\text{NO}\left(\text{g}\right)+2{\text{H}}_{\text{2}}{\text{O+I}}_{\text{2}}\left(\text{aq}\right)\end{array}$

In the overall reaction, charge and elements are balanced on both sides.