Chapter 18, Problem 18.19QP

(a)

Interpretation Introduction

Interpretation: The net ionic equations for the given reactions are to be written with the help of the appropriate half reactions from Appendix 6.

Concept introduction: Voltaic cell is a type of electrochemical cell which converts the chemical energy to electrical energy. This chemical energy generates by processing the redox reaction. Redox reaction is the combination of two half reactions that are oxidation half and reduction half. Oxidation is a process in which loss of electrons take place, while in reduction gain of electron take place.

In voltaic cell during the process of redox reaction electrons are generated on anode and then transfer to the cathode by the external circuit. In this way we can say that the process of oxidation occurs at the anode and the process of reduction occurs at cathode and hence, electron flows.

The species with higher reduction potential always undergoes reduction. The species with lower reduction potential always undergoes oxidation.

In a voltaic cell, for the spontaneous cell reaction the overall cell potential must be positive.

To determine: The net ionic equation for the given reaction between Aluminum metal and ${\text{Fe}}^{3+}$ ions in solution that produces dissolved ${\text{Al}}^{3+}$ and ${\text{Fe}}^{2+}$.

Answer to Problem 18.19QP

Solution

The net ionic equation for the given reaction is given as,

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

Explanation of Solution

Explanation

As given above that the net ionic reaction is the combination of two half reactions that are oxidation half and reduction half.

With the help of reduction values given in Appendix 6, the reactions are written as,

For aluminum,

${\text{Al}}^{3+}\left(aq\right)+3{\text{e}}^{-}\to \text{Al}\left(s\right)E^{\text{ο}}=-1.66$

For ${\text{Fe}}^{3+}$,

${\text{Fe}}^{3+}\left(aq\right)+{\text{e}}^{-}\to {\text{Fe}}^{2+}\left(aq\right)E^{\text{ο}}=+0.77$

The first reaction undergoes oxidation due to lower value of reduction potential. It is represented as,

$\text{Al}\left(s\right)\to {\text{Al}}^{3+}\left(aq\right)+3{\text{e}}^{-}{E}^{\text{ο}}=+1.66$ (1)

The sign of value of reduction potential is revered if the reaction is represented in reverse manner. Therefore, for the first reaction the sigh has been reversed.

In second reaction undergoes reduction due to higher value of reduction potential. It is represented as it is.

${\text{Fe}}^{3+}\left(aq\right)+{\text{e}}^{-}\to {\text{Fe}}^{2+}\left(aq\right)E^{\text{ο}}=+0.77$ (2)

To balance the reactions, equation (2) is multiplied by three. It is represented as,

$3{\text{Fe}}^{3+}\left(aq\right)+3{\text{e}}^{-}\to 3{\text{Fe}}^{2+}\left(aq\right)E^{\text{ο}}=+0.77$ (3)

Now, add the equation (1) and (3). It is represented as,

$\begin{array}{c}\text{Al}\left(s\right)\to {\text{Al}}^{3+}\left(aq\right)+3{\text{e}}^{-}{E}^{\text{ο}}=+1.66\\ 3{\text{Fe}}^{3+}\left(aq\right)+3{\text{e}}^{-}\to 3{\text{Fe}}^{2+}\left(aq\right)E^{\text{ο}}=+0.77\\ \text{Al}\left(s\right)+3{\text{Fe}}^{3+}\left(aq\right)\to {\text{Al}}^{3+}\left(aq\right)+3{\text{Fe}}^{2+}\left(aq\right)\end{array}$ (4)

The three electrons present in both the equations are cancelled with each other. Therefore, the equation (4) is the net ionic balanced equation.

(b)

Interpretation Introduction

To determine: The net ionic equation for the given reaction between ${\text{I}}_{\text{2}}$ and ${\text{NO}}_2^{-}$ ions in an alkaline solutions that produces ${\text{I}}^{-}$ and ${\text{NO}}_3^{-}$ ions.

Answer to Problem 18.19QP

Solution

The net ionic equation for the given reaction is given as,

${\text{NO}}_2^{-}\left(aq\right)+{\text{I}}_{\text{2}}\left(aq\right)+2{\text{OH}}^{-}\to {\text{NO}}_3^{-}\left(aq\right)+{\text{2I}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

Explanation of Solution

Explanation

As given above that the net ionic reaction is the combination of two half reactions that are oxidation half and reduction half.

The molecule of ${\text{I}}_{\text{2}}$ is converted to the ${\text{I}}^{-}$ ion and ${\text{NO}}_2^{-}$ is converted to the ${\text{NO}}_3^{-}$ ions. Therefore, ${\text{I}}_{\text{2}}$ undergoes reduction and ${\text{NO}}_2^{-}$ undergoes oxidation.

With the help of reduction values given in Appendix 6, the reactions are written as,

For ${\text{I}}_{\text{2}}$,

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

For ${\text{NO}}_2^{-}$,

${\text{NO}}_2^{-}\left(aq\right)+2{\text{OH}}^{-}\to {\text{NO}}_3^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)+2{\text{e}}^{-}$ (2)

In both the reactions there are two electrons are involved. Therefore, on the addition of the reactions both the electrons are cancelled with each other.

The net ionic reaction is written as,

${\text{NO}}_2^{-}\left(aq\right)+{\text{I}}_{\text{2}}\left(aq\right)+2{\text{OH}}^{-}\to {\text{NO}}_3^{-}\left(aq\right)+{\text{2I}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

(c)

Interpretation Introduction

To determine: The net ionic equation for the given reaction between ${\text{MnO}}_4^{-}$ and ${\text{Cr}}^{3+}$ ions in an acidic solution that produces ${\text{Mn}}^{2+}$ and ${\text{Cr}}_2{\text{O}}_7^{2-}$ ions.

Answer to Problem 18.19QP

Solution

The net ionic equation for the given reaction is given as,

${\text{6MnO}}_4^{-}\left(aq\right)+10{\text{Cr}}^{3+}\left(aq\right)+11{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 6{\text{Mn}}^{2+}\left(s\right){\text{+5Cr}}_2{\text{O}}_7^{2-}\left(aq\right)+22{\text{H}}^{+}$

Explanation of Solution

Explanation

The molecular ion ${\text{MnO}}_4^{-}$ is converted to ${\text{Mn}}^{2+}$ ions and ${\text{Cr}}^{3+}$ ions are converted to ${\text{Cr}}_2{\text{O}}_7^{2-}$ ions in acidic condition.

The above reaction is a redox reaction. With the help of reduction values given in Appendix 6, the reactions are written as,

$\begin{array}{c}{\text{MnO}}_4^{-}\left(aq\right)\to {\text{Mn}}^{2+}\left(s\right)\\ {\text{Cr}}^{3+}\left(aq\right)\to {\text{Cr}}_2{\text{O}}_7^{2-}\left(aq\right)\end{array}$

The above two reactions are the simplest representation of the complete redox reaction.

Now, to get the balanced and complete reaction, both the half reactions are balanced separately and then added together.

In this method the following steps are followed,

• The atoms other than oxygen and hydrogen are balanced.
• After that to balance oxygen and hydrogen water molecules are added.
• After that in acidic condition hydrogen ions are added to balance the hydrogen.
• After that charges are balanced by the addition of electrons.
• After the balancing of both the half reactions, both are added together.

Now, apply these rules on first half reaction that is reduction half,

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

Both the side same number of manganese atom are present therefore, no coefficient is added either side. The equation is represented as,

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

Now, oxygen atoms are balanced by putting the two water molecule to the right side. Equation is represented as,

${\text{MnO}}_4^{-}\left(aq\right)\to {\text{Mn}}^{2+}\left(s\right)+{\text{4H}}_2\text{O}\left(l\right)$

Now, add eight hydrogen ions to the left side to balance the hydrogen atom. The equation is represented as,

${\text{MnO}}_4^{-}\left(aq\right)+8{\text{H}}^{+}\to {\text{Mn}}^{2+}\left(s\right)+{\text{4H}}_2\text{O}\left(l\right)$

Now, left side has plus seven charge and right side has plus two charge. Therefore, to balance the charge five electrons are added to the left side. The equation is represented as,

${\text{MnO}}_4^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+5{\text{e}}^{-}\to {\text{Mn}}^{2+}\left(s\right)+{\text{4H}}_2\text{O}\left(l\right)$ (1)

Now, apply these rules on second half reaction that is oxidation half,

${\text{Cr}}^{3+}\left(aq\right)\to {\text{Cr}}_2{\text{O}}_7^{2-}\left(aq\right)$

Add the coefficient $2$ on the left side. The equation is represented as,

$2{\text{Cr}}^{3+}\left(aq\right)\to {\text{Cr}}_2{\text{O}}_7^{2-}\left(aq\right)$

Now, oxygen atoms are balanced by the addition of seven water molecules to the left side. Equation is represented as,

$2{\text{Cr}}^{3+}\left(aq\right)+7{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{Cr}}_2{\text{O}}_7^{2-}\left(aq\right)$

Now, add fourteen hydrogen ions to the right side to balance the hydrogen atom. The equation is represented as,

$2{\text{Cr}}^{3+}\left(aq\right)+7{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{Cr}}_2{\text{O}}_7^{2-}\left(aq\right)+14{\text{H}}^{+}$

Now, right side has plus twelve charge and left side has six positive charge. Therefore, to balance the charge six electrons are added to the right side. The equation is represented as,

$2{\text{Cr}}^{3+}\left(aq\right)+7{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{Cr}}_2{\text{O}}_7^{2-}\left(aq\right)+14{\text{H}}^{+}+6{\text{e}}^{-}$ (2)

Now, to balance the electrons in equation (1) and (2), the equation (1) is multiplied by the coefficient six and equation (2) is multiplied by the coefficient five. In combined it is represented as,

$\begin{array}{l}{\text{6MnO}}_4^{-}\left(aq\right)+48{\text{H}}^{+}\left(aq\right)+30{\text{e}}^{-}\to 6{\text{Mn}}^{2+}\left(s\right)+24{\text{H}}_2\text{O}\left(l\right)\\ 10{\text{Cr}}^{3+}\left(aq\right)+35{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 5{\text{Cr}}_2{\text{O}}_7^{2-}\left(aq\right)+70{\text{H}}^{+}+30{\text{e}}^{-}\end{array}$

Now, add both the above equations. All the electrons are cancelled with each other. Overall reaction is represented as,

${\text{6MnO}}_4^{-}\left(aq\right)+10{\text{Cr}}^{3+}\left(aq\right)+11{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 6{\text{Mn}}^{2+}\left(s\right){\text{+5Cr}}_2{\text{O}}_7^{2-}\left(aq\right)+22{\text{H}}^{+}$

This is the net ionic equation.

Conclusion

The net ionic equations for the given reactions are given as,

1. a) $\text{Al}\left(s\right)+3{\text{Fe}}^{3+}\left(aq\right)\to {\text{Al}}^{3+}\left(aq\right)+3{\text{Fe}}^{2+}\left(aq\right)$
2. b) ${\text{NO}}_2^{-}\left(aq\right)+{\text{I}}_{\text{2}}\left(aq\right)+2{\text{OH}}^{-}\to {\text{NO}}_3^{-}\left(aq\right)+{\text{2I}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$
3. c) ${\text{6MnO}}_4^{-}\left(aq\right)+10{\text{Cr}}^{3+}\left(aq\right)+11{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 6{\text{Mn}}^{2+}\left(s\right){\text{+5Cr}}_2{\text{O}}_7^{2-}\left(aq\right)+22{\text{H}}^{+}$