Chapter 18, Problem 18.20QP

(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.

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 tin metal and ${\text{Ag}}^{+}$ ions in solution that produces dissolved ${\text{Sn}}^{2+}$ and $\text{Ag}$ metal.

Answer to Problem 18.20QP

Solution

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

$\text{Sn}\left(s\right)+{\text{2Ag}}^{+}\left(aq\right)\to {\text{Sn}}^{2+}\left(aq\right)+2\text{Ag}\left(s\right)$

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 tin metal,

$\text{Sn}\left(s\right)\to {\text{Sn}}^{2+}\left(aq\right)+2{\text{e}}^{-}{E}^{\text{ο}}=0.14$

For ${\text{Ag}}^{+}$,

${\text{Ag}}^{+}\left(aq\right)+{\text{e}}^{-}\to \text{Ag}\left(s\right)E^{\text{ο}}=0.80$

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

$\text{Sn}\left(s\right)\to {\text{Sn}}^{2+}\left(aq\right)+2{\text{e}}^{-}{E}^{\text{ο}}=0.14$ (1)

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

${\text{Ag}}^{+}\left(aq\right)+{\text{e}}^{-}\to \text{Ag}\left(s\right)E^{\text{ο}}=0.80$ (2)

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

${\text{2Ag}}^{+}\left(aq\right)+2{\text{e}}^{-}\to 2\text{Ag}\left(s\right)E^{\text{ο}}=0.80$ (3)

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

$\begin{array}{c}\text{Sn}\left(s\right)\to {\text{Sn}}^{2+}\left(aq\right)+2{\text{e}}^{-}{E}^{\text{ο}}=0.14\\ {\text{2Ag}}^{+}\left(aq\right)+2{\text{e}}^{-}\to 2\text{Ag}\left(s\right)E^{\text{ο}}=0.80\\ \text{Sn}\left(s\right)+{\text{2Ag}}^{+}\left(aq\right)\to {\text{Sn}}^{2+}\left(aq\right)+2\text{Ag}\left(s\right)\end{array}$ (4)

The two electrons present in both the equations are cancelled with each other.

Therefore, the equation (4) is the net ionic balanced equation and the appropriate two half reactions to write net ionic equation is the equation (1) and (2).

(b)

Interpretation Introduction

To determine: The net ionic equation for the given reaction between copper and ${\text{O}}_2$ in an acidic solution that produces ${\text{Cu}}^{2+}$ ions.

Answer to Problem 18.20QP

Solution

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

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

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 copper is converted to the ${\text{Cu}}^{2+}$ ion and ${\text{O}}_2$ is converted to the ${\text{H}}_2\text{O}$ in acidic conditions. Therefore, with the help of reduction values given in Appendix 6, the reactions are written as,

For ${\text{Cu}}^{2+}$,

$\text{Cu}\left(s\right)\to {\text{Cu}}^{2+}\left(aq\right)+2{\text{e}}^{-}{E}^{\text{ο}}=-0.34$

For ${\text{O}}_2$,

${\text{O}}_2\left(aq\right)+4{\text{H}}^{+}\left(aq\right)+4{\text{e}}^{-}\to 2{\text{H}}_2\text{O}\left(l\right)E^{\text{ο}}=1.23$

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

$\text{Cu}\left(s\right)\to {\text{Cu}}^{2+}\left(aq\right)+2{\text{e}}^{-}{E}^{\text{ο}}=-0.34$ (1)

The second reaction undergoes reduction due to higher value of reduction potential. It is represented as,

${\text{O}}_2\left(aq\right)+4{\text{H}}^{+}\left(aq\right)+4{\text{e}}^{-}\to 2{\text{H}}_2\text{O}\left(l\right)E^{\text{ο}}=1.23$ (2)

To balance the reactions, equation (1) is multiplied by two. It is represented as,

$\text{2Cu}\left(s\right)\to 2{\text{Cu}}^{2+}\left(aq\right)+4{\text{e}}^{-}{E}^{\text{ο}}=-0.34$ (3)

Now, add the equation (2) and (3). It is represented as,

$\begin{array}{c}\text{2Cu}\left(s\right)\to 2{\text{Cu}}^{2+}\left(aq\right)+4{\text{e}}^{-}\\ {\text{O}}_2\left(aq\right)+4{\text{H}}^{+}\left(aq\right)+4{\text{e}}^{-}\to 2{\text{H}}_2\text{O}\left(l\right)\\ \text{2Cu}\left(s\right)+{\text{O}}_2\left(aq\right)+4{\text{H}}^{+}\left(aq\right)\to 2{\text{Cu}}^{2+}\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)\end{array}$ (4)

The four electrons present in both the equations are cancelled with each other. Therefore, the equation (4) is the net ionic balanced equation and the appropriate two half reactions to write net ionic equation is the equation (2) and (3).

(c)

Interpretation Introduction

To determine: The net ionic equation for the given reaction between solid $\text{Cr}{\left(\text{OH}\right)}_3$ and ${\text{O}}_2$ in a basic solution that produces ${\text{CrO}}_4^{2-}$ ions.

Answer to Problem 18.20QP

Solution

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

$\text{4Cr}{\left(\text{OH}\right)}_3\left(s\right)+8{\text{OH}}^{-}+{\text{3O}}_2\left(aq\right)\to 4{\text{CrO}}_4^{2-}\left(aq\right)+{\text{10H}}_2\text{O}\left(l\right)$

Explanation of Solution

Explanation

The solid $\text{Cr}{\left(\text{OH}\right)}_3$ is converted to ${\text{CrO}}_4^{2-}$ ions and ${\text{O}}_2$ is converted to ${\text{OH}}^{-}$ ions in basic condition.

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

$\begin{array}{c}\text{Cr}{\left(\text{OH}\right)}_3\left(s\right)\to {\text{CrO}}_4^{2-}\left(aq\right)\\ {\text{O}}_{\text{2}}\left(aq\right)\to {\text{H}}_2\text{O}\left(l\right)\end{array}$

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,

$\text{Cr}{\left(\text{OH}\right)}_3\left(s\right)\to {\text{CrO}}_4^{2-}\left(aq\right)$

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

$\text{Cr}{\left(\text{OH}\right)}_3\left(s\right)\to {\text{CrO}}_4^{2-}\left(aq\right)$

Now, oxygen atoms are balanced by putting the one hydroxide ion to the left side. Equation is represented as,

$\text{Cr}{\left(\text{OH}\right)}_3\left(s\right)+{\text{OH}}^{-}\left(aq\right)\to {\text{CrO}}_4^{2-}\left(aq\right)$

Now, add four water molecules the right side to balance the hydrogen atom. The equation is represented as,

$\text{Cr}{\left(\text{OH}\right)}_3\left(s\right)+{\text{OH}}^{-}\left(aq\right)\to {\text{CrO}}_4^{2-}\left(aq\right)+{\text{4H}}_2\text{O}\left(l\right)$

Now, add four more hydroxide ions to the left side to balance the rest hydrogen atoms. The equation is represented as,

$\text{Cr}{\left(\text{OH}\right)}_3\left(s\right)+{\text{OH}}^{-}\left(aq\right)+4{\text{OH}}^{-}\left(aq\right)\to {\text{CrO}}_4^{2-}\left(aq\right)+{\text{4H}}_2\text{O}\left(l\right)$

Now, left side has minus five charges and right side has minus two charges. Therefore, to balance the charges three electrons are added to the right side. The equation is represented as,

$\text{Cr}{\left(\text{OH}\right)}_3\left(s\right)+5{\text{OH}}^{-}\left(aq\right)\to {\text{CrO}}_4^{2-}\left(aq\right)+{\text{4H}}_2\text{O}\left(l\right)+3{\text{e}}^{-}$ (1)

Now, for oxygen the basic medium reaction is given as,

${\text{O}}_2\left(aq\right)+4{\text{H}}^{+}\left(aq\right)+4{\text{e}}^{-}\to 2{\text{H}}_2\text{O}\left(l\right)$ .(2)

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

$\begin{array}{l}\text{4Cr}{\left(\text{OH}\right)}_3\left(s\right)+20{\text{OH}}^{-}\left(aq\right)\to 4{\text{CrO}}_4^{2-}\left(aq\right)+{\text{16H}}_2\text{O}\left(l\right)+12{\text{e}}^{-}\\ {\text{3O}}_2\left(aq\right)+12{\text{H}}^{+}\left(aq\right)+12{\text{e}}^{-}\to 6{\text{H}}_2\text{O}\left(l\right)\end{array}$

Now, add both the above equations. The equal number of hydrogen ion and hydroxide ion combines and converted to water molecule. All the electrons are cancelled with each other. Overall reaction is represented as,

$\text{4Cr}{\left(\text{OH}\right)}_3\left(s\right)+8{\text{OH}}^{-}+{\text{3O}}_2\left(aq\right)\to 4{\text{CrO}}_4^{2-}\left(aq\right)+{\text{10H}}_2\text{O}\left(l\right)$

This is the net ionic equation.

Conclusion

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

1. a) $\text{Sn}\left(s\right)+{\text{2Ag}}^{+}\left(aq\right)\to {\text{Sn}}^{2+}\left(aq\right)+2\text{Ag}\left(s\right)$
2. b) $\text{2Cu}\left(s\right)+{\text{O}}_2\left(aq\right)+4{\text{H}}^{+}\left(aq\right)\to 2{\text{Cu}}^{2+}\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$
3. c) $\text{4Cr}{\left(\text{OH}\right)}_3\left(s\right)+8{\text{OH}}^{-}+{\text{3O}}_2\left(aq\right)\to 4{\text{CrO}}_4^{2-}\left(aq\right)+{\text{10H}}_2\text{O}\left(l\right)$