Chapter 8, Problem 8.19E

Interpretation Introduction

(a)

Interpretation:

The values of $E°$ and $\Delta G°$ for the given reaction are to be calculated.

Concept introduction:

Standard Gibbs free energy of a reaction is used to check whether the reaction is spontaneous or not. If the value of $\Delta G°$ is positive, then the reaction is non spontaneous. If the value of $\Delta G°$ is negative, then the reaction is spontaneous.

Answer to Problem 8.19E

The value of $E°$ and $\Delta G°$ for the given reaction is $-0.0005\text{V}$ and $96.485\text{J}$ respectively.

Explanation of Solution

The given reaction is represented as,

${\text{I}}_2+{\text{I}}^{-}\to {\text{I}}_3{}^{-}$

From Table $8.2$, the reduction half reaction of ${\text{I}}_3^{-}$ and the standard reduction potential of ${\text{I}}_3^{-}$ is represented as,

${\text{I}}_3^{-}+2{\text{e}}^{-}\to 3{\text{I}}^{-}E°=0.5360\text{V}$

The above equation is reversed and the value of $E°$ is multiplied by $-1$ to form an oxidation half reaction. The oxidation half reaction is represented as,

$3{\text{I}}^{-}\to {\text{I}}_3^{-}+2{\text{e}}^{-}E°=-0.5360\text{V}$ …(1)

The number of moles of electrons transferred in the above reaction is $2\text{mol}$.

From Table $8.2$, the reduction half reaction of ${\text{I}}_{\text{2}}$ and the standard reduction potential of ${\text{I}}_{\text{2}}$ is represented as,

${\text{I}}_2+2{\text{e}}^{-}\to 2{\text{I}}^{-}E°=0.5355\text{V}$ …(2)

The number of moles of electrons transferred in the above reaction is $2\text{mol}$.

The relation between standard Gibbs free energy and standard electrical potential is represented as,

$\Delta G°=-nFE°$ …(3)

Where,

• $\Delta G°$ represents the standard Gibbs free energy of the reaction.

• $n$ represents the number of moles.

• $F$ represents the Faraday’s constant with value $96,485\text{C}/\text{mol}$.

• $E°$ represents the standard electrical potential.

Substitute the values of the standard oxidation potential of ${\text{I}}^{-}$, $F$ and $n$ in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(-0.5360\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& 103431.92\text{J}\end{array}$

The value $\Delta G°$ for the reaction (1) is $103431.92\text{J}$.

Substitute the values of the standard reduction potential of ${\text{I}}_{\text{2}}$, $F$ and number of moles of electrons transfer in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(0.5355\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& -103335.435\text{J}\end{array}$

The value $\Delta G°$ for the reaction (2) is $-103.335\text{kJ}$.

The value of $\Delta G°$ of overall reaction is calculated by adding the $\Delta G°$ of two half reactions. The standard Gibbs free energy of the given reaction is calculated as,

$\Delta G°=\Delta G{°}_1+\Delta G{°}_2$

Where,

• $\Delta G{°}_1$ represents the standard Gibbs free energy of the reaction (1).

• $\Delta G{°}_2$ represents the standard Gibbs free energy of the reaction (2).

Substitute the value of $\Delta G{°}_1$ and $\Delta G{°}_2$ in the above equation.

$\begin{array}{rll}\Delta G°& =& 103431.92\text{J}+\left(-103335.435\text{J}\right)\\ & =& 96.485\text{J}\end{array}$

Therefore, the value $\Delta G°$ for the given reaction is $96.485\text{J}$.

The number of electrons transferred in the overall reaction is $2\text{ mol}$.

Rearrange the equation (3) for the value of $E°$.

$E°=\frac{-\Delta G°}{nF}$

Substitute the values of $\Delta G°$, $F$ and number of moles of electrons transfer in the equation (3).

$\begin{array}{rll}E°& =& \frac{-\left(96.485\text{J}\right)}{\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)}\\ & =& -0.0005\text{V}\end{array}$

The value of $E°$ for the given reaction is $-0.0005\text{V}$.

Conclusion

The value of $E°$ and $\Delta G°$ for the given reaction is $-0.0005\text{V}$ and $96.485\text{J}$ respectively.

Interpretation Introduction

(b)

Interpretation:

The values of $E°$ and $\Delta G°$ for the given reaction are to be calculated.

Concept introduction:

Standard Gibbs free energy of a reaction is used to check whether the reaction is spontaneous or not. If the value of $\Delta G°$ is positive, then the reaction is non spontaneous. If the value of $\Delta G°$ is negative, then the reaction is spontaneous.

Answer to Problem 8.19E

The value of $E°$ and $\Delta G°$ for the given reaction is $-0.912\text{V}$ and $176.1\text{kJ}$ respectively.

Explanation of Solution

The given reaction is represented as,

${\text{Cr}}^{2+}+2{\text{e}}^{-}\to \text{Cr}$

From Table $8.2$, the reduction half reaction of ${\text{Cr}}^{3+}$ into ${\text{Cr}}^{2+}$ and the standard reduction potential of ${\text{Cr}}^{3+}$ into ${\text{Cr}}^{2+}$ is represented as,

${\text{Cr}}^{3+}+{\text{e}}^{-}\to {\text{Cr}}^{2+}E°=-0.407\text{V}$

The above equation is reversed and the value of $E°$ is multiplied by $-1$ to form an oxidation half reaction. The oxidation half reaction is represented as,

${\text{Cr}}^{2+}\to {\text{Cr}}^{3+}+{\text{e}}^{-}E°=0.407\text{V}$ …(4)

The number of moles of electrons transferred in the above reaction is $1\text{mol}$.

From Table $8.2$, the reduction half reaction of ${\text{Cr}}^{3+}$ into $\text{Cr }$ and the standard reduction potential of ${\text{Cr}}^{3+}$ into $\text{Cr }$ is represented as,

${\text{Cr}}^{3+}+3{\text{e}}^{-}\to \text{Cr }E°=-0.744\text{V}$ …(5)

The number of moles of electrons transferred in the above reaction is $3\text{mol}$.

The relation between standard Gibbs free energy and standard electrical potential is represented as,

$\Delta G°=-nFE°$ …(3)

Where,

• $\Delta G°$ represents the standard Gibbs free energy of the reaction.

• $n$ represents the number of moles.

• $F$ represents the Faraday’s constant with value $96,485\text{C}/\text{mol}$.

• $E°$ represents the standard electrical potential.

Substitute the values of the standard oxidation potential of ${\text{Cr}}^{2+}$, $F$ and number of moles of electrons transfer in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{1 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(0.407\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& -\left(39269.395\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& -39.269\text{kJ}\end{array}$

The value $\Delta G°$ for the reaction (4) is $-39.269\text{kJ}$.

Substitute the values of the standard reduction potential of ${\text{Cr}}^{3+}$, $F$ and number of moles of electrons transfer in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{3 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(-0.744\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& \left(215354.53\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& 215.355\text{kJ}\end{array}$

The value $\Delta G°$ for the reaction (5) is $215.355\text{kJ}$.

The value of $\Delta G°$ of overall reaction is calculated by adding the $\Delta G°$ of two half reactions. The standard Gibbs free energy of the given reaction is calculated as,

$\Delta G°=\Delta G{°}_1+\Delta G{°}_2$

Where,

• $\Delta G{°}_1$ represents the standard Gibbs free energy of the reaction (4).

• $\Delta G{°}_1$ represents the standard Gibbs free energy of the reaction (5).

Substitute the value of ${\Delta }_{\text{rxn}}G{°}_1$ and ${\Delta }_{\text{rxn}}G{°}_1$ in the above equation.

$\begin{array}{rll}\Delta G°& =& -39.269\text{kJ}+215.355\text{kJ}\\ & =& 176.086\text{kJ}\\ & \approx & 176.1\text{kJ}\end{array}$

Therefore, the value $\Delta G°$ for the given reaction is $176.1\text{kJ}$.

The number of electrons transferred in the overall reaction is $2\text{ mol}$.

Rearrange the equation (3) for the value of $E°$.

$E°=\frac{-\Delta G°}{nF}$

Substitute the values of $\Delta G°$, $F$ and number of moles of electrons transferred in the equation (3).

$\begin{array}{rll}E°& =& \frac{-\left(176.1\text{kJ}\right)\left(\frac{1000\text{ J}}{1\text{ kJ}}\right)}{\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)}\\ & =& -0.912\text{V}\end{array}$

The value of $E°$ for the given reaction is $-0.912\text{V}$.

Conclusion

The value of $E°$ and $\Delta G°$ for the given reaction is $-0.912\text{V}$ and $176.1\text{kJ}$ respectively.