Chapter 8, Problem 8.11E

Interpretation Introduction

(a)

Interpretation:

The overall balanced electrochemical reaction and 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 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 8.11E

The overall balanced electrochemical reaction is as follows,

${\text{4MnO}}_2+3{\text{O}}_2+{\text{14H}}_2\text{O}\to 6{\text{OH}}^{-}+4{\text{MnO}}_{\text{4}}{}^{-}+16{\text{H}}^{+}$

The value of $E°$ and $\Delta G°$ for the given reaction is $-1.278\text{V}$ and $1479.697\text{kJ}$ respectively.

Explanation of Solution

The given reaction is represented as,

${\text{MnO}}_2+{\text{O}}_2\to {\text{OH}}^{-}+{\text{MnO}}_{\text{4}}{}^{-}$

From Table $8.2$, the reduction half reaction of ${\text{H}}_{\text{2}}\text{O}$ into ${\text{OH}}^{-}$ and the standard reduction potential of ${\text{H}}_{\text{2}}\text{O}$ into ${\text{OH}}^{-}$ is represented as,

${\text{O}}_{\text{2}}+2{\text{H}}_{\text{2}}\text{O}+4{\text{e}}^{-}\to 2{\text{OH}}^{-}E°=0.401\text{V}$…(1)

The number of moles of electrons transferred in the above reaction is $4\text{mol}$.

From Table $8.2$, the reduction half reaction of ${\text{MnO}}_4{}^{-}$ and the standard reduction potential of ${\text{MnO}}_4{}^{-}$ is represented as,

${\text{MnO}}_4{}^{-}+4{\text{H}}^{+}+3{\text{e}}^{-}\to {\text{MnO}}_2{\text{+2H}}_2\text{O }E°=1.679\text{V}$

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,

${\text{MnO}}_2{\text{+2H}}_2\text{O}\to {\text{MnO}}_4{}^{-}+4{\text{H}}^{+}+3{\text{e}}^{-}E°=-1.679\text{V}$…(2)

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 reduction potential of the reaction (1), $F$ and $n$ in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{4 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(0.401\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& -\left(154761.94\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& -154.761\text{ kJ}\end{array}$

The value $\Delta G°$ for the reaction (1) is $-154.761\text{kJ}$.

Substitute the values of the standard oxidation potential of the reaction (2), $F$ and number of moles of electrons transferred in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{3 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(-1.679\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& \left(485994.945\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& 485.995\text{ kJ}\end{array}$

The value $\Delta G°$ for the reaction (2) is $485.995\text{ kJ}$.

The balanced overall electrochemical reaction is obtained by multiplying chemical equation (1) by $3$ and chemical equation (2) by $4$ and then adding these equations. The formation of overall balanced chemical equation is represented as,

$\begin{array}{lll}{\text{3O}}_{\text{2}}+6{\text{H}}_{\text{2}}\text{O}+12{\text{e}}^{-}\to 6{\text{OH}}^{-}\text{ 3}\times \Delta G°& =& \text{3}\times \left(-154.761\text{ kJ}\right)\\ {\text{4MnO}}_2+{\text{8H}}_2\text{O}\to 4{\text{MnO}}_4{}^{-}+16{\text{H}}^{+}+12{\text{e}}^{-}\text{ 4}\times \Delta G°& =& 4\times 485.995\text{ kJ}\\ {\text{4MnO}}_2+3{\text{O}}_2+{\text{14H}}_2\text{O}\to 6{\text{OH}}^{-}+4{\text{MnO}}_{\text{4}}{}^{-}+16{\text{H}}^{+}\Delta G°& =& 1479.697\text{ kJ}\end{array}$

Therefore, the value $\Delta G°$ for the given reaction is $1479.697\text{ kJ}$.

The number of electrons transferred in the overall reaction is $12\text{ mol}$.

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).

$\begin{array}{rll}E°& =& \frac{-\left(1479.697\text{ kJ}\right)\left(\frac{1000\text{ J}}{1\text{ kJ}}\right)}{\left(12\text{ mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)}\\ & =& -1.278\text{V}\end{array}$

The value of $E°$ for the given reaction is $-1.278\text{V}$.

Conclusion

The overall balanced electrochemical reaction is as follows,

${\text{4MnO}}_2+3{\text{O}}_2+{\text{14H}}_2\text{O}\to 6{\text{OH}}^{-}+4{\text{MnO}}_{\text{4}}{}^{-}+16{\text{H}}^{+}$

The value of $E°$ and $\Delta G°$ for the given reaction is $-1.278\text{V}$ and $1479.697\text{ kJ}$ respectively.

Interpretation Introduction

(b)

Interpretation:

The overall balanced electrochemical reaction and 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 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 8.11E

The overall balanced electrochemical reaction is as follows,

${\text{2Cu}}^{+}\to \text{Cu}+{\text{Cu}}^{2+}$

The value of $E°$ and $\Delta G°$ for the given reaction is $-0.1791\text{V}$ and $34.561\text{ kJ}$ respectively.

Explanation of Solution

The given reaction is represented as,

${\text{Cu}}^{+}\to \text{Cu}+{\text{Cu}}^{2+}$

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

${\text{Cu}}^{2+}+2{\text{e}}^{-}\to \text{Cu }E°=0.3419\text{V}$

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,

$\text{Cu}\to {\text{Cu}}^{2+}+2{\text{e}}^{-}E°=-0.3419\text{V}$…(4)

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

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

${\text{Cu}}^{+}+{\text{e}}^{-}\to \text{Cu }E°=0.521\text{V}$…(5)

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

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 the reaction (4), $F$ and number of moles of electrons transferred in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(-0.3419\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& \left(65976.443\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& 65.9764\text{kJ}\end{array}$

The value $\Delta G°$ for the reaction (4) is $65.9764\text{kJ}$.

Substitute the values of the standard reduction potential of the reaction (5), $F$ and number of moles of electrons transferred in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{1 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(0.521\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& -\left(50268.685\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& -50.2687\text{kJ}\end{array}$

The value $\Delta G°$ for the reaction (5) is $-50.2687\text{kJ}$.

The balanced overall electrochemical reaction is obtained by multiplying chemical equation (5) by $2$ and then the equation (5) is added in equation (4). The formation of overall balanced chemical equation is represented as,

$\begin{array}{lll}\text{Cu}\to {\text{Cu}}^{2+}+2{\text{e}}^{-}\Delta G°& =& 65.9764\text{kJ}\\ {\text{2Cu}}^{+}+2{\text{e}}^{-}\to 2\text{Cu 2}\times \Delta G°& =& 2\times \left(-50.2687\text{kJ}\right)\\ {\text{2Cu}}^{+}\to \text{Cu}+{\text{Cu}}^{2+}\Delta G°& =& 34.561\text{ kJ}\end{array}$

Therefore, the value $\Delta G°$ for the given reaction is $34.561\text{ kJ}$.

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).

$\begin{array}{rll}E°& =& \frac{-\left(34.561\text{ kJ}\right)\left(\frac{1000\text{ J}}{1\text{ kJ}}\right)}{\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)}\\ & =& -0.1791\text{V}\end{array}$

The value of $E°$ for the given reaction is $-0.1791\text{V}$.

Conclusion

The overall balanced electrochemical reaction is as follows,

${\text{2Cu}}^{+}\to \text{Cu}+{\text{Cu}}^{2+}$

The value of $E°$ and $\Delta G°$ for the given reaction is $-0.1791\text{V}$ and $34.561\text{ kJ}$ respectively.

Interpretation Introduction

(c)

Interpretation:

The overall balanced electrochemical reaction and 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 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 8.11E

The overall balanced electrochemical reaction is as follows,

${\text{Br}}_2+2{\text{F}}^{-}\to 2{\text{Br}}^{-}+{\text{F}}_2$

The value of $E°$ and $\Delta G°$ for the given reaction is $-1.7790\text{V}$ and $343.294\text{ kJ}$ respectively.

Explanation of Solution

The given reaction is represented as,

${\text{Br}}_2+{\text{F}}^{-}\to {\text{Br}}^{-}+{\text{F}}_2$

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

${\text{F}}_{\text{2}}+2{\text{e}}^{-}\to 2{\text{F}}^{-}E°=2.866\text{V}$

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,

$2{\text{F}}^{-}\to {\text{F}}_{\text{2}}+2{\text{e}}^{-}E°=-2.866\text{V}$…(6)

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

From Table $8.2$, the reduction half reaction of ${\text{Br}}_{\text{2}}$ and the standard reduction potential of ${\text{Br}}_2$ is represented as,

${\text{Br}}_{\text{2}}+2{\text{e}}^{-}\to 2{\text{Br}}^{-}E°=1.087\text{V}$…(7)

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 reduction potential of reaction (6), $F$ and $n$ in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(-2.866\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& \left(553052.02\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& 553.052\text{ kJ}\end{array}$

The value $\Delta G°$ for the reaction (6) is $553.052\text{ kJ}$.

Substitute the values of the standard reduction potential of reaction (7), $F$ and number of moles of electrons transferred in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(1.087\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& -\left(209758.39\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& -209.758\text{ kJ}\end{array}$

The value $\Delta G°$ for the reaction (7) is $-209.758\text{ kJ}$.

Add chemical equation (6) and chemical equation (7) to obtain the balanced overall electrochemical reaction. The formation of overall balanced chemical equation is represented as,

$\begin{array}{lll}2{\text{F}}^{-}\to {\text{F}}_{\text{2}}+2{\text{e}}^{-}\Delta G°& =& 553.052\text{ kJ}\\ {\text{Br}}_{\text{2}}+2{\text{e}}^{-}\to 2{\text{Br}}^{-}\Delta G°& =& -209.758\text{ kJ}\\ {\text{Br}}_2+2{\text{F}}^{-}\to 2{\text{Br}}^{-}+{\text{F}}_2\Delta G°& =& 343.294\text{ kJ}\end{array}$

Therefore, the value $\Delta G°$ for the given reaction is $343.294\text{ kJ}$.

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).

$\begin{array}{rll}E°& =& \frac{-\left(343.294\text{ kJ}\right)\left(\frac{1000\text{ J}}{1\text{ kJ}}\right)}{\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)}\\ & =& -1.7790\text{V}\end{array}$

The value of $E°$ for the given reaction is $-1.7790\text{V}$.

Conclusion

The overall balanced electrochemical reaction is as follows,

${\text{Br}}_2+2{\text{F}}^{-}\to 2{\text{Br}}^{-}+{\text{F}}_2$

The value of $E°$ and $\Delta G°$ for the given reaction is $-1.7790\text{V}$ and $343.294\text{ kJ}$ respectively.

Interpretation Introduction

(d)

Interpretation:

The overall balanced electrochemical reaction and 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 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 8.11E

The overall balanced electrochemical reaction is as follows,

${\text{H}}_2{\text{O}}_2+2{\text{H}}^{+}+2{\text{Cl}}^{-}\to 2{\text{H}}_2\text{O}+{\text{Cl}}_{\text{2}}$

The value of $E°$ and $\Delta G°$ for the given reaction is $0.418\text{V}$ and $-80.661\text{ kJ}$ respectively.

Explanation of Solution

The given reaction is represented as,

${\text{H}}_2{\text{O}}_2+{\text{H}}^{+}+{\text{Cl}}^{-}\to {\text{H}}_2\text{O}+{\text{Cl}}_{\text{2}}$

From Table $8.2$, the reduction half reaction of ${\text{H}}_2{\text{O}}_2$ into ${\text{H}}_{\text{2}}\text{O}$ and the standard reduction potential of ${\text{H}}_2{\text{O}}_2$ into ${\text{H}}_{\text{2}}\text{O}$ is represented as,

${\text{H}}_2{\text{O}}_2+2{\text{H}}^{+}+2{\text{e}}^{-}\to 2{\text{H}}_2\text{O }E°=1.776\text{V}$…(8)

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

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

${\text{Cl}}_{\text{2}}+2{\text{e}}^{-}\to 2{\text{Cl}}^{-}E°=1.358\text{V}$

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,

$2{\text{Cl}}^{-}\to {\text{Cl}}_{\text{2}}+2{\text{e}}^{-}E°=-1.358\text{V}$…(9)

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 reduction potential of the reaction (8), $F$ and number of moles of electrons transferred in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(1.776\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& -\left(342714.72\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& -342.714\text{kJ}\end{array}$

The value $\Delta G°$ for the reaction (8) is $-342.714\text{kJ}$.

Substitute the values of the standard oxidation potential of the reaction (9), $F$ and number of moles of electrons transferred in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(-1.358\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& \left(262053.26\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& 262.053\text{kJ}\end{array}$

The value $\Delta G°$ for the reaction (9) is $262.053\text{kJ}$.

Add chemical equation (8) and chemical equation (9) to obtain the balanced overall electrochemical reaction. The formation of overall balanced chemical equation is represented as,

$\begin{array}{lll}{\text{H}}_2{\text{O}}_2+2{\text{H}}^{+}+2{\text{e}}^{-}\to 2{\text{H}}_2\text{O }\Delta G°& =& -342.714\text{kJ}\\ 2{\text{Cl}}^{-}\to {\text{Cl}}_{\text{2}}+2{\text{e}}^{-}\Delta G°& =& 262.053\text{kJ}\\ {\text{H}}_2{\text{O}}_2+2{\text{H}}^{+}+2{\text{Cl}}^{-}\to 2{\text{H}}_2\text{O}+{\text{Cl}}_{\text{2}}\Delta G°& =& -80.661\text{ kJ}\end{array}$

The value $\Delta G°$ for the given reaction is $-80.661\text{ kJ}$.

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).

$\begin{array}{rll}E°& =& \frac{-\left(-80.661\text{ kJ}\right)\left(\frac{1000\text{ J}}{1\text{ kJ}}\right)}{\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)}\\ & =& 0.418\text{V}\end{array}$

The value of $E°$ for the given reaction is $0.418\text{V}$.

Conclusion

The overall balanced electrochemical reaction is as follows,

${\text{H}}_2{\text{O}}_2+2{\text{H}}^{+}+2{\text{Cl}}^{-}\to 2{\text{H}}_2\text{O}+{\text{Cl}}_{\text{2}}$

The value of $E°$ and $\Delta G°$ for the given reaction is $0.418\text{V}$ and $-80.661\text{ kJ}$ respectively.