Chapter 18, Problem 18.29QP

(a)

Interpretation Introduction

Interpretation: The values of $\Delta G^{\text{ο}}$ and $E_{\text{cell}}^{\text{ο}}$ for the given reactions are to be calculated.

Concept introduction: The potential of a reduction half cell is called standard reduction potential if reactants and products are in their standard states at $\text{25}{}^{\text{ο}}\text{C}$.

To determine: The values of $\Delta G^{\text{ο}}$ and $E_{\text{cell}}^{\text{ο}}$ for the given reaction.

Answer to Problem 18.29QP

Solution

The values of $\Delta G^{\text{ο}}$ and $E_{\text{cell}}^{\text{ο}}$ for the given reaction are $\underset{\_}{35512J}$ and $\underset{\_}{0.368V}$ respectively.

Explanation of Solution

Explanation

The given reaction is,

${\text{2Cu}}^{\text{+}}\left(aq\right)\to {\text{Cu}}^{2+}\left(aq\right)+\text{Cu}\left(s\right)$ (1)

The oxidation half cell reaction at anode is,

${\text{Cu}}^{\text{2+}}\left(aq\right)+{\text{e}}^{-}\to {\text{Cu}}^{+}\left(aq\right)E_1{}^{\text{ο}}=0.153\text{V}$ (2)

The reduction half cell reaction at cathode is,

${\text{Cu}}^{\text{+}}\left(aq\right)+{\text{e}}^{-}\to \text{Cu}\left(s\right)E_2{}^{\text{ο}}=0.521\text{V}$ (3)

Where,

• $E_1{}^{\text{ο}}$ is the standard electrode potential of equation (2).
• $E_2{}^{\text{ο}}$ is the standard electrode potential of equation (3).

The standard electrode potential of cell ( $E_{\text{cell}}^{\text{ο}}$) is calculated by the formula,

$E_{\text{cell}}^{\text{ο}}=E_2{}^{\text{ο}}-E_1{}^{\text{ο}}$

Substitute the value of $E_1{}^{\text{ο}}$ and $E_2{}^{\text{ο}}$ in the above equation.

$\begin{array}{c}{E}_{\text{cell}}^{\text{ο}}=0.521\text{V}-0.153\text{V}\\ =\underset{\_}{0.368V}\end{array}$

The value of $E_{\text{cell}}^{\text{ο}}$ is $\underset{\_}{0.368V}$.

The standard change in free energy ( $\Delta G^{\text{ο}}$) is calculated by the formula,

$\Delta G^{\text{ο}}=-nF{E}_{\text{cell}}^{\text{ο}}$

Where,

• $n$ is the number of electrons.
• $F$ is the faraday constant ( $96500\text{J}{\text{V}}^{-1}$).

The value of $n$ for the reaction is $1$.

Substitute the value of $n$, $F$ and $E_{\text{cell}}^{\text{ο}}$ in the above equation.

$\begin{array}{c}\Delta G^{\text{ο}}=-1\times 96500\text{J}{\text{V}}^{-1}\times 0.368\text{V}\\ =\underset{\_}{35512J}\end{array}$

The value of $\Delta G^{\text{ο}}$ is $\underset{\_}{35512J}$.

(b)

Interpretation Introduction

To determine: The values of $\Delta G^{\text{ο}}$ and $E_{\text{cell}}^{\text{ο}}$ for the given reaction.

Answer to Problem 18.29QP

Solution

The values of $\Delta G^{\text{ο}}$ and $E_{\text{cell}}^{\text{ο}}$ for the given reaction are $\underset{\_}{82604J}$ and $\underset{\_}{0.428V}$ respectively.

Explanation of Solution

Explanation

The given reaction is,

$\text{Cu}\left(s\right)+{\text{2Fe}}^{\text{3+}}\left(aq\right)\to {\text{Cu}}^{2+}\left(aq\right)+{\text{2Fe}}^{\text{2+}}\left(aq\right)$ (4)

The oxidation half cell reaction at anode is,

$\text{Cu}\left(s\right)+2{\text{e}}^{-}\to {\text{Cu}}^{2+}\left(aq\right)E_3{}^{\text{ο}}=0.342\text{V}$ (5)

The reduction half cell reaction at cathode is,

${\text{Fe}}^{\text{3+}}\left(aq\right)+{\text{e}}^{-}\to {\text{Fe}}^{\text{2+}}\left(aq\right)E_4{}^{\text{ο}}=0.770\text{V}$ (6)

Where,

• $E_3{}^{\text{ο}}$ is the standard electrode potential of equation (5).
• $E_4{}^{\text{ο}}$ is the standard electrode potential of equation (6).

The standard electrode potential of cell ( $E_{\text{cell}}^{\text{ο}}$) is calculated by the formula,

$E_{\text{cell}}^{\text{ο}}=E_4{}^{\text{ο}}-E_3{}^{\text{ο}}$

Substitute the value of $E_3{}^{\text{ο}}$ and $E_4{}^{\text{ο}}$ in the above equation.

$\begin{array}{c}{E}_{\text{cell}}^{\text{ο}}=0.770\text{V}-0.342\text{V}\\ =\underset{\_}{0.428V}\end{array}$

The value of $E_{\text{cell}}^{\text{ο}}$ is $\underset{\_}{0.428V}$.

The standard change in free energy ( $\Delta G^{\text{ο}}$) is calculated by the formula,

$\Delta G^{\text{ο}}=-nF{E}_{\text{cell}}^{\text{ο}}$

Where,

• $n$ is the number of electrons.
• $F$ is the faraday constant ( $96500\text{J}{\text{V}}^{-1}$).

The value of $n$ for the reaction is $2$.

Substitute the value of $n$, $F$ and $E_{\text{cell}}^{\text{ο}}$ in the above equation.

$\begin{array}{c}\Delta G^{\text{ο}}=-2\times 96500\text{J}{\text{V}}^{-1}\times 0.428\text{V}\\ =\underset{\_}{82604J}\end{array}$

The value of $\Delta G^{\text{ο}}$ is $\underset{\_}{82604J}$.

Conclusion

The values of $\Delta G^{\text{ο}}$ and $E_{\text{cell}}^{\text{ο}}$ for the first reaction are $\underset{\_}{35512J}$ and $\underset{\_}{0.368V}$ respectively