Chapter 19, Problem 19.35QP

(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 19.35QP

Solution

The values of $\Delta G^{\text{ο}}$ and $E_{\text{cell}}^{\text{ο}}$ for the given reaction are $\underset{\_}{55970J}$ and $\underset{\_}{0.29V}$ respectively.

Explanation of Solution

Explanation

The given reaction is,

${\text{Cl}}_{\text{2}}\left(g\right)+{\text{2Br}}^{-}\left(aq\right)\to {\text{Br}}_{\text{2}}\left(l\right)+2{\text{Cl}}^{-}\left(aq\right)$ (1)

The oxidation half cell reaction at anode is,

${\text{2Br}}^{-}\left(aq\right)\to {\text{Br}}_{\text{2}}\left(aq\right)+2{\text{e}}^{-}{E}_1{}^{\text{ο}}=-1.07\text{V}$ (2)

The reduction half cell reaction at cathode is,

${\text{Cl}}_2\left(aq\right)+2{\text{e}}^{-}\to 2{\text{Cl}}^{-}\left(aq\right)E_2{}^{\text{ο}}=1.36\text{V}$ (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,

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

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

$\begin{array}{c}{E}_{\text{cell}}^{\text{ο}}=-1.07\text{V}+1.36\text{V}\\ =\underset{\_}{0.29V}\end{array}$

The value of $E_{\text{cell}}^{\text{ο}}$ is $\underset{\_}{0.29V}$.

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.

$\begin{array}{c}\Delta G^{\text{ο}}=2\times 96500\text{J}{\text{V}}^{-1}\times 0.29\text{V}\\ =\underset{\_}{55970J}\end{array}$

The value of $\Delta G^{\text{ο}}$ is $\underset{\_}{55970J}$.

(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 19.35QP

Solution

The values of $\Delta G^{\text{ο}}$ and $E_{\text{cell}}^{\text{ο}}$ for the given reaction are $\underset{\_}{102290J}$ and $\underset{\_}{0.53V}$ respectively.

Explanation of Solution

Explanation

The given reaction is,

$\text{Zn}\left(s\right)+{\text{Ni}}^{\text{2+}}\left(aq\right)\to {\text{Zn}}^{2+}\left(aq\right)+\text{Ni}\left(s\right)$ (4)

The oxidation half cell reaction at anode is,

$\text{Zn}\left(s\right)\to {\text{Zn}}^{2+}\left(aq\right)+{\text{2e}}^{-}{E}_3{}^{\text{ο}}=0.76\text{V}$ (5)

The reduction half cell reaction at cathode is,

${\text{Ni}}^{\text{2+}}\left(aq\right)+2{\text{e}}^{-}\to \text{Ni}\left(s\right)E_4{}^{\text{ο}}=-0.23\text{V}$ (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,

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

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

$\begin{array}{c}{E}_{\text{cell}}^{\text{ο}}=0.76\text{V}-0.23\text{V}\\ =\underset{\_}{0.53V}\end{array}$

The value of $E_{\text{cell}}^{\text{ο}}$ is $\underset{\_}{0.53V}$.

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.

$\begin{array}{c}\Delta G^{\text{ο}}=2\times 96500\text{J}{\text{V}}^{-1}\times 0.53\text{V}\\ =\underset{\_}{102290J}\end{array}$

The value of $\Delta G^{\text{ο}}$ is $\underset{\_}{102290J}$.

Conclusion

The values of $\Delta G^{\text{ο}}$ and $E_{\text{cell}}^{\text{ο}}$ for the first reaction are $\underset{\_}{55970J}$ and $\underset{\_}{0.29V}$ respectively