Chapter 21, Problem 21.138P

a)

Interpretation Introduction

Interpretation:

The value of ${\text{E}}^{\text{o}}{}_{\text{half-cell}}$ for (3) has to be calculated using ${\text{E}}^{\text{o}}{}_{\text{half-cell}}$ of (1) and (2).

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: ${\text{E}}^{\text{o}}{}_{\text{cell}}$ is composed of a contribution from the anode and a contribution from the cathode is given by,

${\text{E}}_{\text{cell}}^{\text{o}}\text{=}{\text{E}}_{\text{cathode}}^{\text{o}}\text{-}{\text{E}}_{\text{anode}}^{\text{o}}$

Where both ${\text{E}}_{\text{cathode}}^{\text{o}}$ and ${\text{E}}_{\text{anode}}^{\text{o}}$ are the standard reduction potentials of the electrodes.

Explanation of Solution

The given half-reactions (1) and (2) are,

$\begin{array}{l}{\text{Fe}}^{\text{3+}}\left(aq\right)+e^{-}\rightleftarrows {\text{Fe}}^{\text{2+}}{\text{E}}_{\text{half-cell}}^{\text{0}}\left(1\right)\text{=}+\text{0}\text{.77}\text{V}\\ \\ {\text{Fe}}^{\text{2+}}\left(\text{aq}\right)\text{+}2{\text{e}}^{\text{-}}\rightleftarrows \text{Fe}\left(\text{S}\right){\text{E}}_{\text{half-cell}}^{\text{0}}\left(2\right)\text{=}\text{-0}\text{.44}\text{V}\end{array}$

Equation (3) is obtained by addition of Equation (1) and (2) gives,

$\underset{\_}{\begin{array}{l}\underset{\_}{\begin{array}{l}{\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\text{+}{\text{e}}^{\text{-}}\rightleftarrows \cancel{{\text{Fe}}^{\text{2+}}\left(\text{aq}\right)}{\text{E}}_{\text{half-cell}}^{\text{0}}\left(\text{1}\right)\text{=}\text{+0}\text{.77}\text{V}\\ \\ \cancel{{\text{Fe}}^{\text{2+}}\left(\text{aq}\right)}\text{+}{\text{2e}}^{\text{-}}\rightleftarrows \text{Fe}\left(\text{S}\right){\text{E}}_{\text{half-cell}}^{\text{0}}\left(\text{2}\right)\text{=}\text{-}\text{0}\text{.44}\text{V}\\ \end{array}}\\ {\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\text{+}{\text{2e}}^{\text{-}}\rightleftarrows \text{Fe}\left(\text{S}\right){\text{E}}_{\text{half-cell}}^{\text{0}}\left(\text{3}\right)\text{=}+0.33V\end{array}}$

Therefore, ${\text{E}}^{\text{o}}{}_{\text{half-cell}}$ for half-reaction (3) is $+0.33V$

b)

Interpretation Introduction

Interpretation:

Gibb’s free energy ${\text{ΔG}}^{\text{o}}$ has to be calculated for equation (1) and (2).

Concept Introduction:

Gibb’s free energy: The energy available to do work and also used to determine the spontaneity of a reaction. The energy released by the overall system.

The Gibb’s free energy is,

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}\text{n}\text{F}{\text{E}}^{\text{o}}\\ \text{where,}\\ \text{n is the number of electrons involved,}\\ \text{F is the Faraday constant,}\\ {\text{E}}^{\text{o}}\text{ is the standard reduction potential}\text{.}\end{array}$

Explanation of Solution

Predict ${\text{E}}^{\text{o}}$ value for both half-reactions:

The given half-reactions (1) and (2) are,

$\begin{array}{l}{\text{Fe}}^{\text{3+}}\left(aq\right)+e^{-}\rightleftarrows {\text{Fe}}^{\text{2+}}{\text{E}}_{\text{half-cell}}^{\text{0}}\left(1\right)\text{=}+\text{0}\text{.77}\text{V}\\ \\ {\text{Fe}}^{\text{2+}}\left(\text{aq}\right)\text{+}2{\text{e}}^{\text{-}}\rightleftarrows \text{Fe}\left(\text{S}\right){\text{E}}_{\text{half-cell}}^{\text{0}}\left(2\right)\text{=}\text{-0}\text{.44}\text{V}\end{array}$

For equation (1),

$\begin{array}{c}{\text{ΔG}}^{\text{o}}\text{=}\text{-n}\text{F}{\text{E}}^{\text{o}}\\ \\ \text{=}\text{-}\left({\text{1 mole e}}^{\text{-}}\right)\left(\text{96,500 }\frac{\text{C}}{{\text{mol e}}^{\text{-}}}\right)\left(+\text{0}\text{.77}\text{V}\right)\left(\frac{\text{J}}{\text{C}\text{.V}}\right)\\ \\ {\text{ΔG}}^{\text{o}}\left(\text{1}\right)\text{= -}\text{74305}\text{J}\text{=}-7.4\times 10^{4}J\end{array}$

For equation (2),

$\begin{array}{c}{\text{ΔG}}^{\text{o}}\text{=}\text{-n}\text{F}{\text{E}}^{\text{o}}\\ \\ \text{=}\text{-}\left({\text{2 mole e}}^{\text{-}}\right)\left(\text{96,500 }\frac{\text{C}}{{\text{mol e}}^{\text{-}}}\right)\left(-0.44\text{V}\right)\left(\frac{\text{J}}{\text{C}\text{.V}}\right)\\ \\ {\text{ΔG}}^{\text{o}}\left(\text{1}\right)\text{= +84920}\text{J}\text{=}+8.5\times 10^{4}J\end{array}$

Therefore, the Gibb’s free energy ${\text{ΔG}}^{\text{o}}$ for equation (1) and (2) are $-7.4\times 10^{4}J$ and $+8.5\times 10^{4}J$, respectively.

c)

Interpretation Introduction

Interpretation:

Gibb’s free energy ${\text{ΔG}}^{\text{o}}$ of equation (3) has to be calculated for equation (1) and (2).

Concept Introduction:

Gibb’s free energy: The energy available to do work and also used to determine the spontaneity of a reaction. The energy released by the overall system.

The Gibb’s free energy is,

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}\text{n}\text{F}{\text{E}}^{\text{o}}\\ \text{where,}\\ \text{n is the number of electrons involved,}\\ \text{F is the Faraday constant,}\\ {\text{E}}^{\text{o}}\text{ is the standard reduction potential}\text{.}\end{array}$

Explanation of Solution

The given half-reactions (1) and (2) are,

$\begin{array}{l}{\text{Fe}}^{\text{3+}}\left(aq\right)+e^{-}\rightleftarrows {\text{Fe}}^{\text{2+}}{\text{E}}_{\text{half-cell}}^{\text{0}}\left(1\right)\text{=}+\text{0}\text{.77}\text{V}\\ \\ {\text{Fe}}^{\text{2+}}\left(\text{aq}\right)\text{+}2{\text{e}}^{\text{-}}\rightleftarrows \text{Fe}\left(\text{S}\right){\text{E}}_{\text{half-cell}}^{\text{0}}\left(2\right)\text{=}\text{-0}\text{.44}\text{V}\end{array}$

As known,

$\begin{array}{c}{\text{ΔG}}_{\text{1}}^{\text{o}}\text{=}\text{-n}\text{F}{\text{E}}_{\text{1}}^{\text{o}}\text{and}{\text{ΔG}}_{\text{2}}^{\text{o}}\text{=}\text{-n}\text{F}{\text{E}}_{\text{2}}^{\text{o}}\\ \\ {\text{ΔG}}_{\text{3}}^{\text{o}}\text{=}\text{-n}\text{F}{\text{E}}_{\text{3}}^{\text{o}}\to (a)\\ \\ \text{where}{\text{E}}_{\text{3}}^{\text{o}}\text{=}{\text{E}}_{\text{1}}^{\text{o}}\text{+}{\text{E}}_{\text{2}}^{\text{o}}\to (b)\end{array}$

By substituting (b) into (a) gives,

$\begin{array}{c}{\text{ΔG}}_{\text{3}}^{\text{o}}\text{=}\text{-n}\text{F}\left({\text{E}}_{\text{1}}^{\text{o}}\text{+}{\text{E}}_{\text{2}}^{\text{o}}\right)\\ \\ {\text{ΔG}}_{\text{3}}^{\text{o}}\text{=}\left(\text{-n}{\text{FE}}_{\text{1}}^{\text{o}}\right)\text{+}\left(\text{-n}{\text{FE}}_{\text{2}}^{\text{o}}\right)\\ \\ {\text{ΔG}}_{\text{3}}^{\text{o}}\text{=}{\text{ΔG}}_{\text{1}}^{\text{o}}\text{+}{\text{ΔG}}_{\text{2}}^{\text{o}}\end{array}$

The ${\text{E}}^{\text{o}}{}_{\text{half-cell}}$ for equation (3) is obtained by addition of (1) and (2), whereas the addition of Gibb’s free energy of each half-reaction (1) and (2) gives, the value of equation of (3).

Thus,

The calculated Gibb’s free energy ${\text{ΔG}}^{\text{o}}$ values for equation (1) and (2) are $-7.4\times 10^{4}J$ and $+8.5\times 10^{4}J$, respectively

Hence,

$\begin{array}{c}{\text{ΔG}}_{\text{3}}^{\text{o}}\text{=}{\text{ΔG}}_{\text{1}}^{\text{o}}\text{+}{\text{ΔG}}_{\text{2}}^{\text{o}}\\ \\ \text{=}\left(\text{-7}\text{.4}{\text{×10}}^{\text{4}}\text{J}\right)\text{+}\left(\text{+8}\text{.5}{\text{×10}}^{\text{4}}\text{J}\right)\\ \\ =1.1\times 10^{4}J\end{array}$

Therefore, the calculated ${\text{ΔG}}_{\text{3}}^{\text{o}}$ value for equation (3) is $1.1\times 10^{4}J$.

d)

Interpretation Introduction

Interpretation:

The ${\text{E}}^{\text{o}}{}_{\text{half-cell}}$ value for equation (3) has to be calculated using its Gibb’s free energy ${\text{ΔG}}^{\text{o}}$.

Concept Introduction:

Gibb’s free energy: The energy available to do work and also used to determine the spontaneity of a reaction. The energy released by the overall system.

The Gibb’s free energy is,

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}\text{n}\text{F}{\text{E}}^{\text{o}}\\ \text{where,}\\ \text{n is the number of electrons involved,}\\ \text{F is the Faraday constant,}\\ {\text{E}}^{\text{o}}\text{ is the standard reduction potential}\text{.}\end{array}$

Explanation of Solution

The given half-reactions (1) and (2) are,

$\begin{array}{l}{\text{Fe}}^{\text{3+}}\left(aq\right)+e^{-}\rightleftarrows {\text{Fe}}^{\text{2+}}{\text{E}}_{\text{half-cell}}^{\text{0}}\left(1\right)\text{=}+\text{0}\text{.77}\text{V}\\ \\ {\text{Fe}}^{\text{2+}}\left(\text{aq}\right)\text{+}2{\text{e}}^{\text{-}}\rightleftarrows \text{Fe}\left(\text{S}\right){\text{E}}_{\text{half-cell}}^{\text{0}}\left(2\right)\text{=}\text{-0}\text{.44}\text{V}\end{array}$

As known,

$\begin{array}{c}{\text{ΔG}}_{\text{1}}^{\text{o}}\text{=}\text{-n}\text{F}{\text{E}}_{\text{1}}^{\text{o}}\text{and}{\text{ΔG}}_{\text{2}}^{\text{o}}\text{=}\text{-n}\text{F}{\text{E}}_{\text{2}}^{\text{o}}\\ \\ {\text{ΔG}}_{\text{3}}^{\text{o}}\text{=}\text{-n}\text{F}{\text{E}}_{\text{3}}^{\text{o}}\to (a)\\ \\ \text{where}{\text{E}}_{\text{3}}^{\text{o}}\text{=}{\text{E}}_{\text{1}}^{\text{o}}\text{+}{\text{E}}_{\text{2}}^{\text{o}}\to (b)\end{array}$

By substituting (b) into (a) gives,

$\begin{array}{c}{\text{ΔG}}_{\text{3}}^{\text{o}}\text{=}\text{-n}\text{F}\left({\text{E}}_{\text{1}}^{\text{o}}\text{+}{\text{E}}_{\text{2}}^{\text{o}}\right)\\ \\ {\text{ΔG}}_{\text{3}}^{\text{o}}\text{=}\left(\text{-n}{\text{FE}}_{\text{1}}^{\text{o}}\right)\text{+}\left(\text{-n}{\text{FE}}_{\text{2}}^{\text{o}}\right)\\ \\ {\text{ΔG}}_{\text{3}}^{\text{o}}\text{=}{\text{ΔG}}_{\text{1}}^{\text{o}}\text{+}{\text{ΔG}}_{\text{2}}^{\text{o}}\end{array}$

The ${\text{E}}^{\text{o}}{}_{\text{half-cell}}$ for equation (3) is obtained by addition of (1) and (2), whereas the addition of Gibb’s free energy of each half-reaction (1) and (2) gives, the value of equation of (3).

Thus,

The calculated Gibb’s free energy ${\text{ΔG}}^{\text{o}}$ values for equation (1) and (2) are $-7.4\times 10^{4}J$ and $+8.5\times 10^{4}J$, respectively

Hence,

$\begin{array}{c}{\text{ΔG}}_{\text{3}}^{\text{o}}\text{=}{\text{ΔG}}_{\text{1}}^{\text{o}}\text{+}{\text{ΔG}}_{\text{2}}^{\text{o}}\\ \\ \text{=}\left(\text{-7}\text{.4}{\text{×10}}^{\text{4}}\text{J}\right)\text{+}\left(\text{+8}\text{.5}{\text{×10}}^{\text{4}}\text{J}\right)\\ \\ =1.1\times 10^{4}J\end{array}$

Thus, the calculated ${\text{ΔG}}_{\text{3}}^{\text{o}}$ value for equation (3) is $1.1\times 10^{4}J$.

Calculation for ${\text{E}}^{\text{o}}{}_{\text{half-cell}}$ equation (3) is,

$\begin{array}{c}{\text{ΔG}}^{\text{o}}\text{=}\text{-n}\text{F}{\text{E}}^{\text{o}}\\ \\ \text{1}{\text{.1×10}}^{\text{4}}\text{J=}\text{-}\left({\text{3 mole e}}^{\text{-}}\right)\left(\text{96,500 }\frac{\text{C}}{{\text{mol e}}^{\text{-}}}\right)\left({\text{E}}^{\text{o}}\right)\left(\frac{\text{J}}{\text{C}\text{.V}}\right)\\ \\ \left(\frac{\text{1}{\text{.1×10}}^{\text{4}}\text{J}}{\text{-}\left({\text{3 mole e}}^{\text{-}}\right)\left(\text{96,500 }\frac{\text{C}}{{\text{mol e}}^{\text{-}}}\right)\left(\frac{\text{J}}{\text{C}\text{.V}}\right)}\right)\text{= }\frac{\text{1}{\text{.1×10}}^{\text{4}}}{2.8{\text{×10}}^5}\text{V}\text{=}-0.039V\end{array}$

Therefore, the calculated value of ${\text{E}}^{\text{o}}{}_{\text{half-cell}}$ value for equation (3) is $-0.039V$

e)

Interpretation Introduction

Interpretation:

The relationship between ${\text{E}}^{\text{o}}{}_{\text{half-cell}}$ values for equation (1) and (2) and the ${\text{E}}^{\text{o}}{}_{\text{half-cell}}$ value for equation (3) has to be interpreted.

Concept Introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: ${\text{E}}^{\text{o}}{}_{\text{cell}}$ is composed of a contribution from the anode and a contribution from the cathode is given by,

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

Where both $E_{cathode}^o$ and $E_{anode}^o$ are the standard reduction potentials of the electrodes.

Explanation of Solution

The given half-reactions (1) and (2) are,

$\begin{array}{l}{\text{Fe}}^{\text{3+}}\left(aq\right)+e^{-}\rightleftarrows {\text{Fe}}^{\text{2+}}{\text{E}}_{\text{half-cell}}^{\text{0}}\left(1\right)\text{=}+\text{0}\text{.77}\text{V}\\ \\ {\text{Fe}}^{\text{2+}}\left(\text{aq}\right)\text{+}2{\text{e}}^{\text{-}}\rightleftarrows \text{Fe}\left(\text{S}\right){\text{E}}_{\text{half-cell}}^{\text{0}}\left(2\right)\text{=}\text{-0}\text{.44}\text{V}\end{array}$

Equation (3) is obtained by addition of Equation (1) and (2) gives,

$\underset{\_}{\begin{array}{l}\underset{\_}{\begin{array}{l}{\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\text{+}{\text{e}}^{\text{-}}\rightleftarrows \cancel{{\text{Fe}}^{\text{2+}}\left(\text{aq}\right)}{\text{E}}_{\text{half-cell}}^{\text{0}}\left(\text{1}\right)\text{=}\text{+0}\text{.77}\text{V}\\ \\ \cancel{{\text{Fe}}^{\text{2+}}\left(\text{aq}\right)}\text{+}{\text{2e}}^{\text{-}}\rightleftarrows \text{Fe}\left(\text{S}\right){\text{E}}_{\text{half-cell}}^{\text{0}}\left(\text{2}\right)\text{=}\text{-}\text{0}\text{.44}\text{V}\\ \end{array}}\\ {\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\text{+}{\text{2e}}^{\text{-}}\rightleftarrows \text{Fe}\left(\text{S}\right){\text{E}}_{\text{half-cell}}^{\text{0}}\left(\text{3}\right)\text{=}+0.33V\end{array}}$

From the calculation method,

The addition of equation (1) and (2) gives the overall reaction of (3), that indicates the ${\text{E}}^{\text{o}}{}_{\text{half-cell}}$ value for (3) is

${\text{E}}^{\text{o}}{}_{\text{half-cell}}\left(3\right)={\text{E}}^{\text{o}}{}_{\text{half-cell}}\left(1\right)+{\text{E}}^{\text{o}}{}_{\text{half-cell}}\left(2\right)$