Chapter 5, Problem 5.73P

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant for the given reaction has to be calculated.

Concept Introduction:

Half–cell reactions: In the electrochemical cell, both oxidation and reduction occur simultaneously. The oxidation occurs at the anode, while the reduction occurs at the cathode. The loss of an electron occurs from the anode by oxidation as well as the gain of an electron occurs from the cathode by reduction.

Standard cell potentials: The standard emf (electromotive force, $\text{E}°$) of the half-cell is measured against the standard hydrogen electrode (SHE) at standard temperature and pressure with the electrolyte concentration of $\text{1}\text{.0 M}$. The $\text{E}°$ values for various materials are available.

Redox reactions: A redox reaction is a chemical reaction where both oxidation and reduction occur simultaneously. In a redox reaction, one of the reactant is oxidized, while the other reactant is reduced at the same time. It can be represented as two half-reactions with the number of transferred electrons. They are as follows:

• Oxidation half-reaction.
• Reduction half-reaction.

Explanation of Solution

The given reaction is $Sn\left(s\right)+S{n}^{4+}\left(aq\right)\rightleftharpoons 2S{n}^{2+}\left(aq\right)$

$\begin{array}{l}Sn\left(s\right)+S{n}^{4+}\left(aq\right)\rightleftharpoons 2S{n}^{2+}\left(aq\right)\\ R:S{n}^{4+}\left(aq\right)+2e^{-}\to S{n}^{2+}\left(aq\right)+0.15V\\ L:S{n}^{2+}\left(aq\right)+2e^{-}\to Sn\left(aq\right)-0.14V\\ ---------------------------\\ E=+0.29V\\ \end{array}$

The equilibrium constant, K is calculated as follows,

$\begin{array}{c}\ln K=\frac{nFE}{RT}\\ \\ =\frac{2\times 0.29V}{25.693\times {10}^{-3}V}\\ \\ \ln K=22.6\\ \\ K=e^{22.6}\\ \\ =6.5\times {10}^9\end{array}$

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant for the given reaction has to be calculated.

Concept Introduction:

Half–cell reactions: In the electrochemical cell, both oxidation and reduction occur simultaneously. The oxidation occurs at the anode, while the reduction occurs at the cathode. The loss of an electron occurs from the anode by oxidation as well as the gain of an electron occurs from the cathode by reduction.

Standard cell potentials: The standard emf (electromotive force, $\text{E}°$) of the half-cell is measured against the standard hydrogen electrode (SHE) at standard temperature and pressure with the electrolyte concentration of $\text{1}\text{.0 M}$. The $\text{E}°$ values for various materials are available.

Redox reactions: A redox reaction is a chemical reaction where both oxidation and reduction occur simultaneously. In a redox reaction, one of the reactant is oxidized, while the other reactant is reduced at the same time. It can be represented as two half-reactions with the number of transferred electrons. They are as follows:

• Oxidation half-reaction.
• Reduction half-reaction.

Explanation of Solution

The given reaction is $Sn\left(s\right)+2AgBr\left(s\right)\left(aq\right)\rightleftharpoons SnB{r}_2\left(aq\right)+2Ag\left(s\right)$

$\begin{array}{l}Sn\left(s\right)+2AgBr\left(s\right)\left(aq\right)\rightleftharpoons SnB{r}_2\left(aq\right)+2Ag\left(s\right)\\ R:AgBr\left(s\right)+e^{-}\to Ag\left(s\right)+B{r}^{-}\left(aq\right)+0.07V\\ L:S{n}^{2+}\left(aq\right)+2e^{-}\to Sn\left(s\right)-0.14V\\ ------------------------------\\ E=+0.21V\\ \end{array}$

The equilibrium constant, K is calculated as follows,

$\begin{array}{c}\ln K=\frac{nFE}{RT}\\ \\ =\frac{2\times 0.21V}{25.693\times {10}^{-3}V}\\ \\ \ln K=16.3\\ \\ K=e^{16.3}\\ \\ =1.2\times {10}^7\end{array}$

(c)

Interpretation Introduction

Interpretation:

The equilibrium constant for the given reaction has to be calculated.

Concept Introduction:

Half–cell reactions: In the electrochemical cell, both oxidation and reduction occur simultaneously. The oxidation occurs at the anode, while the reduction occurs at the cathode. The loss of an electron occurs from the anode by oxidation as well as the gain of an electron occurs from the cathode by reduction.

Standard cell potentials: The standard emf (electromotive force, $\text{E}°$) of the half-cell is measured against the standard hydrogen electrode (SHE) at standard temperature and pressure with the electrolyte concentration of $\text{1}\text{.0 M}$. The $\text{E}°$ values for various materials are available.

Redox reactions: A redox reaction is a chemical reaction where both oxidation and reduction occur simultaneously. In a redox reaction, one of the reactant is oxidized, while the other reactant is reduced at the same time. It can be represented as two half-reactions with the number of transferred electrons. They are as follows:

• Oxidation half-reaction.
• Reduction half-reaction.

Explanation of Solution

The given reaction is $Fe\left(s\right)+Hg{\left(N{O}_3\right)}_2\left(aq\right)\rightleftharpoons Hg\left(l\right)+Fe{\left(N{O}_3\right)}_2\left(aq\right)$

$\begin{array}{l}Fe\left(s\right)+Hg{\left(N{O}_3\right)}_2\left(aq\right)\rightleftharpoons Hg\left(l\right)+Fe{\left(N{O}_3\right)}_2\left(aq\right)\\ R:H{g}^{2+}\left(aq\right)+2e^{-}\to Hg\left(l\right)+0.86V\\ L:F{e}^{2+}\left(aq\right)+2e^{-}\to Fe\left(s\right)-0.44V\\ -----------------------------------\\ E=+1.30V\\ \end{array}$

The equilibrium constant, K is calculated as follows,

$\begin{array}{c}\ln K=\frac{nFE}{RT}\\ \\ =\frac{2\times 1.30V}{25.693\times {10}^{-3}V}\\ \\ \ln K=101\\ \\ K=e^{101}\\ \\ =7.3\times {10}^{43}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The equilibrium constant for the given reaction has to be calculated.

Concept Introduction:

Half–cell reactions: In the electrochemical cell, both oxidation and reduction occur simultaneously. The oxidation occurs at the anode, while the reduction occurs at the cathode. The loss of an electron occurs from the anode by oxidation as well as the gain of an electron occurs from the cathode by reduction.

Standard cell potentials: The standard emf (electromotive force, $\text{E}°$) of the half-cell is measured against the standard hydrogen electrode (SHE) at standard temperature and pressure with the electrolyte concentration of $\text{1}\text{.0 M}$. The $\text{E}°$ values for various materials are available.

Redox reactions: A redox reaction is a chemical reaction where both oxidation and reduction occur simultaneously. In a redox reaction, one of the reactant is oxidized, while the other reactant is reduced at the same time. It can be represented as two half-reactions with the number of transferred electrons. They are as follows:

• Oxidation half-reaction.
• Reduction half-reaction.

Explanation of Solution

The given reaction is $Cd\left(s\right)+CuS{O}_4\left(aq\right)\rightleftharpoons Cu\left(s\right)+CdS{O}_4\left(aq\right)$

$\begin{array}{l}Cd\left(s\right)+CuS{O}_4\left(aq\right)\rightleftharpoons Cu\left(s\right)+CdS{O}_4\left(aq\right)\\ R:C{u}^{2+}\left(aq\right)+2e^{-}\to Cu\left(s\right)+0.34V\\ L:C{d}^{2+}\left(aq\right)+2e^{-}\to Cd\left(s\right)-0.40V\\ ---------------------------\\ E=+0.74V\\ \end{array}$

The equilibrium constant, K is calculated as follows,

$\begin{array}{c}\ln K=\frac{nFE}{RT}\\ \\ =\frac{2\times 0.74V}{25.693\times {10}^{-3}V}\\ \\ \ln K=57.6\\ \\ K=e^{57.6}\\ \\ =1\times {10}^{25}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The equilibrium constant for the given reaction has to be calculated.

Concept Introduction:

Half–cell reactions: In the electrochemical cell, both oxidation and reduction occur simultaneously. The oxidation occurs at the anode, while the reduction occurs at the cathode. The loss of an electron occurs from the anode by oxidation as well as the gain of an electron occurs from the cathode by reduction.

Standard cell potentials: The standard emf (electromotive force, $\text{E}°$) of the half-cell is measured against the standard hydrogen electrode (SHE) at standard temperature and pressure with the electrolyte concentration of $\text{1}\text{.0 M}$. The $\text{E}°$ values for various materials are available.

Redox reactions: A redox reaction is a chemical reaction where both oxidation and reduction occur simultaneously. In a redox reaction, one of the reactant is oxidized, while the other reactant is reduced at the same time. It can be represented as two half-reactions with the number of transferred electrons. They are as follows:

• Oxidation half-reaction.
• Reduction half-reaction.

Explanation of Solution

The given reaction is $C{u}^{2+}\left(aq\right)+Cu\left(s\right)\rightleftharpoons 2C{u}^{2+}\left(aq\right)$

$\begin{array}{l}C{u}^{2+}\left(aq\right)+Cu\left(s\right)\rightleftharpoons 2C{u}^{2+}\left(aq\right)\\ R:C{u}^{2+}\left(aq\right)+e^{-}\to C{u}^{+}\left(aq\right)+0.16V\\ L:C{u}^{+}\left(aq\right)+e^{-}\to Cu\left(s\right)+0.52V\\ ---------------------------\\ E=-0.36V\\ \end{array}$

The equilibrium constant, K is calculated as follows,

$\begin{array}{c}\ln K=\frac{nFE}{RT}\\ \\ =\frac{-0.36V}{25.693\times {10}^{-3}V}\\ \\ \ln K=-14\\ \\ K=e^{-14}\\ \\ =8.3\times {10}^{-7}\end{array}$

(f)

Interpretation Introduction

Interpretation:

The equilibrium constant for the given reaction has to be calculated.

Concept Introduction:

Half–cell reactions: In the electrochemical cell, both oxidation and reduction occur simultaneously. The oxidation occurs at the anode, while the reduction occurs at the cathode. The loss of an electron occurs from the anode by oxidation as well as the gain of an electron occurs from the cathode by reduction.

Standard cell potentials: The standard emf (electromotive force, $\text{E}°$) of the half-cell is measured against the standard hydrogen electrode (SHE) at standard temperature and pressure with the electrolyte concentration of $\text{1}\text{.0 M}$. The $\text{E}°$ values for various materials are available.

Redox reactions: A redox reaction is a chemical reaction where both oxidation and reduction occur simultaneously. In a redox reaction, one of the reactant is oxidized, while the other reactant is reduced at the same time. It can be represented as two half-reactions with the number of transferred electrons. They are as follows:

• Oxidation half-reaction.
• Reduction half-reaction.

Explanation of Solution

The given reaction is $3A{u}^{2+}\left(aq\right)\rightleftharpoons Au\left(s\right)+2A{u}^{3+}\left(aq\right)$

$\begin{array}{l}3A{u}^{2+}\left(aq\right)\rightleftharpoons Au\left(s\right)+2A{u}^{3+}\left(aq\right)\\ R:A{u}^{+}\left(aq\right)+e^{-}\to Au\left(aq\right)+1.69V\\ L:A{u}^{3+}\left(aq\right)+3e^{-}\to Au\left(s\right)+1.50V\\ ---------------------------\\ E=+0.19V\\ \end{array}$

The equilibrium constant, K is calculated as follows,

$\begin{array}{c}\ln K=\frac{nFE}{RT}\\ \\ =\frac{0.19V}{25.693\times {10}^{-3}V}\\ \\ \ln K=7.40\\ \\ K=e^{7.40}\\ \\ =1.6\times {10}^3\end{array}$