Chapter 11, Problem 111AE

a.

Interpretation Introduction

Interpretation: Whether the cell potential of given galvanic cell increase, decrease, or remain the same when ${\text{CuSO}}_{\text{4}}\text{(s)}$ is added to copper half-cell compartment.

Concept Introduction: The measure of energy per unit charge which is available from the redox reactions to carry out the reaction is said to be cell potential.

Nernst equation gives the relationship between standard reduction potential, ${\text{E}}_{\text{cell}}^{\text{o}}$ , reduction potential, ${\text{E}}_{\text{cell}}$ and the activities of species present in the electrochemical cell at temperature, $\text{T}$ as:

  ${\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\text{o}}\text{- }\frac{\text{RT}}{\text{nF}}\text{ln}\left(\text{Q}\right)$

Where $\text{Q}$ is reaction quotient, $\text{F}$ is Faraday constant, $\text{R}$ is universal gas constant and $\text{n}$ is amount in terms of mol (electrons transferred)

This equation is specified at room temperature, $\text{T = 298}\text{.15 K}$ as:

  ${\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\text{o}}\text{- }\frac{\text{0}\text{.0591}}{\text{n}}\text{log}\left(\text{Q}\right)$

Answer to Problem 111AE

The value of ${\text{E}}_{\text{cell}}$ decreases.

Explanation of Solution

Given:

The half reactions of standard galvanic cell as:

  $\begin{array}{l}{\text{Cu}}^{\text{+2}}+2{\text{e}}^{-}\to \text{Cu}\\ {\text{Ag}}^{+}+{\text{e}}^{-}\to \text{Ag}\end{array}$

The reduction potential values for the half reactions of standard galvanic cell is:

  $\begin{array}{l}{\text{Cu}}^{\text{+2}}+2{\text{e}}^{-}\to {\text{Cu E}}^{\text{o}}=0.34\text{ V}\\ {\text{Ag}}^{+}+{\text{e}}^{-}\to {\text{Ag E}}^{\text{o}}=0.80\text{ V}\end{array}$

Since, the reduction potential value of silver is greater than copper so, the silver will undergo reduction and copper will undergo oxidation so, the half-reactions are written as:

  $\begin{array}{l}\text{Cu}\to {\text{Cu}}^{+2}+2{\text{e}}^{-}{\text{ E}}^{\text{o}}=-0.34\text{ V}\\ {\text{Ag}}^{+}+{\text{e}}^{-}\to {\text{Ag E}}^{\text{o}}=0.80\text{ V}\end{array}$

To balance the charge on both sides, the reduction reaction, $\left({\text{Ag}}^{+}+{\text{e}}^{-}\to \text{Ag}\right)$ is multiplied by 2 so,

  $\begin{array}{l}\text{Cu}\to {\text{Cu}}^{+2}+2{\text{e}}^{-}{\text{ E}}^{\text{o}}=-0.34\text{ V}\\ {\text{2Ag}}^{+}+2{\text{e}}^{-}\to 2{\text{Ag E}}^{\text{o}}=0.80\text{ V}\end{array}$

Adding both the reactions to get the overall reactions as:

  $\begin{array}{l}\text{Cu}\to {\text{Cu}}^{+2}+2{\text{e}}^{-}{\text{ E}}^{\text{o}}=-0.34\text{ V}\\ \underset{\_}{{\text{2Ag}}^{+}+2{\text{e}}^{-}\to 2{\text{Ag E}}^{\text{o}}=0.80\text{ V}}\\ {\text{Cu+2Ag}}^{+}\to {\text{Cu}}^{+2}+2{\text{Ag E}}_{\text{cell}}^{\text{o}}=0.46\text{ V}\end{array}$

So, the overall balanced reaction for the galvanic cell is:

  ${\text{Cu+2Ag}}^{+}\to {\text{Cu}}^{+2}+2{\text{Ag E}}_{\text{cell}}^{\text{o}}=0.46\text{ V}$

Now, according to Nernst equation at room temperature, $\text{T = 298}\text{.15 K}$ ,

  ${\text{E}}_{\text{cell}}=\text{0}\text{.46}-\frac{0.0591}{\text{n}}\text{log}\frac{\left[{\text{Cu}}^{+2}\right]}{{\left[{\text{Ag}}^{+}\right]}^2}$ - (1)

Now, on adding ${\text{CuSO}}_{\text{4}}\text{(s)}$ to copper half-cell compartment it will result in increasing the concentration of copper ion, ${\text{Cu}}^{+2}$ which leads to increase in negative term of equation (1) and hence, the value of ${\text{E}}_{\text{cell}}$ decreases.

b.

Interpretation Introduction

Interpretation: Whether the cell potential of given galvanic cell increase, decrease, or remain the same when ${\text{NH}}_3\text{(aq)}$ is added to copper half-cell compartment.

Concept Introduction: The measure of energy per unit charge which is available from the redox reactions to carry out the reaction is said to be cell potential.

Nernst equation gives the relationship between standard reduction potential, ${\text{E}}_{\text{cell}}^{\text{o}}$ , reduction potential, ${\text{E}}_{\text{cell}}$ and the activities of species present in the electrochemical cell at temperature, $\text{T}$ as:

  ${\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\text{o}}\text{- }\frac{\text{RT}}{\text{nF}}\text{ln}\left(\text{Q}\right)$

Where $\text{Q}$ is reaction quotient, $\text{F}$ is Faraday constant, $\text{R}$ is universal gas constant and $\text{n}$ is amount in terms of mol (electrons transferred)

This equation is specified at room temperature, $\text{T = 298}\text{.15 K}$ as:

  ${\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\text{o}}\text{- }\frac{\text{0}\text{.0591}}{\text{n}}\text{log}\left(\text{Q}\right)$

Answer to Problem 111AE

The value of ${\text{E}}_{\text{cell}}$ increases.

Explanation of Solution

According to Nernst equation at room temperature, $\text{T = 298}\text{.15 K}$ ,

  ${\text{E}}_{\text{cell}}=\text{0}\text{.46}-\frac{0.0591}{\text{n}}\text{log}\frac{\left[{\text{Cu}}^{+2}\right]}{{\left[{\text{Ag}}^{+}\right]}^2}$ - (1)

Now, on adding ${\text{NH}}_3\text{(aq)}$ to copper half-cell compartment it will result in decreasing the concentration of copper ion, ${\text{Cu}}^{+2}$ as ${\text{Cu}}^{+2}$ ions reacts with ${\text{NH}}_3\text{(aq)}$ and results in the formation of ${\text{Cu(NH}}_3{\text{)}}_4^{2+}$ . The decrease in concentration of ${\text{Cu}}^{+2}$ leads to decrease in negative term of equation (1) and hence, the value of ${\text{E}}_{\text{cell}}$ increases.

c.

Interpretation Introduction

Interpretation: Whether the cell potential of given galvanic cell increase, decrease, or remain the same when $\text{NaCl(s)}$ is added to silver half-cell compartment.

Concept Introduction: The measure of energy per unit charge which is available from the redox reactions to carry out the reaction is said to be cell potential.

Nernst equation gives the relationship between standard reduction potential, ${\text{E}}_{\text{cell}}^{\text{o}}$ , reduction potential, ${\text{E}}_{\text{cell}}$ and the activities of species present in the electrochemical cell at temperature, $\text{T}$ as:

  ${\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\text{o}}\text{- }\frac{\text{RT}}{\text{nF}}\text{ln}\left(\text{Q}\right)$

Where $\text{Q}$ is reaction quotient, $\text{F}$ is Faraday constant, $\text{R}$ is universal gas constant and $\text{n}$ is amount in terms of mol (electrons transferred)

This equation is specified at room temperature, $\text{T = 298}\text{.15 K}$ as:

  ${\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\text{o}}\text{- }\frac{\text{0}\text{.0591}}{\text{n}}\text{log}\left(\text{Q}\right)$

Answer to Problem 111AE

The value of ${\text{E}}_{\text{cell}}$ decreases.

Explanation of Solution

According to Nernst equation at room temperature, $\text{T = 298}\text{.15 K}$ ,

  ${\text{E}}_{\text{cell}}=\text{0}\text{.46}-\frac{0.0591}{\text{n}}\text{log}\frac{\left[{\text{Cu}}^{+2}\right]}{{\left[{\text{Ag}}^{+}\right]}^2}$ - (1)

Now, on adding $\text{NaCl(s)}$ to silver half-cell compartment it will result in decreasing the concentration of silver ion, ${\text{Ag}}^{+}$ as ${\text{Ag}}^{+}$ ions reacts with ${\text{Cl}}^{-}$ ion (obtained from dissociation of $\text{NaCl}$ ) and results in the formation of $\text{AgCl}$ . The decrease in concentration of ${\text{Ag}}^{+}$ leads to increase in negative term of equation (1) and hence, the value of ${\text{E}}_{\text{cell}}$ decreases.

d.

Interpretation Introduction

Interpretation: Whether the cell potential of given galvanic cell increase, decrease, or remain the same when water is added to both half-cell compartment until the volume of solution is doubled.

Concept Introduction: The measure of energy per unit charge which is available from the redox reactions to carry out the reaction is said to be cell potential.

Nernst equation gives the relationship between standard reduction potential, ${\text{E}}_{\text{cell}}^{\text{o}}$ , reduction potential, ${\text{E}}_{\text{cell}}$ and the activities of species present in the electrochemical cell at temperature, $\text{T}$ as:

  ${\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\text{o}}\text{- }\frac{\text{RT}}{\text{nF}}\text{ln}\left(\text{Q}\right)$

Where $\text{Q}$ is reaction quotient, $\text{F}$ is Faraday constant, $\text{R}$ is universal gas constant and $\text{n}$ is amount in terms of mol (electrons transferred)

This equation is specified at room temperature, $\text{T = 298}\text{.15 K}$ as:

  ${\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\text{o}}\text{- }\frac{\text{0}\text{.0591}}{\text{n}}\text{log}\left(\text{Q}\right)$

Answer to Problem 111AE

The value of ${\text{E}}_{\text{cell}}$ decreases.

Explanation of Solution

According to Nernst equation at room temperature, $\text{T = 298}\text{.15 K}$ ,

  ${\text{E}}_{\text{cell}}=\text{0}\text{.46}-\frac{0.0591}{\text{n}}\text{log}\frac{\left[{\text{Cu}}^{+2}\right]}{{\left[{\text{Ag}}^{+}\right]}^2}$ - (1)

Now, on adding water to both half-cell compartment until the volume of solution gets doubled it will result in decreasing the concentration of both the ions, silver ion, ${\text{Ag}}^{+}$ and copper ion, ${\text{Cu}}^{\text{2+}}$ . this will lead to increase the value of ratio of both the ions, $\frac{\left[{\text{Cu}}^{+2}\right]}{{\left[{\text{Ag}}^{+}\right]}^2}$ . The increase in ratio value leads to increase in negative term of equation (1) and hence, the value of ${\text{E}}_{\text{cell}}$ decreases.

e.

Interpretation Introduction

Interpretation: Whether the cell potential of given galvanic cell increase, decrease, or remain the same when silver electrode is replaced with platinum electrode:

  ${\text{Pt}}^{\text{+2}}{\text{+2e}}^{\text{-}}\to {\text{Pt E}}^{\text{o}}\text{= 1}\text{.19 V}$

Concept Introduction: The measure of energy per unit charge which is available from the redox reactions to carry out the reaction is said to be cell potential.

Nernst equation gives the relationship between standard reduction potential, ${\text{E}}_{\text{cell}}^{\text{o}}$ , reduction potential, ${\text{E}}_{\text{cell}}$ and the activities of species present in the electrochemical cell at temperature, $\text{T}$ as:

  ${\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\text{o}}\text{- }\frac{\text{RT}}{\text{nF}}\text{ln}\left(\text{Q}\right)$

Where $\text{Q}$ is reaction quotient, $\text{F}$ is Faraday constant, $\text{R}$ is universal gas constant and $\text{n}$ is amount in terms of mol (electrons transferred)

This equation is specified at room temperature, $\text{T = 298}\text{.15 K}$ as:

  ${\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\text{o}}\text{- }\frac{\text{0}\text{.0591}}{\text{n}}\text{log}\left(\text{Q}\right)$

Answer to Problem 111AE

The value of ${\text{E}}_{\text{cell}}$ remain the same.

Explanation of Solution

Given:

The half reactions of standard galvanic cell as:

  $\begin{array}{l}{\text{Cu}}^{\text{+2}}+2{\text{e}}^{-}\to \text{Cu}\\ {\text{Ag}}^{+}+{\text{e}}^{-}\to \text{Ag}\end{array}$

The reduction potential values for the half reactions of standard galvanic cell is:

  $\begin{array}{l}{\text{Cu}}^{\text{+2}}+2{\text{e}}^{-}\to {\text{Cu E}}^{\text{o}}=0.34\text{ V}\\ {\text{Pt}}^{\text{+2}}{\text{+2e}}^{\text{-}}\to {\text{Pt E}}^{\text{o}}\text{= 1}\text{.19 V}\end{array}$

Since, the reduction potential value of platinum is greater than copper so, the platinum will undergo reduction and copper will undergo oxidation so, the half-reactions are written as:

  $\begin{array}{l}\text{Cu}\to {\text{Cu}}^{+2}+2{\text{e}}^{-}{\text{ E}}^{\text{o}}=-0.34\text{ V}\\ {\text{Pt}}^{\text{+2}}{\text{+2e}}^{\text{-}}\to {\text{Pt E}}^{\text{o}}\text{= 1}\text{.19 V}\end{array}$

Adding both the reactions to get the overall reactions as:

  $\begin{array}{l}\text{Cu}\to {\text{Cu}}^{+2}+2{\text{e}}^{-}{\text{ E}}^{\text{o}}=-0.34\text{ V}\\ \underset{\_}{{\text{Pt}}^{\text{+2}}{\text{+2e}}^{\text{-}}\to {\text{Pt E}}^{\text{o}}\text{= 1}\text{.19 V }}\\ {\text{Cu+Pt}}^{\text{+2}}\to {\text{Cu}}^{+2}+{\text{Pt E}}_{\text{cell}}^{\text{o}}=0.85\text{ V}\end{array}$

So, the overall balanced reaction for the galvanic cell is:

  ${\text{Cu+Pt}}^{\text{+2}}\to {\text{Cu}}^{+2}+{\text{Pt E}}_{\text{cell}}^{\text{o}}=0.85\text{ V}$

Now, according to Nernst equation at room temperature, $\text{T = 298}\text{.15 K}$ ,

  ${\text{E}}_{\text{cell}}=\text{ 0}\text{.85}-\frac{0.0591}{2}\text{log}\frac{\left[{\text{Cu}}^{+2}\right]}{\left[{\text{Pt}}^{2+}\right]}$

So, on replacing the silver electrode by platinum electrode, the ${\text{E}}_{\text{cell}}$ will remain same as there is no effect in the ion’s concentrations.