Chapter 17, Problem 117AE

Consider the standard galvanic cell based on the following half-reactions:

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

The electrodes in this cell are Ag(s) and Cu(s). Does the cell potential increase. decrease. or remain the same when the following changes occur to the standard cell?

a. CuSO₄(s) is added to the copper half-cell compartment (assume no volume change).

b. NH₃(aq) is added to the copper half-cell compartment (Hint: Cu²⁺ reacts with NH₃, to form Cu(NH₃)₄²⁺(aq).]

c. NaCl(s) is added to the silver half .cell compartment (Hint: Ag⁺ reacts with Cl⁻ to form AgCl(s).]

d. Water is added to both half-cell compartments until the volume or solution is doubled.

e. The silver electrode is replaced with a platinum electrode

Interpretation Introduction

Interpretation:

A galvanic cell having Silver and Copper electrodes is given. The change in the standard cell when any change is carried out in any compartment for the given reactions is to be stated.

Concept introduction:

The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation.

The value of $E_{\text{cell}}$ is calculated using Nernst formula,

$E=E°-\left(\frac{RT}{nF}\right)\ln \left(Q\right)$

At room temperature, the above equation is specified as,

$E=E°-\left(\frac{0.0591}{n}\right)\log \left(Q\right)$

This relation is further used to determine the relation between $\Delta G°$ and $K$ , $\Delta G°$ and $E{°}_{\text{cell}}$.

To determine: The change in the standard cell when ${\text{CuSO}}_4\left(s\right)$ is added to the Copper half-cell compartment.

When ${\text{CuSO}}_4\left(s\right)$ is added to the Copper half-cell compartment, the value of $E_{\text{cell}}$ decreases.

Explanation of Solution

Given,

The reactions taking place in the cell are,

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

On the basis of reduction potential values, the reaction taking place at each electrode is given as,

The reaction taking place at the cathode is,

${\text{Ag}}^{+}+{\text{e}}^{-}\to \text{Ag}E{°}_{\text{red}}=0.80\text{V}$

The reaction taking place at the anode is,

$\text{Cu}\to {\text{Cu}}^{2+}+2{\text{e}}^{-}E°=-0.34\text{V}$

Multiply the reduction half-reaction with a coefficient of $2$ and then add both the oxidation and reduction half-reaction.

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

The final equation is,

$\text{Cu}+{\text{2Ag}}^{+}\to {\text{Cu}}^{2+}+2\text{Ag}$

The value of $E{°}_{\text{cell}}$ is given as,

$\begin{array}{c}E°cell=E{°}_{\text{ox}}+E{°}_{\text{red}}\\ =-0.34\text{V}+0.80\text{V}\\ =\text{0}\text{.46}\text{V}\end{array}$

The cell potential is calculated using the Nernst equation,

$E=E°-\left(\frac{0.0591}{n}\right)\log \left(Q\right)$

Where,

• $E$ is the cell potential.

• $E°$ is the cell potential at the standard conditions.

• Q is the activity of the species in the cell.

• $n$ is the number of electrons involved in the reaction.

The reaction involves the transfer of $2$ moles of electrons.

Substitute the value of $E{°}_{\text{cell}}$, $n$ and $Q$ in the above equation as,

$E=0.46-\left(\frac{0.0591}{2}\right)\log \left(\frac{\left[{\text{Cu}}^{2+}\right]}{{\left[{\text{Ag}}^{+}\right]}^2}\right)$

When ${\text{CuSO}}_4\left(s\right)$ is added to the Copper half-cell compartment, the concentration of ${\text{Cu}}^{2+}$ increases. Due to this, the value of negative term will increase and therefore, this will lead to the decrease in the value of $E_{\text{cell}}$.

Interpretation Introduction

Interpretation:

A galvanic cell having Silver and Copper electrodes is given. The change in the standard cell when any change is carried out in any compartment for the given reactions is to be stated.

Concept introduction:

The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation.

The value of $E_{\text{cell}}$ is calculated using Nernst formula,

$E=E°-\left(\frac{RT}{nF}\right)\ln \left(Q\right)$

At room temperature, the above equation is specified as,

$E=E°-\left(\frac{0.0591}{n}\right)\log \left(Q\right)$

This relation is further used to determine the relation between $\Delta G°$ and $K$ , $\Delta G°$ and $E{°}_{\text{cell}}$.

To determine: The change in the standard cell when ${\text{NH}}_3\left(aq\right)$ is added to the Copper half-cell compartment.

When ${\text{NH}}_3\left(aq\right)$ is added to the Copper half-cell compartment, the value of $E_{\text{cell}}$ increases.

Explanation of Solution

Given,

The reactions taking place in the cell are,

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

On the basis of reduction potential values, the reaction taking place at each electrode is given as,

The reaction taking place at the cathode is,

${\text{Ag}}^{+}+{\text{e}}^{-}\to \text{Ag}E{°}_{\text{red}}=0.80\text{V}$

The reaction taking place at the anode is,

$\text{Cu}\to {\text{Cu}}^{2+}+2{\text{e}}^{-}E°=-0.34\text{V}$

Multiply the reduction half-reaction with a coefficient of $2$ and then add both the oxidation and reduction half-reaction.

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

The final equation is,

$\text{Cu}+{\text{2Ag}}^{+}\to {\text{Cu}}^{2+}+2\text{Ag}$

The value of $E{°}_{\text{cell}}$ is given as,

$\begin{array}{c}E°cell=E{°}_{\text{ox}}+E{°}_{\text{red}}\\ =-0.34\text{V}+0.80\text{V}\\ =\text{0}\text{.46}\text{V}\end{array}$

The cell potential is calculated using the Nernst equation,

$E=E°-\left(\frac{0.0591}{n}\right)\log \left(Q\right)$

Where,

• $E$ is the cell potential.

• $E°$ is the cell potential at the standard conditions.

• Q is the activity of the species in the cell.

• $n$ is the number of electrons involved in the reaction.

The reaction involves the transfer of $2$ moles of electrons.

Substitute the value of $E{°}_{\text{cell}}$, $n$ and $Q$ in the above equation as,

$E=0.46-\left(\frac{0.0591}{2}\right)\log \left(\frac{\left[{\text{Cu}}^{2+}\right]}{{\left[{\text{Ag}}^{+}\right]}^2}\right)$

When ${\text{NH}}_3\left(aq\right)$ is added to the Copper half-cell, ${\text{Cu}}^{2+}$ ions react with ${\text{NH}}_3\left(aq\right)$ and gets converted to ${\left[\text{Cu}{\left({\text{NH}}_3\right)}_4\right]}^{2+}$. Therefore, the concentration of ${\text{Cu}}^{2+}$ ions decreases. Due to this, the value of negative term will decrease and therefore, this will lead to the increase in the value of $E_{\text{cell}}$.

Interpretation Introduction

Interpretation:

A galvanic cell having Silver and Copper electrodes is given. The change in the standard cell when any change is carried out in any compartment for the given reactions is to be stated.

Concept introduction:

The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation.

The value of $E_{\text{cell}}$ is calculated using Nernst formula,

$E=E°-\left(\frac{RT}{nF}\right)\ln \left(Q\right)$

At room temperature, the above equation is specified as,

$E=E°-\left(\frac{0.0591}{n}\right)\log \left(Q\right)$

This relation is further used to determine the relation between $\Delta G°$ and $K$ , $\Delta G°$ and $E{°}_{\text{cell}}$.

To determine: The change in the standard cell when $\text{NaCl}\left(s\right)$ is added to the Silver half-cell compartment.

When $\text{NaCl}\left(s\right)$ is added to the Silver half-cell compartment, the value of $E_{\text{cell}}$ decreases.

Explanation of Solution

Given,

The reactions taking place in the cell are,

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

On the basis of reduction potential values, the reaction taking place at each electrode is given as,

The reaction taking place at the cathode is,

${\text{Ag}}^{+}+{\text{e}}^{-}\to \text{Ag}E{°}_{\text{red}}=0.80\text{V}$

The reaction taking place at the anode is,

$\text{Cu}\to {\text{Cu}}^{2+}+2{\text{e}}^{-}E°=-0.34\text{V}$

Multiply the reduction half-reaction with a coefficient of $2$ and then add both the oxidation and reduction half-reaction.

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

The final equation is,

$\text{Cu}+{\text{2Ag}}^{+}\to {\text{Cu}}^{2+}+2\text{Ag}$

The value of $E{°}_{\text{cell}}$ is given as,

$\begin{array}{c}E°cell=E{°}_{\text{ox}}+E{°}_{\text{red}}\\ =-0.34\text{V}+0.80\text{V}\\ =\text{0}\text{.46}\text{V}\end{array}$

The cell potential is calculated using the Nernst equation,

$E=E°-\left(\frac{0.0591}{n}\right)\log \left(Q\right)$

Where,

• $E$ is the cell potential.

• $E°$ is the cell potential at the standard conditions.

• Q is the activity of the species in the cell.

• $n$ is the number of electrons involved in the reaction.

The reaction involves the transfer of $2$ moles of electrons.

Substitute the value of $E{°}_{\text{cell}}$, $n$ and $Q$ in the above equation as,

$E=0.46-\left(\frac{0.0591}{2}\right)\log \left(\frac{\left[{\text{Cu}}^{2+}\right]}{{\left[{\text{Ag}}^{+}\right]}^2}\right)$

When $\text{NaCl}\left(s\right)$ is added to the Silver half-cell, ${\text{Ag}}^{+}$ ions react with ${\text{Cl}}^{-}$ and get converted to $\left[\text{AgCl}\right]$. Therefore, the concentration of ${\text{Ag}}^{+}$ ions decreases. Due to this, the value of negative term will increase and therefore, this will lead to the decrease in the value of $E_{\text{cell}}$.

Interpretation Introduction

Interpretation:

A galvanic cell having Silver and Copper electrodes is given. The change in the standard cell when any change is carried out in any compartment for the given reactions is to be stated.

Concept introduction:

The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation.

The value of $E_{\text{cell}}$ is calculated using Nernst formula,

$E=E°-\left(\frac{RT}{nF}\right)\ln \left(Q\right)$

At room temperature, the above equation is specified as,

$E=E°-\left(\frac{0.0591}{n}\right)\log \left(Q\right)$

This relation is further used to determine the relation between $\Delta G°$ and $K$ , $\Delta G°$ and $E{°}_{\text{cell}}$.

To determine: The change in the standard cell when water is added to both half-cell compartments until the volume of solution is doubled.

When water is added to the both the half-cells, there occurs a decrease in the value of $E_{\text{cell}}$.

Explanation of Solution

Given,

The reactions taking place in the cell are,

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

On the basis of reduction potential values, the reaction taking place at each electrode is given as,

The reaction taking place at the cathode is,

${\text{Ag}}^{+}+{\text{e}}^{-}\to \text{Ag}E{°}_{\text{red}}=0.80\text{V}$

The reaction taking place at the anode is,

$\text{Cu}\to {\text{Cu}}^{2+}+2{\text{e}}^{-}E°=-0.34\text{V}$

Multiply the reduction half-reaction with a coefficient of $2$ and then add both the oxidation and reduction half-reaction.

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

The final equation is,

$\text{Cu}+{\text{2Ag}}^{+}\to {\text{Cu}}^{2+}+2\text{Ag}$

The value of $E{°}_{\text{cell}}$ is given as,

$\begin{array}{c}E°cell=E{°}_{\text{ox}}+E{°}_{\text{red}}\\ =-0.34\text{V}+0.80\text{V}\\ =\text{0}\text{.46}\text{V}\end{array}$

The cell potential is calculated using the Nernst equation,

$E=E°-\left(\frac{0.0591}{n}\right)\log \left(Q\right)$

Where,

• $E$ is the cell potential.

• $E°$ is the cell potential at the standard conditions.

• Q is the activity of the species in the cell.

• $n$ is the number of electrons involved in the reaction.

The reaction involves the transfer of $2$ moles of electrons.

Substitute the value of $E{°}_{\text{cell}}$, $n$ and $Q$ in the above equation as,

$E=0.46-\left(\frac{0.0591}{2}\right)\log \left(\frac{\left[{\text{Cu}}^{2+}\right]}{{\left[{\text{Ag}}^{+}\right]}^2}\right)$

When water is added to the both the half-cells, the concentration of each ion will decrease. But there will occur an overall increase in the ratio of $\left(\frac{\left[{\text{Cu}}^{2+}\right]}{{\left[{\text{Ag}}^{+}\right]}^2}\right)$. Due to this, the value of negative term will increase and therefore, this will lead to the decrease in the value of $E_{\text{cell}}$.

Interpretation Introduction

Interpretation:

A galvanic cell having Silver and Copper electrodes is given. The change in the standard cell when any change is carried out in any compartment for the given reactions is to be stated.

Concept introduction:

The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation.

The value of $E_{\text{cell}}$ is calculated using Nernst formula,

$E=E°-\left(\frac{RT}{nF}\right)\ln \left(Q\right)$

At room temperature, the above equation is specified as,

$E=E°-\left(\frac{0.0591}{n}\right)\log \left(Q\right)$

This relation is further used to determine the relation between $\Delta G°$ and $K$ , $\Delta G°$ and $E{°}_{\text{cell}}$.

To determine: The change in the standard cell when Silver electrode is replaced with Platinum electrode.

Step: 5

The value of $E_{\text{cell}}$ will remain same when Silver electrode is replaced with Platinum electrode.

Explanation of Solution

Explanation

Given

The reactions taking place in the cell are,

$\begin{array}{c}{\text{Cu}}^{2+}+2{\text{e}}^{-}\to \text{Cu}\\ {\text{Pt}}^{2+}+2{\text{e}}^{-}\to \text{Pt}\end{array}$

On the basis of reduction potential values, the reaction taking place at each electrode is given as,

The reaction taking place at the cathode is,

${\text{Pt}}^{2+}+2{\text{e}}^{-}\to \text{Pt}E{°}_{\text{red}}=1.19\text{V}$

The reaction taking place at the anode is,

$\text{Cu}\to {\text{Cu}}^{2+}+2{\text{e}}^{-}E°=-0.34\text{V}$

Add both the oxidation and reduction half-reaction.

$\begin{array}{c}{\text{Pt}}^{2+}+2{\text{e}}^{-}\to \text{Pt}\\ \text{Cu}\to {\text{Cu}}^{2+}+2{\text{e}}^{-}\end{array}$

The final equation is,

$\text{Cu}+{\text{Pt}}^{2+}\to {\text{Cu}}^{2+}+\text{Pt}$

The value of $E{°}_{\text{cell}}$ is given as,

$\begin{array}{c}E°cell=E{°}_{\text{ox}}+E{°}_{\text{red}}\\ =-0.34\text{V}+1.19\text{V}\\ =\text{0}\text{.85}\text{V}\end{array}$

The cell potential is calculated using the Nernst equation,

$E=E°-\left(\frac{0.0591}{n}\right)\log \left(Q\right)$

Where,

• $E$ is the cell potential.

• $E°$ is the cell potential at the standard conditions.

• Q is the activity of the species in the cell.

• $n$ is the number of electrons involved in the reaction.

The reaction involves the transfer of $2$ moles of electrons.

Substitute the value of $E{°}_{\text{cell}}$, $n$ and $Q$ in the above equation as,

$E=0.85-\left(\frac{0.0591}{2}\right)\log \left(\frac{\left[{\text{Cu}}^{2+}\right]}{\left[{\text{Pt}}^{2+}\right]}\right)$

When Silver electrode is replaced with the Platinum electrode, the $E_{\text{cell}}$ will not be affected because there is no change in the concentration of ions.