Chapter 19, Problem 19.146QP

 (a)

Interpretation Introduction

Interpretation:

The cell potential (EMF) of given cell should be calculated at standard conditions by using Nernst equation.

Concept introduction:

Cell potential (EMF):

The maximum potential difference between two electrodes of voltaic cell is known as cell potential.

If standard reduction potentials of electrodes are given the cell potential (EMF) is given by,

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

Where,

$\begin{array}{l}{\text{E}}_{\text{cathode}}\text{is}\text{the}\text{reduction}\text{half}\text{cell potential}\\ {\text{E}}_{\text{anode}}\text{is}\text{the}\text{oxidation}\text{half}\text{cell potential}\end{array}$

Nernst equation:

The relationship between standard cell potential and cell potential at non standard conditions and the reaction quotient are given by Nernst equation it is,

${\text{E}}_{\text{cell}}\text{=}{\text{E°}}_{\text{cell}}\text{-}\frac{\text{2}\text{.303}\text{RT}}{\text{nF}}\text{logQ}$

Where,

$\begin{array}{l}{\text{E}}_{\text{cell}}\text{ is}\text{cell potential}\\ {\text{E°}}_{\text{cell}}\text{is}\text{standard cell potential}\\ \text{R}\text{is}\text{gas}\text{constant}\\ \text{T}\text{is}\text{temperature}\\ \text{Q}\text{is}\text{reaction quotient}\end{array}$

Answer to Problem 19.146QP

The cell potential (EMF) of given voltaic cell is $\text{1}\text{.56}\text{V}$

Explanation of Solution

To calculate the cell potential (EMF) of given cell

The standard reduction potentials of (SRP) of Zinc and Silver are record from standard reduction potentials table and they are,

$\begin{array}{l}{\text{Zn}}_{\text{(s)}}\stackrel{}{\to }{\text{Zn}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2e}}^{\text{-}}\text{-}\text{E°}\text{=}\text{0}\text{.76}\text{V}\\ {\text{2Ag}}^{\text{+}}{}_{\text{(aq)}}{\text{+2e}}^{\text{-}}\stackrel{}{\to }{\text{2Ag}}_{\text{(s)}}\text{E°=}\text{0}\text{.80}\text{V}\end{array}$

The most positive SQR is considering as cathode potential.

The SRP of electrodes are plugged in the bellow equation to give cell potential of given voltaic cell.

$\begin{array}{l}{\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{cathode}}{\text{-E}}_{\text{anode}}\\ \text{=}\text{0}\text{.80-(-0}\text{.76)}\\ \text{=}\text{1}\text{.56}\text{V}\end{array}$

The cell potential (EMF) of given voltaic cell is $\text{1}\text{.56}\text{V}$

To calculate the cell potential (EMF) of given cell

Given:

Mass of ${\text{ZnNO}}_{\text{3}}$ is $\text{105}\text{g}$

Mass of ${\text{AgNO}}_{\text{3}}$ is $\text{145}\text{g}$

Final volume of ${\text{ZnNO}}_{\text{3}}$ solution is $\text{1}\text{.1}\text{L}$

Final volume of ${\text{AgNO}}_{\text{3}}$ solution is $\text{1 L}$

To calculate the concentrations of ${\text{ZnNO}}_{\text{3}}$ and ${\text{AgNO}}_{\text{3}}$ solutions

Molar mass of ${\text{AgNO}}_{\text{3}}$ is $\text{169}\text{.99}\text{g}$

$\begin{array}{l}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\frac{\text{145}\text{g}{\text{AgNO}}_{\text{3}}}{\text{1}\text{.0}\text{L}}\text{×}\frac{\text{1}\text{mol}{\text{AgNO}}_{\text{3}}}{\text{169}\text{.99}\text{g}{\text{AgNO}}_{\text{3}}}\text{×}\frac{\text{1}\text{mol}{\text{Ag}}^{\text{+}}}{\text{1}\text{mol}{\text{AgNO}}_{\text{3}}}\\ \text{=}\text{0}\text{.854}\text{M}{\text{Ag}}^{\text{+}}\end{array}$

Molar mass of ${\text{ZnNO}}_{\text{3}}$ is $\text{189}\text{.40}\text{g}$

$\begin{array}{l}{\text{[Zn}}^{\text{2+}}\text{]}\text{=}\frac{\text{105}\text{g}{\text{ZnNO}}_{\text{3}}}{\text{1}\text{.1}\text{L}}\text{×}\frac{\text{1}\text{mol}{\text{ZnNO}}_{\text{3}}}{\text{189}\text{.40}\text{g}{\text{ZnNO}}_{\text{3}}}\text{×}\frac{\text{1}\text{mol}{\text{Zn}}^{\text{2+}}}{\text{1}\text{mol}{\text{ZnNO}}_{\text{3}}}\\ \text{=}\text{0}\text{.504}\text{M}{\text{Zn}}^{\text{2+}}\end{array}$

Taken mass and molar mass of ${\text{ZnNO}}_{\text{3}}$, ${\text{AgNO}}_{\text{3}}$ and volumes of solutions are plugged in above equations to give concentrations of ${\text{Zn}}^{\text{2+}}$ and ${\text{Ag}}^{\text{+}}$ ions.

In this cell reaction, number of electrons transferred are 2

Faraday constant is $\text{96485}{\text{C mol}}^{-1}$

$\begin{array}{l}{\text{E}}_{\text{cell}}\text{=}{\text{E°}}_{\text{cell}}\text{-}\frac{\text{2}\text{.303}\text{RT}}{\text{nF}}\text{logQ}\\ \text{=}{\text{E°}}_{\text{cell}}\frac{\text{0}\text{.0592}}{\text{n}}\text{logQ}\\ \text{We}\text{know,}\\ \text{Q}\text{=}\frac{\text{products}}{\text{reactants}}\\ \text{=}\text{1}\text{.56}\text{V-}\frac{\text{0}\text{.0592}}{\text{2}}\text{log}\frac{{\text{[Zn}}^{\text{2+}}\text{]}}{{{\text{[Ag}}^{\text{+}}\text{]}}^{\text{2}}}\\ \text{=}\text{1}\text{.56}\text{V-}\frac{\text{0}\text{.0592}}{\text{2}}\text{log}\frac{\text{[0}\text{.504]}}{{\text{[0}\text{.854]}}^{\text{2}}}\\ \text{=}\text{1}\text{.56}\text{V-}\text{(-0}\text{.00475}\text{V)}\\ \text{=}\text{1}\text{.56475}\text{V}\\ \text{=}\text{1}\text{.56}\text{V}\end{array}$

The calculated standard cell potential (EMF) of given voltaic cell, number of electron transferred in cell reaction, Faraday constant and calculated concentrations of ions are plugged in the above equation to give a cell potential (EMF) of given cell at non standard conditions.

The cell potential (EMF) of given cell at non-standard conditions is $\text{1}\text{.56}\text{V}$

Conclusion

The cell potential (EMF) of given cell was calculated at standard condition by using Nernst equation and it was found to be $\text{1}\text{.56}\text{V}$

(b)

Interpretation Introduction

Interpretation:

The free energy change of given voltaic cell should be calculated by using standard reduction potentials and cell potential of the cell should be explained, when doing given operations to the cell.

Concept introduction:

Free energy change:

In thermodynamics the cell potential is known as  maximum work of the cell and it is equal to free energy change of the cell and it is given by,

$\text{ΔG}\text{=}{\text{-nFE}}_{\text{cell}}$

Where,

$\begin{array}{l}\text{ΔG}\text{is}\text{free}\text{energy}\text{change}\\ \text{n}\text{is}\text{number}\text{of}\text{electron}\text{transferred}\\ \text{F}\text{is}\text{faraday constant}\\ {\text{E}}_{\text{cell}}\text{is}\text{cell potential}\end{array}$

Answer to Problem 19.146QP

Solution:

The free energy change of voltaic cell is $\text{-301}\text{.0}\text{KJ}$

Explanation of Solution

To calculate the free energy change of given cell:

In this cell reaction number of electron transferred are 2

Faraday constant is $\text{96485}{\text{C mol}}^{-1}$

$\begin{array}{l}\text{ΔG}\text{=}{\text{-nFE}}_{\text{cell}}\\ \text{=}\text{-2}\text{mole}\text{×96485}\text{C}{\text{mole}}^{\text{-1}}\text{×}1.56\text{V}\\ \text{=}\text{-3}{\text{.010×10}}^{\text{5}}\text{C}\text{V}\\ \text{=}\text{-3}{\text{.010×10}}^{\text{5}}\text{J}\\ \text{=}\text{-301}\text{.0}\text{KJ}\end{array}$

The calculated cell potential (EMF) of given voltaic cell and number of electron transferred in cell reaction and Faraday constant are plugged in the above equation to give a free energy change of given cell.

The free energy change of given cell reaction is $\text{-301}\text{.0}\text{KJ}$

(c)

Interpretation Introduction

Interpretation:

The cell potential (EMF) of given cell should be explained in terms of concentration of ${\text{AgNO}}_{\text{3}}$ solution.

Concept introduction:

Nernst equation:

The relationship between standard cell potential and cell potential at standard conditions and the reaction quotient are given by Nernst equation it is,

${\text{E}}_{\text{cell}}\text{=}{\text{E°}}_{\text{cell}}\text{-}\frac{\text{2}\text{.303}\text{RT}}{\text{nF}}\text{logQ}$

Where,

$\begin{array}{l}{\text{E}}_{\text{cell}}\text{ is}\text{cell potential}\\ {\text{E°}}_{\text{cell}}\text{is}\text{standard cell potential}\\ \text{R}\text{is}\text{gas}\text{constant}\\ \text{T}\text{is}\text{temperature}\\ \text{Q}\text{is}\text{reaction quotient}\end{array}$

Answer to Problem 19.146QP

The concentration of ${\text{Ag}}^{\text{+}}$ is same therefore the cell potential is remains same and it is $\text{1}\text{.56}\text{V}$

Explanation of Solution

According to the Nernst equation, the concentration of makes changes in cell potential but in addition of $\text{25}\text{.0 mL}$ ${\text{AgNO}}_{\text{3}}$ is also being a same concentration so there is no concentration changes offered, when addition of $\text{25}\text{.0 mL}$ ${\text{AgNO}}_{\text{3}}$.

Hence, the concentration of ${\text{Ag}}^{\text{+}}$ is same therefore the cell potential is remains same and it is $\text{1}\text{.56}\text{V}$

(d)

Interpretation Introduction

Interpretation:

The cell potential (EMF) of given cell should be explained in terms of electrode concentration.

Concept introduction:

Nernst equation:

The relationship between standard cell potential and cell potential at standard conditions and the reaction quotient are given by Nernst equation it is,

${\text{E}}_{\text{cell}}\text{=}{\text{E°}}_{\text{cell}}\text{-}\frac{\text{2}\text{.303}\text{RT}}{\text{nF}}\text{logQ}$

Where,

$\begin{array}{l}{\text{E}}_{\text{cell}}\text{ is}\text{cell potential}\\ {\text{E°}}_{\text{cell}}\text{is}\text{standard cell potential}\\ \text{R}\text{is}\text{gas}\text{constant}\\ \text{T}\text{is}\text{temperature}\\ \text{Q}\text{is}\text{reaction quotient}\end{array}$

Answer to Problem 19.146QP

The cell potential would not affected by increasing mass of zinc electrode.

Explanation of Solution

The cell potential does not depended on mass of electrodes in voltaic cell.

Hence, the cell potential would not affected by increasing mass of zinc electrode.

(e)

Interpretation Introduction

Interpretation:

The cell potential (EMF) of given cell should be explained, when the addition of $\text{NaCl}$ into ${\text{AgNO}}_{\text{3}}$ half cell.

Concept introduction:

Nernst equation:

The relationship between standard cell potential and cell potential at standard conditions and the reaction quotient are given by Nernst equation it is,

${\text{E}}_{\text{cell}}\text{=}{\text{E°}}_{\text{cell}}\text{-}\frac{\text{2}\text{.303}\text{RT}}{\text{nF}}\text{logQ}$

Where,

$\begin{array}{l}{\text{E}}_{\text{cell}}\text{ is}\text{cell potential}\\ {\text{E°}}_{\text{cell}}\text{is}\text{standard cell potential}\\ \text{R}\text{is}\text{gas}\text{constant}\\ \text{T}\text{is}\text{temperature}\\ \text{Q}\text{is}\text{reaction quotient}\end{array}$

Answer to Problem 19.146QP

The addition of $\text{NaCl}$ is decreases the cell potential of give cell.

Explanation of Solution

According to the Nernst equation, the addition of $\text{NaCl}$ into ${\text{AgNO}}_{\text{3}}$ half cell will causes $\text{AgCl}$ to precipitated, result of this the concentration of ${\text{Ag}}^{\text{+}}$ ion is decreases, in order reaction quotient value also decreases.