Chapter 17, Problem 144CP

(a)

Interpretation Introduction

Interpretation: The cell potential $E_{\text{cell}}$ of the given cell, the mass of each electrode after $10.\text{0 h}$ and the time of the given battery to deliver a current of $10.\text{0 A}$ before it goes dead is to be calculated.

Concept introduction: Cell potential is defined as the measure of energy per unit charge available from the redox reaction to carry out the reaction. The electrochemical equivalent is defined as the mass of the element transported when electricity of one coulomb charge is passed.

To determine: The cell potential of the given cell when the battery is connected.

Answer to Problem 144CP

Answer

The cell potential of the given cell when the battery is connected is $\underset{\_}{1.141V}$.

Explanation of Solution

Explanation

Given

The molar concentration of ${\text{Zn}}^{2+}$ is $\left(0.1\text{0 M}\right)$.

The molar concentration of ${\text{Cu}}^{2+}$ is $\left(2.5\text{0 M}\right)$.

The half cell reactions for the cell are,

${\text{ Zn}}^{+2}+2{\text{e}}^{-}\stackrel{}{\to }\text{Zn }{E}_1{}^{\text{ο}}=-0.76\text{ V}$ (1)

${\text{ Cu}}^{+2}+2{\text{e}}^{-}\stackrel{}{\to }\text{Cu }{E}_2{}^{\text{ο}}=0.34\text{ V}$ (2)

Where,

• $E_1{}^{\text{ο}}$ is the standard electrode potential of equation (1).
• $E_2{}^{\text{ο}}$ is the standard electrode potential of equation (2).

The overall cell reaction is,

$\text{Zn}+{\text{Cu}}^{+2}\stackrel{}{\to }{\text{Zn}}^{2+}+\text{Cu}$ (3)

The standard electrode potential of equation (3) is calculated by the formula,

$E_{\text{cell}}{}^{\text{ο}}=E_2{}^{\text{ο}}-E_1{}^{\text{ο}}$

Where,

• $E_{\text{cell}}{}^{\text{ο}}$ is the standard electrode potential of equation (3).

Substitute the values of $E_1{}^{\text{ο}}$ and $E_2{}^{\text{ο}}$ in the above formula.

$\begin{array}{c}{E}_{\text{cell}}{}^{\text{ο}}=0.3\text{4 V}-\left(-0.7\text{6 V}\right)\\ =1.1\text{0 V}\end{array}$

The cell potential after connecting the battery is calculated by using Nernst equation.

$E_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{0.0591}{n}\log \left(\frac{\left[{\text{Zn}}^{+2}\right]}{\left[{\text{Cu}}^{+2}\right]}\right)$

Where,

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

Substitute the values of $n$, molar concentrations of anode and cathode and $E_{\text{cell}}^{\text{ο}}$ in the above expression.

$\begin{array}{c}{E}_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{0.0591}{n}\log \left(\frac{\left[{\text{Zn}}^{+2}\right]}{\left[{\text{Cu}}^{+2}\right]}\right)\\ =1.1\text{0 V}-\frac{0.0591}{2}\log \left(\frac{\left[0.10\right]}{\left[2.50\right]}\right)\\ =1.1\text{0 V}-\left(-0.041\right)\\ =\underset{\_}{1.141V}\end{array}$

Hence, the cell potential of the given cell when the battery is connected is $\underset{\_}{1.141V}$.

(b)

Interpretation Introduction

Interpretation: The cell potential $E_{\text{cell}}$ of the given cell, the mass of each electrode after $10.\text{0 h}$ and the time of the given battery to deliver a current of $10.\text{0 A}$ before it goes dead is to be calculated.

Concept introduction: Cell potential is defined as the measure of energy per unit charge available from the redox reaction to carry out the reaction. The electrochemical equivalent is defined as the mass of the element transported when electricity of one coulomb charge is passed.

To determine: The cell potential after $10.\text{0 A}$ of current has flowed for $10.\text{0 h}$.

Answer to Problem 144CP

Answer

The cell potential after $10.\text{0 A}$ of current has flowed for $10.\text{0 h}$ is $\underset{\_}{1.09V}$.

Explanation of Solution

Explanation

Given

The initial concentration of zinc is $\left(0.1\text{0 M}\right)$.

The atomic weight of zinc is $65.4\text{1 g/mol}$.

The volume of solution is $1.0\text{0 L}$.

The time is $\text{1 hr}$.

The conversion of 1 hr to sec is done as,

$\begin{array}{c}\text{1 hr}=6\text{0 min}\\ =60\times 6\text{0 sec}\\ =360\text{0 sec}\end{array}$

Therefore the conversion of 10 hr to sec is done as,

$\begin{array}{c}\text{1 hr}=360\text{0 sec}\\ \text{10 hr}=10\times 360\text{0 sec}\end{array}$

The first step is to calculate the new concentrations of ${\text{Zn}}^{\text{2+}}$ and ${\text{Cu}}^{\text{2+}}$. This is done by calculating the change in the number of moles of ${\text{Zn}}^{\text{2+}}$ and ${\text{Cu}}^{\text{2+}}$ that result from passing a current of 10.0 A for 10 hours by using the given expression.

$\text{Q}\left(\text{C}\right)=\text{I}\left(\text{A}\right)\times \text{t}\left(\text{s}\right)$.

Where,

• $\text{Q}$ is the total charge.
• I is the current.
• $\text{t}$ is the time.

Substitute the value of current and time in the above expression.

$\begin{array}{c}\text{Q}\left(\text{C}\right)=\text{I}\left(\text{A}\right)\times \text{t}\left(\text{s}\right)\\ =10.\text{0 A}\times 10.0\times 360\text{0 s}\\ =3.60\times 1{\text{0}}^{\text{5}}\text{ C}\end{array}$

Thus, the number of moles of charge passed is calculated by using the expression,

$\text{Number of moles}=\frac{\text{Given charge}}{\text{F}}$.

Where,

• F is the Faradays constant $\left(\text{F}=\text{96,485 C/mol}\right)$.

Substitute the value of charge and F in the above expression to calculate the number of moles of charge passed.

$\begin{array}{c}\text{Number of moles}=\frac{\text{Given charge}}{\text{F}}\\ =\frac{3.60\times {10}^{\text{5}}\text{ C}}{96,48\text{5 C/mol}}\\ =3.7\text{3 mol}\end{array}$

Since two moles of electrons are involved in the reaction. The number of moles of ${\text{Cu}}^{\text{2+}}$ consumed is half of the number of moles of ${\text{Zn}}^{\text{2+}}$ produced.

$\begin{array}{c}{\text{Number of moles of Cu}}^{\text{2+}}=\frac{{\text{Number of moles of Zn}}^{\text{2+}}}{\text{2}}\\ =\frac{3.\text{73 mol}}{2}\\ =1.8\text{65 mol}\end{array}$.

Initially in a $1.0\text{0 L}$ solution, $\left(2.5\text{0 M}\right)$ moles of ${\text{Cu}}^{\text{2+}}$ and $\left(0.1\text{0 M}\right)$ moles of ${\text{Zn}}^{\text{2+}}$ are given. After 10 hours, the concentration of ${\text{Cu}}^{\text{2+}}$ and ${\text{Zn}}^{\text{2+}}$ is calculated as,

$\begin{array}{c}\left[{\text{Cu}}^{\text{2+}}\right]=2.5\text{0 M}-1.86\text{5 M}\\ =0.6\text{35 M}\end{array}$

$\begin{array}{c}\left[{\text{Zn}}^{\text{2+}}\right]=0.10\text{ M+}1.86\text{5 M}\\ =1.965\text{ M}\end{array}$.

Thus, after $10.\text{0 hours}$ $\begin{array}{c}\text{Q}=\left(\frac{\left[{\text{Zn}}^{\text{2+}}\right]}{{\text{Cu}}^{\text{2+}}}\right)\\ =\left(\frac{1.965}{0.635}\right)\\ =\left(3.\text{09 M}\right)\end{array}$.

The cell potential after $10.\text{0 A}$ of current is calculated by using Nernst equation.

$\begin{array}{c}{E}_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{0.0591}{n}\log \left(\frac{\left[{\text{Zn}}^{+2}\right]}{\left[{\text{Cu}}^{+2}\right]}\right)\\ =1.1\text{0 V}-\frac{0.0591}{2}\log \left(3.09\right)\\ =1.1\text{0 V}-1.45\times {10}^{-2}\text{ V}\\ =\underset{\_}{1.09V}\end{array}$

Hence, the cell potential after $10.\text{0 A}$ of current has flowed for $10.\text{0 h}$ is $\underset{\_}{1.09V}$.

(c)

Interpretation Introduction

Interpretation: The cell potential $E_{\text{cell}}$ of the given cell, the mass of each electrode after $10.\text{0 h}$ and the time of the given battery to deliver a current of $10.\text{0 A}$ before it goes dead is to be calculated.

Concept introduction: Cell potential is defined as the measure of energy per unit charge available from the redox reaction to carry out the reaction. The electrochemical equivalent is defined as the mass of the element transported when electricity of one coulomb charge is passed.

To determine: The mass of each electrode after $10.\text{0 h}$.

Answer to Problem 144CP

Answer

The mass of copper electrode after $10.\text{0 h}$ is $\underset{\_}{82.5g}$.

The mass of zinc electrode after $10.\text{0 h}$ is $\underset{\_}{78.32g}$.

Explanation of Solution

Explanation

Given

Current (C) is $10.\text{0 A}$.

Time is $10.\text{0 h}$.

The conversion of 1 hr to sec is done as,

$\begin{array}{c}\text{1 hr}=6\text{0 min}\\ =60\times 6\text{0 sec}\\ =360\text{0 sec}\end{array}$

Therefore, the conversion of 10 hr to sec is done as,

$\begin{array}{c}\text{1 hr}=360\text{0 sec}\\ \text{10 hr}=10\times 360\text{0 sec}\end{array}$

The equivalent weight of copper is calculated by using the given expression,

$\text{Equivalent weight}=\frac{\text{Atomic weight}}{\text{Valancy}}$.

The atomic weight of Copper is $6\text{3 amu}$.

Substitute the value of atomic weight and valancy of copper in the above expression to calculate its equivalent weight,

$\begin{array}{c}\text{Equivalent weight}=\frac{\text{Atomic weight}}{\text{Valancy}}\\ =\frac{63}{2}\\ =31.5\end{array}$

Hence, electrochemical equivalent $\left(\text{Z}\right)$ is calculated as,

$\begin{array}{c}\text{Z}=\frac{\text{Equivalent weight}}{\text{F}}\\ =\frac{31.5}{96485}\\ =3.2\times {10}^{-4}\end{array}$.

The weight of copper deposited is calculated by using the given expression,

$\text{W}=\text{Z}\times \text{c}\times \text{t}$.

Where,

• $\text{Z}$ is the electrochemical equivalent.
• $\text{c}$ is the charge.
• $\text{t}$ is the time in sec.

Substitute the value of $\text{Z}$, c and t in the above expression to calculate the weight of copper deposited.

$\begin{array}{c}\text{W}=\text{Z}\times \text{c}\times \text{t}\\ =3.2\times {10}^{-4}\times 10\times 3600\\ =117.\text{5 g}\end{array}$

If the weight of copper deposited in 10 hr is $117.\text{5 g}$. Then the weight of copper electrode after 10 hr is calculated as,

$\begin{array}{c}\text{Weight of Cu electrode}=\text{Total mass of electrode}-\text{Weight of Cu in 10hr}\\ =200-117.5\\ =\underset{\_}{82.5g}\end{array}$

Hence, the mass of copper electrode after $10.\text{0 h}$ is $\underset{\_}{82.5g}$.

Similarly, the equivalent weight of zinc is calculated by using the given expression,

$\text{Equivalent weight}=\frac{\text{Atomic weight}}{\text{Valancy}}$.

The atomic weight of zinc is $65.41\text{ amu}$.

Substitute the value of atomic weight and valancy of zinc in the above expression to calculate its equivalent weight,

$\begin{array}{c}\text{Equivalent weight}=\frac{\text{Atomic weight}}{\text{Valancy}}\\ =\frac{65.41}{2}\\ =32.705\end{array}$

Hence, electrochemical equivalent $\left(\text{Z}\right)$ is calculated as,

$\begin{array}{c}\text{Z}=\frac{\text{Equivalent weight}}{\text{F}}\\ =\frac{32.705}{96485}\\ =3.38\times {10}^{-4}\end{array}$.

The weight of zinc deposited is calculated by using the given expression,

$\text{W}=\text{Z}\times \text{c}\times \text{t}$.

Where,

• $\text{Z}$ is the electrochemical equivalent.
• $\text{c}$ is the charge.
• $\text{t}$ is the time in sec.

Substitute the value of $\text{Z}$, c and t in the above expression to calculate the weight of zinc deposited.

$\begin{array}{c}\text{W}=\text{Z}\times \text{c}\times \text{t}\\ =3.38\times {10}^{-4}\times 10\times 3600\\ =121.68\text{ g}\end{array}$

If the weight of zinc deposited in 10 hr is $121.68\text{ g}$. Then the weight of zinc electrode after 10 hr is calculated as,

$\begin{array}{c}\text{Weight of Zn electrode}=\text{Total mass of electrode}-\text{Weight of Zn in 10hr}\\ =200-121.68\\ =\underset{\_}{78.32g}\end{array}$

Hence, the mass of zinc electrode after $10.\text{0 h}$ is $\underset{\_}{78.32g}$.

(d)

Interpretation Introduction

Interpretation: The cell potential $E_{\text{cell}}$ of the given cell, the mass of each electrode after $10.\text{0 h}$ and the time of the given battery to deliver a current of $10.\text{0 A}$ before it goes dead is to be calculated.

Concept introduction: Cell potential is defined as the measure of energy per unit charge available from the redox reaction to carry out the reaction. The electrochemical equivalent is defined as the mass of the element transported when electricity of one coulomb charge is passed.

To determine: The time of the given battery to deliver a current of $10.\text{0 A}$ before it goes dead.

Answer to Problem 144CP

Answer

The time of the given battery to deliver a current of $10.\text{0 A}$ before it goes dead is $\underset{\_}{17.36hr}$.

Explanation of Solution

Explanation

Given

The battery delivers current till all the copper is deposited

The weight of copper to be deposited is $\text{W}=20\text{0 g}$.

Current (C) is $10.\text{0 A}$.

Electrochemical equivalent of copper $\left(\text{Z}\right)=3.2\times {10}^{-4}$.

The time is calculated by using the expression,

$\begin{array}{c}\text{W}=\text{Z}\times \text{c}\times \text{t}\\ \text{t}=\frac{\text{W}}{\text{Z}\times \text{c}}\end{array}$

Substitute the value of W, Z and c in the above expression to calculate the time of the given battery.

$\begin{array}{c}\text{t}=\frac{\text{W}}{\text{Z}\times \text{c}}\\ \text{t}=\frac{200}{3.2\times {10}^{-4}\times 10}\\ =62.5\times {10}^{\text{3}}\text{ sec}\end{array}$

The conversion of sec to hr is done as,

$\begin{array}{c}\text{1 sec}=\frac{1}{60}\text{ min}\\ =\frac{1}{3600}\text{hr}\end{array}$

Therefore, the conversion of $62.5\times {10}^{\text{3}}\text{ sec}$ to hr is done as,

$\begin{array}{c}\text{1 sec}=\frac{1}{60}\text{ min}\\ =\frac{1}{3600}\text{hr}\\ 62.5\times {10}^{\text{3}}\text{ sec}=\frac{62.5\times {10}^{\text{3}}}{3600}\text{hr}\\ =\underset{\_}{17.36hr}\end{array}$

Hence, the time of the given battery to deliver a current of $10.\text{0 A}$ before it goes dead is $\underset{\_}{17.36hr}$.