Chapter 11, Problem 67E

(a)

Interpretation Introduction

Interpretation:Electrode potential for a copper electrode in solution is to be calculated.

Concept introduction:Study of interchange between electrical and chemical energy falls under branch of chemistry called electrochemistry. It includes occurrence of oxidation and reduction reactions. This includes production of electric current from chemical reaction and vice-versa.

Reactions that occur in galvanic cells can be broken down into two half-cell reactions. One half-cell reaction is that of reduction whereas another half-cell reaction is that of oxidation.

Nernst equation represents relation between potential of electrochemical reaction, standard cell potential, activities of species and temperature.

Expression of Nernst equation at room temperature is as follows:

  $E_{\text{cell}}=E_{\text{cell}}^0-\frac{0.0591}{n}\log \left(Q\right)$

Where,

• $E_{\text{cell}}$ is the cell potential.
• $E_{\text{cell}}^0$ is the standard cell potential.
• $n$ is total electrons transferred.
• $Q$ is reaction quotient.

Answer to Problem 67E

  $E_{\text{cell}}$ of electrochemical cell is $0.23\text{ V}$ .

Explanation of Solution

The half cell reaction for standard hydrogen electrode at anode isas follows:

  ${\text{H}}_{\text{2}}\to 2{\text{H}}^{\text{+}}+2{\text{e}}^{-}$

The half cell reaction for copper electrode at cathode isas follows:

  ${\text{Cu}}^{+2}+2{\text{e}}^{-}\to \text{Cu}$

Overall cell reaction is as follows:

  ${\text{Cu}}^{+2}+{\text{H}}_{\text{2}}\to \text{Cu}+2{\text{H}}^{\text{+}}$

The expression to calculate $E{°}_{\text{cell}}$ is as follows:

  $E_{\text{cell}}^0=E_{\text{oxd}}^{°}+E_{\text{red}}^{°}$

Where,

• $E{°}_{\text{cell}}$ is standard electrode potential for cell.
• $E_{\text{red}}^{°}$ is standard reduction electrode potential for cell.
• $E_{\text{oxd}}^{°}$ is standard oxidation electrode potential for cell.

Value of $E_{\text{red}}^{°}$ is $0.34\text{ V}$ .

Value of $E_{\text{oxd}}^{°}$ is $0\text{ V}$ .

Substitute values in above equation.

  $\begin{array}{c}{E}_{\text{cell}}^0=E_{\text{oxd}}^{°}+E_{\text{red}}^{°}\\ =\left(0\text{ V}\right)+\left(0.34\text{ V}\right)\\ =0.34\text{ V}\end{array}$

Hence, $E_{\text{red}}^{°}$ for half cell reaction is $0.34\text{ V}$ .

Expression of Nernst equation for above net reaction of electrochemical cell at room temperature is as follows:

  $E_{\text{cell}}=E_{\text{cell}}^0-\frac{0.0591}{n}\log \frac{{\left[{\text{H}}^{\text{+}}\right]}^2}{\left[{\text{Cu}}^{+2}\right]}$

Where,

• $E_{\text{cell}}$ is the cell potential.
• $E_{\text{cell}}^0$ is the standard cell potential.
• $n$ is total electrons transferred.
• $\left[{\text{H}}^{+}\right]$ is concentration of ${\text{H}}^{+}$ .
• $\left[{\text{Cu}}^{+2}\right]$ is concentration of ${\text{Cu}}^{+2}$ .

Value of $E_{\text{cell}}^0$ is 0.34.

Value of $n$ is 2.

Value of $\left[{\text{H}}^{+}\right]$ is 1.

Value of $\left[{\text{Cu}}^{+2}\right]$ is $2.5\times {10}^{-4}$ .

Substitute the value in above equation.

  $\begin{array}{c}{E}_{\text{cell}}=E_{\text{cell}}^0-\frac{0.0591}{n}\log \frac{{\left[{\text{H}}^{\text{+}}\right]}^2}{\left[{\text{Cu}}^{+2}\right]}\\ =0.34-\frac{0.0591}{2}\log \frac{{\left[\text{1}\right]}^2}{\left[2.5\times {10}^{-4}\right]}\\ =0.23\end{array}$

Hence $E_{\text{cell}}$ of electrochemical cell is $0.23\text{ V}$ .

(b)

Interpretation Introduction

Interpretation:Electrode potential for a copper electrode in sodium hydroxide solution is to be calculated.

Concept introduction:Study of interchange between electrical and chemical energy falls under branch of chemistry called electrochemistry. It includes occurrence of oxidation and reduction reactions. This includes production of electric current from chemical reaction and vice-versa.

Reactions that occur in galvanic cells can be broken down into two half-cell reactions. One half-cell reaction is that of reduction whereas another half-cell reaction is that of oxidation.

Nernst equation represents relation between potential of electrochemical reaction, standard cell potential, activities of species and temperature.

Expression of Nernst equation at room temperature is as follows:

  $E_{\text{cell}}=E_{\text{cell}}^0-\frac{0.0591}{n}\log \left(Q\right)$

Where,

• $E_{\text{cell}}$ is the cell potential.
• $E_{\text{cell}}^0$ is the standard cell potential.
• $n$ is total electrons transferred.
• $Q$ is reaction quotient.

Answer to Problem 67E

  $E_{\text{cell}}$ of electrochemical cell is $0.15\text{ V}$ .

Explanation of Solution

The half cell reaction for standard hydrogen electrode at anode isas follows:

  ${\text{H}}_{\text{2}}\to 2{\text{H}}^{\text{+}}+2{\text{e}}^{-}$

The half cell reaction for copper electrode at cathode isas follows:

  ${\text{Cu}}^{+2}+2{\text{e}}^{-}\to \text{Cu}$

Overall cell reaction is as follows:

  ${\text{Cu}}^{+2}+{\text{H}}_{\text{2}}\to \text{Cu}+2{\text{H}}^{\text{+}}$

The expression to calculate $E{°}_{\text{cell}}$ is as follows:

  $E_{\text{cell}}^0=E_{\text{oxd}}^{°}+E_{\text{red}}^{°}$

Where,

• $E{°}_{\text{cell}}$ is standard electrode potential for cell.
• $E_{\text{red}}^{°}$ is standard reduction electrode potential for cell.
• $E_{\text{oxd}}^{°}$ is standard oxidation electrode potential for cell.

Value of $E_{\text{red}}^{°}$ is $0.34\text{ V}$ .

Value of $E_{\text{oxd}}^{°}$ is $0\text{ V}$ .

Substitute values in above equation.

  $\begin{array}{c}{E}_{\text{cell}}^0=E_{\text{oxd}}^{°}+E_{\text{red}}^{°}\\ =\left(0\text{ V}\right)+\left(0.34\text{ V}\right)\\ =0.34\text{ V}\end{array}$

Hence, $E_{\text{red}}^{°}$ for half cell reaction is $0.34\text{ V}$ .

Copper electrode present in $\text{NaOH}$ solution produces $\text{Cu}{\left(\text{OH}\right)}_{\text{2}}$ . Reaction of $\text{Cu}{\left(\text{OH}\right)}_{\text{2}}$ produces one mole of ${\text{Cu}}^{+2}$ and two moles of ${\text{OH}}^{-}$ . Hence, consider concentration of ${\text{Cu}}^{+2}$ as $s$ . Thus, reaction is as follows:

  $\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)\to {\text{Cu}}^{+2}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)$

Therefore, expression of $K_{\text{sp}}$ for $\text{Cu}{\left(\text{OH}\right)}_{\text{2}}$ is as follows:

  $K_{\text{sp}}=\left[{\text{Cu}}^{+2}\right]{\left[{\text{OH}}^{-}\right]}^2$

Rearrange above equation for $\left[{\text{Cu}}^{+2}\right]$ .

  $\left[{\text{Cu}}^{+2}\right]=\frac{K_{\text{sp}}}{{\left[{\text{OH}}^{-}\right]}^2}$

Where,

• $\left[{\text{Cu}}^{+2}\right]$ is concentration of ${\text{Cu}}^{+2}$ .
• $K_{\text{sp}}$ is solubility product constant.
• $\left[{\text{OH}}^{-}\right]$ is concentration of ${\text{OH}}^{-}$ .

Value of $\left[{\text{Cu}}^{+2}\right]$ is $s$ .

Value of $\left[{\text{OH}}^{-}\right]$ is $2s$ .

Value of $K_{\text{sp}}$ is $1.6\times {10}^{-19}$ .

Substitute values in above equation.

  $\begin{array}{c}\left[{\text{Cu}}^{+2}\right]=\frac{K_{\text{sp}}}{{\left[{\text{OH}}^{-}\right]}^2}\\ s=\frac{1.6\times {10}^{-19}}{{\left(2s\right)}^2}\\ 4s^3=1.6\times {10}^{-19}\end{array}$

Above equation is solved for $s$ .

  $s=3.41\times {10}^{-7}$

Hence, concentration of ${\text{Cu}}^{+2}$ is $3.41\times {10}^{-7}M$ .

Expression of Nernst equation for above net reaction of electrochemical cell at room temperature is as follows:

  $E_{\text{cell}}=E_{\text{cell}}^0-\frac{0.0591}{n}\log \frac{{\left[{\text{H}}^{\text{+}}\right]}^2}{\left[{\text{Cu}}^{+2}\right]}$

Where,

• $E_{\text{cell}}$ is the cell potential.
• $E_{\text{cell}}^0$ is the standard cell potential.
• $n$ is total electrons transferred.
• $\left[{\text{H}}^{+}\right]$ is concentration of ${\text{H}}^{+}$ .
• $\left[{\text{Cu}}^{+2}\right]$ is concentration of ${\text{Cu}}^{+2}$ .

Value of $E_{\text{cell}}^0$ is 0.34.

Value of $n$ is 2.

Value of $\left[{\text{H}}^{+}\right]$ is 1.

Value of $\left[{\text{Cu}}^{+2}\right]$ is $3.41\times {10}^{-7}$ .

Substitute the value in above equation.

  $\begin{array}{c}{E}_{\text{cell}}=E_{\text{cell}}^0-\frac{0.0591}{n}\log \frac{{\left[{\text{H}}^{\text{+}}\right]}^2}{\left[{\text{Cu}}^{+2}\right]}\\ =0.34-\frac{0.0591}{2}\log \frac{{\left[\text{1}\right]}^2}{\left[3.41\times {10}^{-7}\right]}\\ =0.15\end{array}$

Hence $E_{\text{cell}}$ of electrochemical cell is $0.15\text{ V}$ .

(c)

Interpretation Introduction

Interpretation: Concentration of ${\text{Cu}}^{+2}$ in an unknown solution is to be calculated.

Concept introduction:Study of interchange between electrical and chemical energy falls under branch of chemistry called electrochemistry. It includes occurrence of oxidation and reduction reactions. This includes production of electric current from chemical reaction and vice-versa.

Reactions that occur in galvanic cells can be broken down into two half-cell reactions. One half-cell reaction is that of reduction whereas another half-cell reaction is that of oxidation.

Nernst equation represents relation between potential of electrochemical reaction, standard cell potential, activities of species and temperature.

Expression of Nernst equation at room temperature is as follows:

  $E_{\text{cell}}=E_{\text{cell}}^0-\frac{0.0591}{n}\log \left(Q\right)$

Where,

• $E_{\text{cell}}$ is the cell potential.
• $E_{\text{cell}}^0$ is the standard cell potential.
• $n$ is total electrons transferred.
• $Q$ is reaction quotient.

Answer to Problem 67E

Concentration of ${\text{Cu}}^{+2}$ is $1.2\times {10}^{-5}M$ .

Explanation of Solution

The half cell reaction for standard hydrogen electrode at anode is as follows:

  ${\text{H}}_{\text{2}}\to 2{\text{H}}^{\text{+}}+2{\text{e}}^{-}$

The half cell reaction for copper electrode at cathode isas follows:

  ${\text{Cu}}^{+2}+2{\text{e}}^{-}\to \text{Cu}$

Overall cell reaction is as follows:

  ${\text{Cu}}^{+2}+{\text{H}}_{\text{2}}\to \text{Cu}+2{\text{H}}^{\text{+}}$

The expression to calculate $E{°}_{\text{cell}}$ is as follows:

  $E_{\text{cell}}^0=E_{\text{oxd}}^{°}+E_{\text{red}}^{°}$

Where,

• $E{°}_{\text{cell}}$ is standard electrode potential for cell.
• $E_{\text{red}}^{°}$ is standard reduction electrode potential for cell.
• $E_{\text{oxd}}^{°}$ is standard oxidation electrode potential for cell.

Value of $E_{\text{red}}^{°}$ is $0.34\text{ V}$ .

Value of $E_{\text{oxd}}^{°}$ is $0\text{ V}$ .

Substitute values in above equation.

  $\begin{array}{c}{E}_{\text{cell}}^0=E_{\text{oxd}}^{°}+E_{\text{red}}^{°}\\ =\left(0\text{ V}\right)+\left(0.34\text{ V}\right)\\ =0.34\text{ V}\end{array}$

Hence, $E_{\text{red}}^{°}$ for half cell reaction is $0.34\text{ V}$ .

Expression of Nernst equation for above net reaction of electrochemical cell at room temperature is as follows:

  $E_{\text{cell}}=E_{\text{cell}}^0-\frac{0.0591}{n}\log \frac{{\left[{\text{H}}^{\text{+}}\right]}^2}{\left[{\text{Cu}}^{+2}\right]}$

Rearrange above equation for $\log \left[{\text{Cu}}^{+2}\right]$ .

  $\log \left[{\text{Cu}}^{+2}\right]=2\log \left[{\text{H}}^{\text{+}}\right]-\frac{n\left(E_{\text{cell}}^0-E_{\text{cell}}\right)}{0.0591}$

Where,

• $E_{\text{cell}}$ is the cell potential.
• $E_{\text{cell}}^0$ is the standard cell potential.
• $n$ is total electrons transferred.
• $\left[{\text{H}}^{+}\right]$ is concentration of ${\text{H}}^{+}$ .
• $\left[{\text{Cu}}^{+2}\right]$ is concentration of ${\text{Cu}}^{+2}$ .

Value of $E_{\text{cell}}$ is 0.195.

Value of $E_{\text{cell}}^0$ is 0.34.

Value of $n$ is 2.

Value of $\left[{\text{H}}^{+}\right]$ is 1.

Substitute the value in above equation.

  $\begin{array}{c}\log \left[{\text{Cu}}^{+2}\right]=2\log \left[{\text{H}}^{\text{+}}\right]-\frac{n\left(E_{\text{cell}}^0-E_{\text{cell}}\right)}{0.0591}\\ =2\log \left[\text{1}\right]-\frac{2\left(0.34-0.195\right)}{0.0591}\\ =-4.92\end{array}$

Value of $\left[{\text{Cu}}^{+2}\right]$ is as follows:

  $\left[{\text{Cu}}^{+2}\right]=1.2\times {10}^{-5}$

Hence, concentration of ${\text{Cu}}^{+2}$ is $1.2\times {10}^{-5}M$ .

(d)

Interpretation Introduction

Interpretation:A plot of electrode potential and concentration of ${\text{Cu}}^{+2}$ that yield straight line should be determined. Also, slope is to be calculated.

Concept introduction:Expression of Nernst equation at room temperature is as follows:

  $E_{\text{cell}}=E_{\text{cell}}^0-\frac{0.0591}{n}\log \left(Q\right)$

Where,

• $E_{\text{cell}}$ is the cell potential.
• $E_{\text{cell}}^0$ is the standard cell potential.
• $n$ is total electrons transferred.
• $Q$ is reaction quotient.

Explanation of Solution

The graph between electrode potential that is $E_{\text{cell}}$ at $y-axis$ and $\log \left[{\text{Cu}}^{+2}\right]$ at $x-axis$ gives straight line. The equation of line is as follows:

  $y=mx+c$

The half cell reaction for copper electrode at cathode is as follows:

  ${\text{Cu}}^{+2}+2{\text{e}}^{-}\to \text{Cu}$

Expression of Nernst equation of electrochemical cell at room temperature is as follows:

  $E_{\text{cell}}=E_{\text{cell}}^0-\frac{0.0591}{n}\log \frac{1}{\left[{\text{Cu}}^{+2}\right]}$

Rearrange above equation as equation of line.

  $E_{\text{cell}}=\frac{0.0591}{n}\log \left[{\text{Cu}}^{+2}\right]+E_{\text{cell}}^0$

Slope in above equation is $\frac{0.0591}{n}$ and value of $n$ is 2. Therefore, slope is calculated as follows:

  $\begin{array}{c}\text{slope}=\frac{0.0591}{2}\\ =0.0296\end{array}$

Hence, slope is $0.0296\text{ V}$ .