Chapter 11, Problem 60E

(a)

Interpretation Introduction

Interpretation:Reduction potential for $\text{Cu}$ at $0.10M$ is to be determined.

Concept introduction: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 60E

Value of $E_{\text{red}}$ for half cell reaction is $0.31\text{ V}$ .

Explanation of Solution

The reduction half cell reaction for $\text{Cu}$ is as follows:

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

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

  $E_{\text{red}}=E_{\text{red}}^{°}-\frac{0.0591}{n}\log \left(\frac{1}{\left[{\text{Cu}}^{\text{+2}}\right]}\right)$

Where,

• $E_{\text{red}}$ is reduction electrode potential for cell.
• $E_{\text{red}}^{°}$ is standard reduction electrode potential for cell.
• $n$ is total electrons transferred.
• $Q$ is reaction quotient.

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

Value of $n$ is $2$ .

Value of $\left[{\text{Cu}}^{\text{+2}}\right]$ is 0.10.

Substitute the value in above equation.

  $\begin{array}{c}{E}_{\text{red}}=E_{\text{red}}^{°}-\frac{0.0591}{n}\log \left(\frac{1}{\left[{\text{Cu}}^{\text{+2}}\right]}\right)\\ =0.34-\frac{0.0591}{2}\log \left(\frac{1}{0.1}\right)\\ =0.34-0.03\\ =0.31\end{array}$

Hence, $E_{\text{red}}$ for half cell reactionis $0.31\text{ V}$ .

(b)

Interpretation Introduction

Interpretation:Reduction potential for $\text{Cu}$ at $2M$ is to be determined.

Concept introduction: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 60E

Value of $E_{\text{red}}$ for half cell reaction is $0.35\text{ V}$ .

Explanation of Solution

The reduction half cell reaction for $\text{Cu}$ is as follows:

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

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

  $E_{\text{red}}=E_{\text{red}}^{°}-\frac{0.0591}{n}\log \left(\frac{1}{\left[{\text{Cu}}^{\text{+2}}\right]}\right)$

Where,

• $E_{\text{red}}$ is reduction electrode potential for cell.
• $E_{\text{red}}^{°}$ is standard reduction electrode potential for cell.
• $n$ is total electrons transferred.
• $Q$ is reaction quotient.

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

Value of $n$ is $2$ .

Value of $\left[{\text{Cu}}^{\text{+2}}\right]$ is 2.0.

Substitute the value in above equation.

  $\begin{array}{c}{E}_{\text{red}}=E_{\text{red}}^{°}-\frac{0.0591}{n}\log \left(\frac{1}{\left[{\text{Cu}}^{\text{+2}}\right]}\right)\\ =0.34-\frac{0.0591}{2}\log \left(\frac{1}{2}\right)\\ =0.34+0.0088\\ =0.35\end{array}$

Hence, $E_{\text{red}}$ for half cell reaction is $0.35\text{ V}$ .

(c)

Interpretation Introduction

Interpretation:Reduction potential for $\text{Cu}$ at $1\times {10}^{-4}M$ is to be determined.

Concept introduction: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 60E

Value of $E_{\text{red}}$ for half cell reaction is $0.22\text{ V}$ .

Explanation of Solution

The reduction half cell reaction for $\text{Cu}$ is as follows:

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

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

  $E_{\text{red}}=E_{\text{red}}^{°}-\frac{0.0591}{n}\log \left(\frac{1}{\left[{\text{Cu}}^{\text{+2}}\right]}\right)$

Where,

• $E_{\text{red}}$ is reduction electrode potential for cell.
• $E_{\text{red}}^{°}$ is standard reduction electrode potential for cell.
• $n$ is total electrons transferred.
• $Q$ is reaction quotient.

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

Value of $n$ is $2$ .

Value of $\left[{\text{Cu}}^{\text{+2}}\right]$ is $1\times {10}^{-4}$ .

Substitute the value in above equation.

  $\begin{array}{c}{E}_{\text{red}}=E_{\text{red}}^{°}-\frac{0.0591}{n}\log \left(\frac{1}{\left[{\text{Cu}}^{\text{+2}}\right]}\right)\\ =0.34-\frac{0.0591}{2}\log \left(\frac{1}{1\times {10}^{-4}}\right)\\ =0.34-0.03\left(4\log 10\right)\\ =0.22\end{array}$

Hence, $E_{\text{red}}$ for half cell reaction is $0.22\text{ V}$ .

(d)

Interpretation Introduction

Interpretation:Reduction potential for below half cell reaction is at $\text{pH}$ equals to 3 to be determined.

  ${\text{MnO}}_{\text{4}}^{-}+8{\text{H}}^{+}+5{\text{e}}^{-}\to {\text{Mn}}^{+2}+4{\text{H}}_{\text{2}}\text{O}$

Concept introduction: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 60E

Value of $E_{\text{red}}$ for half cell reaction is $1.24\text{ V}$ .

Explanation of Solution

The reduction half cell reaction is as follows:

  ${\text{MnO}}_{\text{4}}^{-}+8{\text{H}}^{+}+5{\text{e}}^{-}\to {\text{Mn}}^{+2}+4{\text{H}}_{\text{2}}\text{O}$

The expression used for calculation of $\text{pH}$ is as follows:

  $\text{pH}=-\log \left[{\text{H}}^{+}\right]$

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

  $E_{\text{red}}=E_{\text{red}}^{°}-\frac{0.0591}{n}\log \left(\frac{\left[{\text{Mn}}^{+2}\right]}{\left[{\text{MnO}}_{\text{4}}^{-}\right]{\left[{\text{H}}^{+}\right]}^8}\right)$

Above equation is simplified as follows:

  $\begin{array}{c}{E}_{\text{red}}=E_{\text{red}}^{°}-\frac{0.0591}{n}\left(\log \left(\frac{\left[{\text{Mn}}^{+2}\right]}{\left[{\text{MnO}}_{\text{4}}^{-}\right]}\right)-8\log \left[{\text{H}}^{+}\right]\right)\\ =E_{\text{red}}^{°}-\frac{0.0591}{n}\left(\log \left(\frac{\left[{\text{Mn}}^{+2}\right]}{\left[{\text{MnO}}_{\text{4}}^{-}\right]}\right)+8\text{pH}\right)\end{array}$

Where,

• $E_{\text{red}}$ is reduction electrode potential for cell.
• $E_{\text{red}}^{°}$ is standard reduction electrode potential for cell.
• $n$ is total electrons transferred.
• $\left[{\text{Mn}}^{+2}\right]$ is concentration of ${\text{Mn}}^{+2}$ .
• $\left[{\text{MnO}}_{\text{4}}^{-}\right]$ is concentration of ${\text{MnO}}_{\text{4}}^{-}$ .

Value of $E_{\text{red}}^{°}$ is 1.51.

Value of $n$ is 5.

Value of $\left[{\text{Mn}}^{+2}\right]$ is 0.01.

Value of $\left[{\text{MnO}}_{\text{4}}^{-}\right]$ is 0.10.

Value of $\text{pH}$ is 3.

Substitute the value in above equation.

  $\begin{array}{c}{E}_{\text{red}}=E_{\text{red}}^{°}-\frac{0.0591}{n}\left(\log \left(\frac{\left[{\text{Mn}}^{+2}\right]}{\left[{\text{MnO}}_{\text{4}}^{-}\right]}\right)+8\text{pH}\right)\\ =1.51-\frac{0.0591}{5}\left(\log \left(\frac{0.01}{0.10}\right)+8\left(3\right)\right)\\ =1.51-0.27\\ =1.24\end{array}$

Hence, $E_{\text{red}}$ for half cell reaction is $1.24\text{ V}$ .

(e)

Interpretation Introduction

Interpretation:Reduction potential for below half cell reaction at $\text{pH}$ equals to 1 is to be determined.

  ${\text{MnO}}_{\text{4}}^{-}+8{\text{H}}^{+}+5{\text{e}}^{-}\to {\text{Mn}}^{+2}+4{\text{H}}_{\text{2}}\text{O}$

Concept introduction: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 60E

Value of $E_{\text{red}}$ for half cell reaction is $1.43\text{ V}$ .

Explanation of Solution

The reduction half cell reaction is as follows:

  ${\text{MnO}}_{\text{4}}^{-}+8{\text{H}}^{+}+5{\text{e}}^{-}\to {\text{Mn}}^{+2}+4{\text{H}}_{\text{2}}\text{O}$

The expression used for calculation of $\text{pH}$ is as follows:

  $\text{pH}=-\log \left[{\text{H}}^{+}\right]$

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

  $E_{\text{red}}=E_{\text{red}}^{°}-\frac{0.0591}{n}\log \left(\frac{\left[{\text{Mn}}^{+2}\right]}{\left[{\text{MnO}}_{\text{4}}^{-}\right]{\left[{\text{H}}^{+}\right]}^8}\right)$

Above equation is simplified as follows:

  $\begin{array}{c}{E}_{\text{red}}=E_{\text{red}}^{°}-\frac{0.0591}{n}\left(\log \left(\frac{\left[{\text{Mn}}^{+2}\right]}{\left[{\text{MnO}}_{\text{4}}^{-}\right]}\right)-8\log \left[{\text{H}}^{+}\right]\right)\\ =E_{\text{red}}^{°}-\frac{0.0591}{n}\left(\log \left(\frac{\left[{\text{Mn}}^{+2}\right]}{\left[{\text{MnO}}_{\text{4}}^{-}\right]}\right)+8\text{pH}\right)\end{array}$

Where,

• $E_{\text{red}}$ is reduction electrode potential for cell.
• $E_{\text{red}}^{°}$ is standard reduction electrode potential for cell.
• $n$ is total electrons transferred.
• $\left[{\text{Mn}}^{+2}\right]$ is concentration of ${\text{Mn}}^{+2}$ .
• $\left[{\text{MnO}}_{\text{4}}^{-}\right]$ is concentration of ${\text{MnO}}_{\text{4}}^{-}$ .

Value of $E_{\text{red}}^{°}$ is 1.51.

Value of $n$ is 5.

Value of $\left[{\text{Mn}}^{+2}\right]$ is 0.01.

Value of $\left[{\text{MnO}}_{\text{4}}^{-}\right]$ is 0.10.

Value of $\text{pH}$ is 1.

Substitute the value in above equation.

  $\begin{array}{c}{E}_{\text{red}}=E_{\text{red}}^{°}-\frac{0.0591}{n}\left(\log \left(\frac{\left[{\text{Mn}}^{+2}\right]}{\left[{\text{MnO}}_{\text{4}}^{-}\right]}\right)+8\text{pH}\right)\\ =1.51-\frac{0.0591}{5}\left(\log \left(\frac{0.01}{0.10}\right)+8\left(1\right)\right)\\ =1.51-0.08\\ =1.43\end{array}$

Hence, $E_{\text{red}}$ for half cell reaction is $1.43\text{ V}$ .