Chapter 14, Problem 14.10P

(a)

Interpretation Introduction

Interpretation:

Cell voltage from below half reaction has to be calculated.

  $\begin{array}{c}\text{AgCl}+2{\text{e}}^{-}\to 2\text{Ag}+2{\text{Cl}}^{-}\\ {\text{PbF}}_2+2{\text{e}}^{-}\to \text{Pb}+2{\text{F}}^{-}\end{array}$

Concept Introduction:

Expression of Nernst equation at room temperature is as follows:

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

Here,

$E_{\text{cell}}$ 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 14.10P

Cell voltage for complete reaction is $0.572\text{V}$.

Explanation of Solution

The half cell reaction at cathode is as follows:

  $\text{AgCl}+2{\text{e}}^{-}\to 2\text{Ag}+2{\text{Cl}}^{-}$

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

  ${\left(E_{\text{cell}}\right)}_{\text{Cathode}}=E_{\text{cell}}^0-\frac{0.0591}{n}\log {\left[{\text{Cl}}^{-}\right]}^2$        (1)

Substitute $0.222\text{V}$ for $E_{\text{cell}}^0$, 2 for $n$ and $0.10\text{M}$ for $\left[{\text{Cl}}^{-}\right]$ in equation (1).

  $\begin{array}{c}{\left(E_{\text{cell}}\right)}_{\text{Cathode}}=\left(0.222\text{V}\right)-\frac{0.0591}{2}\log {\left(0.10\text{M}\right)}^2\\ =0.2816\text{V}\end{array}$

The half cell reaction at anode is as follows:

  ${\text{PbF}}_2+2{\text{e}}^{-}\to \text{Pb}+2{\text{F}}^{-}$

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

  ${\left(E_{\text{cell}}\right)}_{\text{Anode}}=E_{\text{cell}}^0-\frac{0.0591}{n}\log {\left[{\text{F}}^{-}\right]}^2$        (2)

Substitute $-0.350\text{V}$ for $E_{\text{cell}}^0$, 2 for $n$ and $0.10\text{M}$ for $\left[{\text{F}}^{-}\right]$ in equation (2).

  $\begin{array}{c}{\left(E_{\text{cell}}\right)}_{\text{Anode}}=\left(-0.350\text{V}\right)-\frac{0.0591}{2}\log {\left(0.10\text{M}\right)}^2\\ =-0.29084\text{V}\end{array}$

$E{°}_{\text{cell}}$ for complete cell is calculated as follows:

  $\begin{array}{c}{E}_{\text{cell}}={\left(E_{\text{cell}}\right)}_{\text{Cathode}}-{\left(E_{\text{cell}}\right)}_{\text{Anode}}\\ =0.2816\text{V}-\left(-0.29084\text{V}\right)\\ =0.572\text{V}\end{array}$

Hence, cell voltage for complete reaction is $0.572\text{V}$.

(b)

Interpretation Introduction

Interpretation:

Cell voltage from below half reaction has to be calculated.

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

Concept Introduction:

Refer to part (a).

Answer to Problem 14.10P

Cell voltage for complete reaction is $0.569\text{V}$.

Explanation of Solution

Concentration of ${\text{Ag}}^{+}$ is calculated as follows:

  $\begin{array}{c}\left[{\text{Ag}}^{+}\right]=\frac{K_{\text{sp}}}{\left[{\text{Cl}}^{-}\right]}\\ =\frac{1.8\times {10}^{-10}}{0.10}\\ =1.8\times {10}^{-9}\end{array}$

Concentration of ${\text{Pb}}^{2+}$ is calculated as follows:

  $\begin{array}{c}\left[{\text{Pb}}^{2+}\right]=\frac{K_{\text{sp}}}{{\left[{\text{Cl}}^{-}\right]}^2}\\ =\frac{3.6\times {10}^{-8}}{{\left(0.10\right)}^2}\\ =3.60\times {10}^{-6}\end{array}$

The half cell reaction at cathode is as follows:

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

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

  ${\left(E_{\text{cell}}\right)}_{\text{Cathode}}=E_{\text{cell}}^0-\frac{0.0591}{n}\log \left(\frac{1}{{\left[{\text{Ag}}^{+}\right]}^2}\right)$        (3)

Substitute $0.799\text{V}$ for $E_{\text{cell}}^0$, 2 for $n$ and $1.8\times {10}^{-9}$ for $\left[{\text{Ag}}^{+}\right]$ in equation (3).

  $\begin{array}{c}{\left(E_{\text{cell}}\right)}_{\text{Cathode}}=\left(0.799\text{V}\right)-\frac{0.0591}{2}\mathrm{l}\log \left(\frac{1}{{\left(1.8\times {10}^{-9}\right)}^2}\right)\\ =0.282\text{V}\end{array}$

The half cell reaction at anode is as follows:

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

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

  ${\left(E_{\text{cell}}\right)}_{\text{Anode}}=E_{\text{cell}}^0-\frac{0.0591}{n}\log \left(\frac{1}{\left[{\text{Pb}}^{2+}\right]}\right)$        (4)

Substitute $-0.126\text{V}$ for $E_{\text{cell}}^0$, 2 for $n$ and $3.60\times {10}^{-6}$ for $\left[{\text{Pb}}^{2+}\right]$ in equation (4).

  $\begin{array}{c}{\left(E_{\text{cell}}\right)}_{\text{Anode}}=\left(-0.126\text{V}\right)-\frac{0.0591}{2}\log \left(\frac{1}{3.60\times {10}^{-6}}\right)\\ =-0.287\text{V}\end{array}$

$E{°}_{\text{cell}}$ for complete cell is calculated as follows:

  $\begin{array}{c}{E}_{\text{cell}}={\left(E_{\text{cell}}\right)}_{\text{Cathode}}-{\left(E_{\text{cell}}\right)}_{\text{Anode}}\\ =0.282\text{V}-\left(-0.287\text{V}\right)\\ =0.569\text{V}\end{array}$

Hence, cell voltage for complete reaction is $0.569\text{V}$.