Chapter 24, Problem 24.1QAP

(a)

Interpretation Introduction

Interpretation: The thermodynamic potential of the cell should be determined.

Concept Introduction:The electrode potential of the cell is defined as the potential of cell consisting of two electrodes. Therefore at anode the oxidation occurs and at cathode reduction occurs.

Answer to Problem 24.1QAP

The thermodynamic potential of the cell is $-1.409\text{V}$ .

Explanation of Solution

The solution is $0.150\text{M}$ in ${\text{Pb}}^{2+}$ and $0.215\text{M}$ in ${\text{HClO}}_4$ .

The expression for electrode potential is:

  $E_{\text{cell}}=E_{\text{cell}}^0-\frac{0.0592}{n}\log \frac{\left[\mathrm{Re}\text{d}\right]}{\left[\text{Ox}\right]}$

Here, standard electrode potential is $E_{\text{cell}}^0$ , number of electrons is $n$ , reduction is $\text{Red}$ and oxidation is $\text{Ox}$ .

The electrode potential of the cell is equal to the difference between electrode potential of cathode and electrode potential of anode.

  $E_{\text{cell}}=E_{\text{right}}-E_{\text{left}}$ ...... (I)

Here, the electrode potential of cathode is $E_{\text{right}}$ andelectrode potential of anode is $E_{\text{left}}$ .

The overall reaction is:

  $2{\text{Pb}}^{2+}+2{\text{H}}_2\text{O}\stackrel{}{\to }\text{2Pb}\left(\text{s}\right)+{\text{O}}_2+4{\text{H}}^{+}$

Write the reaction at cathode.

  $2{\text{Pb}}^{2+}+4\text{e}\stackrel{}{\to }\text{2Pb}\left(\text{s}\right)$

The reaction at anode is:

  ${\text{O}}_2+4{\text{H}}^{+}+4{\text{e}}^{-}\stackrel{}{\to }{\text{2H}}_2\text{O}$

The expression for the electrode potential of cathode is:

  $E_{\text{right}}=E_{\text{cell}}^0-\frac{0.0592}{n}\log \frac{1}{{\left[{\text{Pb}}^{2+}\right]}^2}$ ...... (II)

The expression for the electrode potential of anode is:

  $E_{\text{left}}=E_{\text{cell}}^0-\frac{0.0592}{n}\log \frac{1}{\left[{\text{pO}}_2\right]{\left[{\text{H}}^{+}\right]}^4}$ ...... (III)

Substitute $-0.126$ for $E_{\text{cell}}^0$ , $2$ for $n$ and $0.150$ for ${\text{Pb}}^{2+}$ in equation (II).

  $\begin{array}{c}{E}_{\text{right}}=-0.126-\frac{0.0592}{4}\log \frac{1}{{\left[0.150\right]}^2}\\ =-0.1504\text{V}\end{array}$

Substitute $1.299$ for $E_{\text{cell}}^0$ , $0.850$ for ${\text{pO}}_2$ , $4$ for $n$ and $0.215$ for ${\text{H}}^{+}$ in equation (III).

  $\begin{array}{c}{E}_{\text{left}}=1.299-\frac{0.0592}{4}\log \frac{1}{\left[0.850\right]{\left[0.215\right]}^4}\\ =1.2584\text{V}\end{array}$

Substitute $-0.1504\text{V}$ for $E_{\text{right}}$ and $1.2584\text{V}$ for $E_{\text{left}}$ .

  $\begin{array}{c}{E}_{\text{cell}}=\left(-0.1504\text{V}\right)-\left(1.2584\text{V}\right)\\ =-1.409\text{V}\end{array}$

Therefore, the thermodynamic potential of the cell is $-1.409\text{V}$ .

(b)

Interpretation Introduction

Interpretation:The $\text{IR}$ drop when the current is $0.220\text{A}$ should be determined.

Concept Introduction:The current is directly proportional to voltage.

Answer to Problem 24.1QAP

The $\text{IR}$ drop when the current is $0.220\text{A}$ is $0.198\text{V}$ .

Explanation of Solution

The oxygen is evolved at pressure of $0.850\text{atm}$ at a $30{\text{cm}}^2$ platinum anode. And resistance of cell is $0.900\Omega$

The expression for $IR$ drop is:

  $V=IR$ ...... (IV)

Here, voltage of cell is $V$ , the current is $I$ and resistance of the cell is $R$ .

Substitute $0.220\text{A}$ for $I$ and $0.900\Omega$ for $R$ in Equation (IV).

  $\begin{array}{c}V=\left(0.220\text{A}\right)\left(0.900\Omega \right)\\ =0.198\text{V}\end{array}$

Therefore, the $IR$ drop is $0.198\text{V}$ .