Lead is to be deposited at a cathode from a solution that is 0.150 M in Pb²⁺ and 0.215 M in HClO₄. Oxygen is evolved at a pressure of 0.850 atm at a 30-cm²platinum anode. The cell has a resistance of 0.900 V.

(a) Calculate the thermodynamic potential of the cell.

(b) Calculate the IR drop if a current of 0.220 A is to be used.

The thermodynamic potential of the cell should be determined.

The solution is 0.150 M in Pb²⁺ and 0.215 M in HClO₄.

The expression for electrode potential is:

${\mathrm{E}}_{\mathrm{cell}}={{\mathrm{E}}^0}_{\mathrm{cell}}-\frac{0.0592}{\mathrm{n}}\log \frac{\left[\mathrm{Red}\right]}{\left[\mathrm{Ox}\right]}$

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

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

${\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{right}}-{\mathrm{E}}_{\mathrm{left}}$…… (I)

Here, the electrode potential of cathode is ${\mathrm{E}}_{\mathrm{right}}$and electrode potential of anode is ${\mathrm{E}}_{\mathrm{left}}$.

The overall reaction is:

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

Write the reaction at cathode.

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

The reaction at anode is:

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

The expression for the electrode potential of cathode is:

${\mathrm{E}}_{\mathrm{right}}={{\mathrm{E}}^0}_{\mathrm{cell}}-\frac{0.0592}{\mathrm{n}}\log \frac{1}{{\left[{\mathrm{Pb}}^{2+}\right]}^2}$…… (II)

The expression for the electrode potential of anode is:

${\mathrm{E}}_{\mathrm{left}}={{\mathrm{E}}^0}_{\mathrm{cell}}-\frac{0.0592}{\mathrm{n}}\log \frac{1}{\left[{\mathrm{pO}}_2\right]{\left[{\mathrm{H}}^{+}\right]}^4}$…… (III)

Substitute -0.126 for , 2 for n and 0.150 for Pb²⁺ in equation (II).

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

Substitute 1.299 for ${{\mathrm{E}}^0}_{\mathrm{cell}}$, 0.850 for pO₂, 4 for n and 0.215 for H⁺ in equation (III).

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

Substitute -0.1504 V for ${\mathrm{E}}_{\mathrm{right}}$and 1.2584 V for ${\mathrm{E}}_{left}$.

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}=(-0.1504\mathrm{V})-(1.2584\mathrm{V}) \\ =-1.409\mathrm{V} \end{gathered}$$

Therefore, the thermodynamic potential of the cell is -1.409 V..

The IR drop when the current is 0.220 A should be determined.

The oxygen is evolved at pressure of 0.850 atm at a 30 cm² platinum anode. And resistance of cell is 0.900 ohm.

The expression for IR drop is:

$\mathrm{V}=\mathrm{IR}$ …… (IV)

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

Substitute 0.220 A for I and 0.900 ohm for R in Equation (IV).

$$\begin{gathered} \mathrm{V}=(0.220\mathrm{A})(0.900\mathrm{\Omega }) \\ =0.198\mathrm{V} \end{gathered}$$

Therefore, the IR drop is 0.198 V.