A dilute ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$solution is to be electrolyzed with a pair of smooth Pt electrodes at a current density of $\mathbf{100}\mathbf{A}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$and a current of 0.100A. The products are${\mathbf{H}}_{\mathbf{2}}\left(g\right)\mathbf{}\mathbf{and}\mathbf{}{\mathbf{O}}_{\mathbf{2}}\left(g\right)\mathbf{}\mathbf{at}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{bar}$andatbar. Calculate the required voltage if the cell resistance is$\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\Omega }$and there is no concentration polarization. What voltage would be required if the Pt electrodes were replaced by Au electrodes?

The voltage of a cell when the electric current is too small,

$\mathbf{E}\mathbf{=}\mathbf{E}\left(\mathrm{cathode}\right)\mathbf{-}\mathbf{E}\left(\mathrm{anode}\right)$

E is the electrode's potential that is connected to the current source's negative terminal.

The electrode's potential is E(anode), which is connected to the positive terminal of the current source.

Overpotential: Voltage can override the activation energy of a process at an electrode, resulting in overpotential. Overpotential is the needed voltage to apply.

Ohmic potential: In an electrochemical cell, voltage can overcome the electrical resistance of a solution while currentflows. The ohmic potential is the voltage that must be applied.

${\mathbf{E}}_{\mathbf{ohmic}}\mathbf{=}\mathbf{IR}$

Concentration Polarization: Polarization is defined as a change in product and reactant concentrations at the electrode's surface, although they are the same in solution.

At cathode:$2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_{2\left(\mathrm{g}\right)}{\mathrm{E}}^{\mathrm{o}}=0\mathrm{V}$

At anode:$0.5{\mathrm{O}}_{2\left(\mathrm{g}\right)}+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2\mathrm{O}\mathrm{E}°=1.229\mathrm{V}$

E (cathode)$=0-\frac{0.05916}{2}\log \frac{{\mathrm{P}}_{{\mathrm{H}}_2}}{[{\mathrm{H}}^{+}{|}^2}$

E (anode) $=1.229-\frac{0.05916}{2}\log \frac{1}{{\left|{\mathrm{H}}^{+}\right|}^2{{\mathrm{P}}_{{\mathrm{e}}_2}}^{09}}$

$$\begin{gathered} \mathrm{E}\left(\mathrm{cell}\right)=\mathrm{E}\left(\mathrm{cathode}\right)-\mathrm{E}\left(\mathrm{anode}\right) \\ =-1.229-\frac{0.05916}{2}{\mathrm{logP}}_{{\mathrm{H}}_2}{{\mathrm{P}}_{{\mathrm{o}}_2}}^{0.5} \\ =-1.229\mathrm{V} \\ \mathrm{E}=\mathrm{E}\left(\mathrm{cell}\right)-\mathrm{IR}-\mathrm{overpotentials} \\ =1.229-\left(0.100\mathrm{A}\right)\left(2.00\mathrm{\Omega }\right)-0.85\mathrm{V}-0.068\mathrm{V} \\ =2.35\mathrm{V} \end{gathered}$$

For gold electrodes, the voltage is,

$$\begin{gathered} \mathrm{E}=\mathrm{E}\left(\mathrm{cell}\right)-\mathrm{IR}-\mathrm{overpotentials} \\ =1.229-\left(0.100\mathrm{A}\right)\left(2.00\mathrm{\Omega }\right)-0.963\mathrm{V}-0.390\mathrm{V} \\ =-2.78\mathrm{V} \end{gathered}$$