Question

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?

Short answer

The voltage required to electrolyze sodium sulfate with given current and current density is$-2.35\mathrm{V}$ The voltage when gold electrodes replaces platinum electrodes is $-2.78\mathrm{V}$.

Step-by-step solution

Concept used.

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.

Step 2: The voltage required to electrolyze sodium sulfate with given current and current density.

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,

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