$\mathit{T}{\mathit{i}}^{\mathbf{3}\mathbf{+}}$ is to be generated in $\mathbf{0}\mathbf{.}\mathbf{10}\mathit{M}\mathit{H}\mathit{C}\mathit{l}{\mathit{O}}_{\mathbf{4}}$ for coulometric reduction of azobenzene.

$\begin{array}{r}{\mathbf{TiO}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{+}{\mathbf{e}}^{\mathbf{-}}\mathbf{\rightleftharpoons }{\mathbf{Ti}}^{\mathbf{3}\mathbf{+}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}{\mathbf{E}}^{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.100}\mathbf{V}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{4}{\mathbf{Ti}}^{\mathbf{3}\mathbf{+}}\mathbf{+}{\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}\mathbf{N}\mathbf{=}{\mathbf{NC}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}\mathbf{+}\mathbf{4}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\to }\mathbf{2}{\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{NH}}_{\mathbf{2}}\mathbf{+}\mathbf{4}{\mathbf{TiO}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathbf{4}{\mathbf{H}}^{\mathbf{+}}\end{array}$

At the counter electrode, water is oxidized, and \(\mathrm{O}_{2}\) is liberated at a pressure of \(0.20\) bar. Both electrodes are made of smooth Pt, and each has a total surface area of $\mathbf{1}\mathbf{.}\mathbf{00}\mathit{c}{\mathit{m}}^{\mathbf{2}}$. The rate of reduction of the azobenzene is 25.9nmol/s , and the resistance of the solution between the generator electrodes is $\mathbf{52}\mathbf{.}\mathbf{4}\mathit{\Omega }$.

1. Calculate the current density $\mathbf{(}\mathit{A}\mathbf{/}{\mathit{m}}^{\mathbf{2}}\mathbf{)}$at the electrode surface. Use Table 17-1 to estimate the overpotential for ${\mathit{O}}_{\mathbf{2}}$liberation.
2. Calculate the cathode potential (versus S.H.E.) assuming that role="math" localid="1668356673323" $\mathbf{\left[}\mathbf{Ti}{{\mathbf{O}}^{\mathbf{2}\mathbf{+}}}_{\mathbf{surface}}\mathbf{\right]}\mathbf{=}{\mathbf{\left[}{\mathbf{TiO}}^{\mathbf{2}\mathbf{+}}\mathbf{\right]}}_{\mathbf{bulk}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{050}\mathit{M}$and $\mathbf{[}\mathit{T}{\mathit{i}}^{\mathbf{3}\mathbf{+}}{\mathbf{]}}_{\mathbf{s}\mathbf{u}\mathbf{r}\mathbf{f}\mathbf{a}\mathbf{c}\mathbf{e}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}\mathit{M}$.
3. Calculate the anode potential (versus S.H.E.).
4. What should the applied voltage be?

When the electric current is too modest, the voltage of a cell is expressed as follows.

$E=E\left(cathode\right)-E\left(anode\right)$

The electrode potential E (cathode) is connected to the current source's negative terminal.

E(anode) is the potential of an electrode attached to a current source's positive terminal.

Overpotential occurs when the activation energy of a process at an electrode is overridden by voltage. Overpotential is the required voltage to apply.

Voltage can overcome the electrical resistance of a solution in an electrochemical cell while current I flows. The ohmic potential is the voltage that must be applied.

role="math" localid="1668356277863" $E_{ohmic}=IR$

Concentration Polarization is the difference in concentration of products and reactants at the electrode's surface, as opposed to the same concentration in solution.

Determing the current density and overpotential of reduction of azobenzene.

The reduction of azobenzene is stated as

$\begin{array}{r}{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}={\mathrm{NC}}_6{\mathrm{H}}_5+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}\to 2{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2\\ \\ \text{Electron flow}=\left(4\frac{{\mathrm{e}}^{-\mathrm{s}}}{{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}={\mathrm{NC}}_6{\mathrm{H}}_5}\right)\left(25.9\frac{\mathrm{nmol}}{\mathrm{s}}\right)(96485\mathrm{C}/\mathrm{mol})\\ =1.00\times {10}^{-2}\mathrm{C}/\mathrm{s}\end{array}$

The current density of reduction of azobenzene $=\frac{1.00\times {10}^2\mathrm{A}}{1.00\times {10}^4{\mathrm{m}}^2}=100\mathrm{A}/{\mathrm{m}}^2$

For smooth platinum electrode current density of $100A/m^2=0.85V$

Over potential is 0.85 V.

Determing the cathode potential at which azobenzene get reduced.

Using Nernst eqn the Cathode potential is determined as

$\begin{array}{r}\mathrm{E}\left(\text{cathode}\right)=0.100-0.05916\log \frac{{\left[{\mathrm{Ti}}^{3+}\right]}_s}{{\left[{\mathrm{TiO}}^{2+}\right]}_{\mathrm{s}}{\left[{\mathrm{H}}^{+}\right]}^2}\\ =0.100-0.05916\log \frac{\left[0.10\right]}{\left[0.050\right][0.10{]}^2}\\ =0.100\mathrm{V}-0.136\mathrm{V}\\ =-0.036\mathrm{V}\end{array}$

Determing the anode potential at which azobenzene get reduced.

Using Nernst equation, the anode potential is determined as

$\begin{array}{r}E\text{(anode)}=1.229-\frac{0.05916}{4}\log \frac{1}{P_{{\mathrm{O}}_2}{\left|{\mathrm{H}}^{+}\right|}^4}\\ =1.229-\frac{0.05916}{4}\log \frac{1}{\left(0.20\right)(0.10{)}^4}\\ =1.229\mathrm{V}-0.069\mathrm{V}\\ =1.160\mathrm{V}\end{array}$

Determing the voltage at which azobenzene reduced to aniline.

The reduction of azobenzene is stated as

$\begin{array}{r}{\mathrm{C}}_8{\mathrm{H}}_5\mathrm{N}={\mathrm{NC}}_6{\mathrm{H}}_5+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}\to 2{\mathrm{C}}_8{\mathrm{H}}_5{\mathrm{NH}}_2\\ \\ E=E\text{(cathode)}-E\text{(anode)}-IR-\text{Overpotential}\\ =-0.036-1.160-\left(1.00\times {10}^{-2}\mathrm{A}\right)\left(52.4\mathrm{\Omega }\right)-0.85\\ =-2.57\mathrm{V}\end{array}$