Consider the following electrolysis reactions.

Cathode:${\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{l}\mathbf{\right)}\mathbf{+}{\mathit{e}}^{\mathbf{-}}\mathbf{\rightleftharpoons }\frac{\mathbf{1}}{\mathbf{2}}{\mathit{H}}_{\mathbf{2}}\mathbf{(}\mathit{g}\mathbf{,}\mathbf{1}\mathbf{.}\mathbf{0}\mathit{b}\mathit{a}\mathit{r}\mathbf{)}\mathbf{+}\mathit{O}{\mathit{H}}^{\mathbf{-}}\mathbf{(}\mathit{a}\mathit{q}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{10}\mathit{M}\mathbf{)}$

Anode:$\mathit{B}{\mathit{r}}^{\mathbf{-}}\mathbf{(}\mathit{a}\mathit{q}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{10}\mathit{M}\mathbf{)}\mathbf{\rightleftharpoons }\frac{\mathbf{1}}{\mathbf{2}}\mathit{B}{\mathit{r}}_{\mathbf{2}}\mathbf{\left(}\mathit{l}\mathbf{\right)}\mathbf{+}{\mathit{e}}^{\mathbf{-}}$

1. Calculate the voltage needed to drive the net reaction if current is negligible.
2. Suppose that the cell has a resistance of$\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{\Omega }$ and a current of 100 mA. How much voltage is needed to overcome the cell resistance? This is the ohmic potential.
3. Suppose that the anode reaction has an overpotential of 0.20 V and that the cathode overpotential is 0.40 V. What voltage is needed to overcome these effects combined with those of parts (a) and (b)?
4. Suppose that concentration polarization occurs $\mathbf{[}\mathit{O}{\mathit{H}}^{\mathbf{-}}{\mathbf{]}}_{\mathbf{s}}$. at the cathode surface increases to 1.0 M and$\mathbf{[}\mathit{B}{\mathit{r}}^{\mathbf{-}}{\mathbf{]}}_{\mathbf{s}}$ at the anode surface decreases to 0.010 M. What voltage is needed to overcome these effects combined with those of (b) and (c)?

The voltage of a cell is stated as below when the electric current is too small.

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

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

E(anode) is the potential of an electrode that is connected to the positive terminal of a current source.

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

Ohmic potential: In an electrochemical cell, voltage can overcome the electrical resistance of a solution while current I flows. The required voltage to apply is called ohmic potential.

role="math" localid="1668357739331" $E_{ohmic}=IR$

Concentration Polarization: It is the change in concentration of products and reactants at electrode's surface unlike they are same in solution.

When the electric current is negligible, the voltage of net reaction is given as

$\begin{array}{r}E=E\left(\text{cathode}\right)-E\left(\text{anode}\right)\\ \left.=\left\{{\mathrm{E}}^{\circ }\text{(cathode}\right)0.05916\log {\mathrm{P}}_{{\mathrm{H}}_2}^{0.5}\left[{\mathrm{OH}}^{-}\right]\right\}-\left\{{\mathrm{E}}^{\circ }\text{(anode)}-0.05916\log \left[{\mathrm{Br}}^{-}\right]\right\}\\ =\left\{0.828-0.05916\log \left(1_{0.5}^{00}\left[0.10\right]\right\}-\{1.078-0.05916\log [0.10\left]\right\}\right.\\ =-1.906\mathrm{V}\end{array}$

Determining ohmic potential of the cell is calculated as follows

$\begin{array}{r}{E}_{\text{ohmic}}=IR\\ =\left(0.100\mathrm{A}\right)\left(2.0\mathrm{\Omega }\right)\\ =0.20\mathrm{V}\end{array}$

Determining voltage required to overcome the ohmic potential effect and electrolysis effect.

$\begin{array}{r}\mathrm{E}=\mathrm{E}\left(\text{cathode}\right)-\mathrm{E}\left(\text{anode}\right)-\mathrm{I}\cdot \mathrm{R}-\text{overpotential}\\ =\mathrm{1.9060.20}(0.20+0.10)\\ =-2.17\mathrm{V}\end{array}$

Determining voltage required to overcome the ohmic potential effect and anodic reaction overpotential

$\begin{array}{r}\mathrm{E}\left(\text{cathode}\right)={\mathrm{E}}^{\circ }\left(\text{cathode}\right)-0.05916\log {\mathrm{P}}_{{\mu }_{-2}}^{0.5}\left[{\mathrm{OH}}^{-}\right]\\ =0.828-0.05916\log \left(1_{0.5}^{0)}\left[1.0\right]\right.\\ =-0.828\mathrm{V}\\ \mathrm{E}\left(\text{anode}\right)={\mathrm{E}}^{\mathrm{\prime }}\left(\text{anode}\right)-0.05916\log {\left[{\mathrm{Br}}^{-}\right]}_{\mathrm{s}}\\ =1.078-0.05916\log \left(0.010\right)\\ =-1.196\mathrm{V}\end{array}$

The voltage required to overcome the ohmic potential effect and anodic reaction overpotential is calculated using the anode and cathode voltages.

$\begin{array}{r}\mathrm{E}=\mathrm{E}\left(\text{cathode}\right)-\mathrm{E}\left(\text{anode}\right)-\mathrm{I}.\mathrm{R}-\text{overpotential}\\ =-0.828-1.196-0.20-(0.20+0.40)\\ =-2.82\mathrm{V}\end{array}$