Suppose we wish to electrolyze ${\mathbf{I}}^{\mathbf{-}}$to ${\mathbf{I}}_{\mathbf{3}}^{\mathbf{-}}$in a 0.10M Klsolution containing $\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathbf{M}$${\mathbf{I}}_{\mathbf{3}}^{\mathbf{-}}$at pH 10.00with ${\mathbf{P}}_{{\mathbf{H}}_{\mathbf{2}}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{bar}$.

$\mathbf{3}{\mathbf{I}}^{\mathbf{-}}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\to }{\mathbf{I}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{OH}}^{\mathbf{-}}$

(a) Find the cell voltage if no current is flowing.

(b) Then suppose that electrolysis increases $\left[I_3^{-}\right]$to$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathbf{M}$, but other concentrations are unaffected. Suppose that the cell resistance is$\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\Omega }$, the current is 63 mA, the cathode overpotential is 0.382V, and the anode overpotential is 0.025 V. What voltage is needed to drive the reaction?

Find the voltage in part (b) ifrole="math" localid="1663645169401" ${\mathbf{\left[}{\mathbf{I}}^{\mathbf{-}}\mathbf{\right]}}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}\mathbf{M}$.

When the electric current is too small, the voltage of cell is given as

E=E (cathode)-E(anode)

E( cathode) is electrode's potential which is attached to the negative terminal of the current source.

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

Overpotential:The activation energy of a reaction at an electrode can be overcome by voltage. The required voltage to apply is called overpotential.

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

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

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

Given,

The concentration of${\mathrm{I}}^{-}$ is 0.01 M.

E( anode$)0.535-\frac{0.05916}{2}\log \frac{0.{01}^3}{0.0003}=0.608\mathrm{V}$

E=E( cathode ) -E( anode )IR-overpoetentials

$$\begin{gathered} =-0.591\mathrm{V}-0.608\mathrm{V}-(2.0\mathrm{\Omega })(0.063\mathrm{A})-0.382\mathrm{V}-0.025\mathrm{V} \\ =-1.732\mathrm{V} \end{gathered}$$

In order to drive the electrolysis of ${\mathrm{I}}^{-}$to role="math" localid="1663645612720" ${\mathrm{I}}_3^{-},-1.732\mathrm{V}$is required.