Question

What product forms at each electrode in the aqueous electrolysis of the following salts: (a) ${\mathbf{FeI}}_{\mathbf{2}}$ ; (b) ${\mathbf{K}}_{\mathbf{3}}{\mathbf{PO}}_{\mathbf{4}}$?

Short answer

(a) Product forms at each electrode in the aqueous electrolysis of the following salts ${\mathrm{FeI}}_2$is $\mathrm{Anode}:{\mathrm{I}}_2\left(\mathrm{g}\right)\mathrm{and}\mathrm{Cathode}:\mathrm{F}{\mathrm{e}}_{\left(\mathrm{s}\right)}$.

(b) Product forms at each electrode in the aqueous electrolysis of the following salts ${\mathrm{K}}_3{\mathrm{PO}}_4$is$\mathrm{Anode}:{\mathrm{O}}_2{\left(\mathrm{g}\right)+4\mathrm{H}}_{\left(\mathrm{aq}\right)}^{+}\mathrm{and}\mathrm{Cathode}:{\mathrm{H}}_2{\left(\mathrm{g}\right)+2\mathrm{OH}}_{\left(\mathrm{aq}\right)}^{-}$

Step-by-step solution

Definition

It's an electrochemical process that involves passing current between two electrodes through an ionized solution (the electrolyte) to deposit positive ions (anions) on the negative electrode (cathode) and negative ions (cations) on the positive electrode (cathode) (anode).

Determining the electrolysis

Electrolysis cannot convert all ions into their free element form due to overvoltage, including the following:

- Metal cations from Group $1\mathrm{A}$

- Metal cations from Group $2\mathrm{A}$and Aluminum cations from Group $3\mathrm{A}$

- While the fluoride ion cannot be oxidized, the other halides may. - Common oxyanions such as ${\mathrm{SO}}_4^{2-}{,\mathrm{CO}}_3^{2-}{,\mathrm{NO}}_3^{-}{,\mathrm{PO}}_4^{3-}$cannot be oxidized because they are at their highest oxidation states as they are. Water is oxidized to ${\mathrm{O}}_2,{\mathrm{H}}^{+}$instead.

There are particular guidelines to follow when forecasting the responses at each cell:

- At the cathode, a greater positive standard potential arises.

- At the anode, a greater negative standard potential develops.

Determining the reduction potentials

Given these reduction potentials, one can determine whether the ions or water will be the species reacting:

Reduction:

${\mathrm{E}}_{\mathrm{half}-\mathrm{cell}}^{\mathrm{o}}=-0.42\mathrm{V}(+-0.6\mathrm{duetoovervoltage},\mathrm{becomes}-1.02\mathrm{V})$

$2{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2{\left(\mathrm{g}\right)+2\mathrm{OH}}_{\left(\mathrm{aq}\right)}^{-}$Oxidation

${\mathrm{E}}_{\mathrm{half}-\mathrm{cell}}^{\mathrm{o}}=0.82\mathrm{V}(+0.6\mathrm{due}\mathrm{too}\mathrm{ver}\mathrm{voltage},\mathrm{becomes}1.42\mathrm{V})$

$2{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}\to {\mathrm{O}}_{2\left(\mathrm{g}\right)}{+4\mathrm{H}}_{\left(\mathrm{aq}\right)}^{+}+4{\mathrm{e}}^{-}$

Calculating for FeI2

a)

Let’s start,

${\mathrm{FeI}}_2$

$\begin{array}{l}{\mathrm{Fe}}_{\left(\mathrm{aq}\right)}^{2+}+2{\mathrm{e}}^{-}\to {\mathrm{Fe}}_{\left(\mathrm{s}\right)}{\mathrm{with}\mathrm{E}}_{\mathrm{half}-\mathrm{cell}}^{\mathrm{o}}=-0.44\\ {2\mathrm{I}}_{\left(\mathrm{aq}\right)}^{-}\to {\mathrm{I}}_{2\left(\mathrm{g}\right)}+2{\mathrm{e}}^{-}{\mathrm{with}\mathrm{E}}_{\mathrm{half}-\mathrm{cell}}^{\mathrm{o}}=0.53\end{array}$

Fe is created at the cathode due to its higher standard potential, whereas iodine is formed at the anode due to its lower standard potential.

Anode ${2\mathrm{I}}_{\left(\mathrm{aq}\right)}^{-}\to {\mathrm{I}}_{2\left(\mathrm{g}\right)}+2{\mathrm{e}}^{-}$

Cathode: ${\mathrm{Fe}}_{\left(\mathrm{aq}\right)}^{2+}+2{\mathrm{e}}^{-}\to {\mathrm{Fe}}_{\left(\mathrm{s}\right)}$

Therefore, the solution is$\mathrm{Anode}:{\mathrm{I}}_2\left(\mathrm{g}\right)\mathrm{and}\mathrm{Cathode}:\mathrm{F}{\mathrm{e}}_{\left(\mathrm{s}\right)}$

Calculating for  

b)

Let’s start,

${\mathrm{K}}_3{\mathrm{PO}}_4$

${\mathrm{K}}_{\left(\mathrm{aq}\right)}^{+}+{\mathrm{e}}^{-}\to {\mathrm{K}}_{\left(\mathrm{s}\right)}$with ${\mathrm{E}}_{\mathrm{half}-\mathrm{cell}}^{\mathrm{o}}=-2.93$

${\mathrm{PO}}_4^{3-}$Can no longer be oxidized;

Water is oxidised at the anode because phosphate ions can no longer be oxidised.

Water is created near the cathode because it has a greater standard potential.

Anode: $2{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}\to {\mathrm{O}}_{2\left(\mathrm{g}\right)}{+4\mathrm{H}}_{\left(\mathrm{aq}\right)}^{+}+4{\mathrm{e}}^{-}$

Cathode : $2{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_{2\left(\mathrm{g}\right)}{+2\mathrm{OH}}_{\left(\mathrm{aq}\right)}^{-}$

Therefore, the solution is $\mathrm{Anode}:{\mathrm{O}}_2{\left(\mathrm{g}\right)+4\mathrm{H}}_{\left(\mathrm{aq}\right)}^{+}\mathrm{and}\mathrm{Cathode}:{\mathrm{H}}_2{\left(\mathrm{g}\right)+2\mathrm{OH}}_{\left(\mathrm{aq}\right)}^{-}$