(a) What are the three commercial products formed in the chlor-alkali process?

(b) State an advantage and a disadvantage of the mercury-cell method for this process.

The chlor-alkali process is a commercial electrolysis method for sodium chloride solutions. It's the method for making chlorine and sodium hydroxide, both of which are common industrial chemicals.

a)The chlor-alkali process involves electrolyzing concentrated aqueous $\text{NaCl}$to generate ${\text{Cl}}_{\text{2}}$and other by-products.

We've already mentioned that electrolyzing an aqueous solution of $\text{NaCl}$would not result in sodium on the cathode since the reduction of ${\text{H}}^{+}$ions from water is preferable because it has a higher half-cell potential. The following reactions can be used to describe the electrolysis of an aqueous solution of $\text{NaCl}$:

$$\begin{gathered} \frac{2{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)\to {\mathrm{Cl}}_2\left(\mathrm{g}\right)+2{\mathrm{e}}^{-} {\mathrm{E}}^0=1.36\mathrm{V} \mathrm{oxidation} \\ 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+2{\mathrm{e}}^{-}\to 2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right) \mathrm{E}=-1.0\mathrm{V} \mathrm{reduction}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+2{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)\to 2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right) {\mathrm{E}}_{\mathrm{cell}}=-2.4\mathrm{V}} \end{gathered}$$

It's important to remember that the solution also contains${\text{Na}}^{+}$ions, therefore the whole ionic equation is:

${\text{2Na}}^{\text{+}}{\text{(aq) + 2Cl}}^{\text{-}}{\text{(aq) + 2H}}_{\text{2}}\text{O(l)}\to {\text{2Na}}^{\text{+}}{\text{(aq) + 2OH}}^{\text{-}}{\text{(aq) +H}}_{\text{2}}{\text{(g) + Cl}}_{\text{2}}\text{(g)}$

Thus, in the process of chlor-alkali, the electrolysis of $\text{NaCl}$yields three important products:

${\text{Cl}}_{\text{2}}{\text{,H}}_{\text{2}}\text{, NaOH.}$

b) A mercury-cell technique is used to obtain high-purity $\text{NaOH}$. The cathode is mercury, which produces such a high overvoltage for water to hydrogen reduction that the process favours reduction of ${\text{Na}}^{\text{+}}$. Sodium amalgam is formed when sodium dissolves in mercury, $\text{Na(Hg)}$. The sodium amalgam is pumped out of the system and treated with sodium-depleted water.

$$\begin{gathered} \frac{2{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)\to {\mathrm{Cl}}_2\left(\mathrm{g}\right)+2{\mathrm{e}}^{-} \\ 2{\mathrm{Na}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to 2\mathrm{Na}\left(\mathrm{Hg}\right)}{2\mathrm{Na}\left(\mathrm{Hg}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to 2{\mathrm{Na}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)} \end{gathered}$$

The mercury that is liberated during this process is returned to the electrolysis bath. The main benefit of this method is that it produces high purity $\text{NaOH}$. The main downside of this approach is that some mercury is lost in the industrial wastewater as it is recycled.