Question

(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.

Short answer

a) The three commercial products formed in the chlor-alkali process are, ${\text{Cl}}_{\text{2}}{\text{,H}}_{\text{2}}\text{, NaOH.}$

b) Advantage of the mercury-cell method for this process is obtaining high-purity $\text{NaOH}$and the disadvantage is release of mercury in the industrial wastewater.

Step-by-step solution

Concept Introduction

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.

Finding the Three Commercial Products formed in the Chlor-Alkali Process

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:

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

Advantage and Disadvantage of the Mercury-Cell Method

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.