If a chlor-alkali cell used a current of $\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{A}$, how many how many pounds of${\mathbf{\text{Cl}}}_{\mathbf{\text{2}}}$would be produced in a typical$\mathbf{\text{8 - h}}$operating day?

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

Let us first write the reaction of chloride ions electrolysis:

${\text{2Cl}}^{\text{-}}\text{(aq)}\to {\text{Cl}}_{\text{2}}{\text{(g) + 2e}}^{\text{-}}$

In order to produce$1\mathrm{mol}\mathrm{of}{\mathrm{Cl}}_2,2\mathrm{mol}$of electrons is usually required and the Faraday constant is$\text{96485 C/mol}$of electron.

It is known that the current$\left(\mathrm{I}\right)\mathrm{is}3\times {10}^4\mathrm{A}$and it is passed through the cell for 8-h through the electrodes and the molar mass of${\text{Cl}}_{\text{2}}$ gas is $70.906\mathrm{g}/\mathrm{mol}$.

$$\begin{gathered} \mathrm{n}\left({\mathrm{Cl}}_2\right)=\frac{\mathrm{I}\times \mathrm{t}}{\mathrm{F}\times 2} \\ \mathrm{m}\left({\mathrm{Cl}}_2\right)=\mathrm{n}\left({\mathrm{Cl}}_2\right)\times {\mathrm{M}}_{\mathrm{R}} \end{gathered}$$

Since$1\mathrm{A}=1\mathrm{C}/\mathrm{s}$, the time should be converted to:

$$\begin{gathered} \mathrm{n}\left({\mathrm{Cl}}_2\right)=\frac{3\times {10}^4\frac{\mathrm{C}}{\mathrm{s}}\times (8\times 3600\mathrm{s})}{96485\frac{\mathrm{C}}{\mathrm{s}}\times 2\mathrm{mol}} \\ \mathrm{n}\left({\mathrm{Cl}}_2\right)=4477.38 \end{gathered}$$

Now, let us calculate the mass of${\text{Cl}}_{\text{2}}$.

role="math" localid="1663334259406" $\begin{array}{rcl}\mathrm{m}\left({\mathrm{Cl}}_2\right)& =& 4477.38\mathrm{mol}\times 70.906\mathrm{g}/\mathrm{mol}\\ \mathrm{m}\left({\mathrm{Cl}}_2\right)& =& 317473.1063\mathrm{g}\\ & =& 317.4731063\mathrm{kg}\end{array}$

Converting the$\mathrm{kg}\mathrm{to}\mathrm{lb}\left(\mathrm{pounds}\right),$

$$\begin{gathered} \mathrm{m}=\mathrm{m}\left({\mathrm{Cl}}_2\right)\times 2.205 \\ \mathrm{m}=317.4731063\mathrm{kg}\times 2.205\mathrm{lb}/\mathrm{kg} \\ \mathrm{m}=700.028\mathrm{lb} \end{gathered}$$

The mass of ${\text{Cl}}_{\text{2}}$produced is $\text{700.028 lb}$.