Q22.53P - Chemistry: Molecular Nature Of Matter And Change

Question

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?

Step-by-Step Solution

Answer

The mass of ${\text{Cl}}_{\text{2}}$ produced is data-custom-editor="chemistry" $\text{700.028 lb}$.

Step 1: Concept Introduction

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

Step 2: Calculation

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 :

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

Step 3: Calculating the Mass of Cl 2

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

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

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 $\text{700.028 lb}$.