Question

The products of the chlor-alkali process, $\mathit{C}{\mathit{I}}_{\mathbf{2}}$and $\mathbf{NaOH}$, are kept separated.

(a) Why is this necessary whenrole="math" localid="1663308994449" ${\mathbf{\text{CI}}}_{\mathbf{2}}$ is the desired product?

(b) role="math" localid="1663309062193" ${\mathbf{\text{ClO}}}_{}^{\mathbf{-}}$orrole="math" localid="1663309079234" ${\mathbf{\text{CIO}}}_3^{-}$ may form by disproportionation of${\mathbf{\text{CI}}}_{\mathbf{2}}$ in basic solution. What determines which product forms?

(c) What mole ratio of ${\mathbf{\text{CI}}}_{\mathbf{2}}$to ${\mathbf{\text{OH}}}^{-}$will produce${\mathbf{\text{ClO}}}_{}^{\mathbf{-}}$ ?${\mathbf{\text{CIO}}}_3^{-}$ ?

Short answer

(a) If ${\text{CI}}_2$is the desired product, it is kept separately from the $\text{NaOH}$solution to avoid its disproportionation into Cl-containing species.

(b) The product created would be determined by the ratio of ${\text{CI}}_2$gas to available.

(c) If the ratio is 1:2 sodium hypochlorite $\text{NaOCI}$ would be formed. If the ratio is 1:6 Sodium chlorate ${\text{NaOCI}}_3$would be formed.

Step-by-step solution

Definition of Elements in nature and Industry

Chemically, elements are substances that cannot be broken down into simpler things. Hydrogen (H), with an atomic number of 1, is one of nature's elements. This element gave origin to all others and makes up 75% of the mass of the cosmos. Carbon (C) is a six-atomic element.

Why is this necessary when  is the desired product

(a)

The chlor-alkali process' total reaction can be stated as

${\text{2NaCl(s)+H}}_{2}O\left(l\right)\to 2NaOH\left(aq\right)+H_2\left(g\right)+C{l}_2\left(g\right)$

When${\text{Cl}}_{\text{2}}$ is an undesirable product, the membrane fails to prevent the ${\text{OH}}^{\text{-}}$ions produced at the cathode from diffusing into the electrolyte, allowing them to diffuse throughout the electrolyte. The electrolyte becomes more basic, resulting in less ${\text{Cl}}_{\text{2}}$being created as the ${\text{Cl}}_{\text{2}}$is disproportionate to form chloride and hypochlorite ions at the anode compartment, a mixture known as bleach:

${\text{Cl}}_2\left(g\right)+2NaOH\left(aq\right)\to NaCl\left(aq\right)+NaCl{O}_3\left(aq\right)+H_{2}O\left(l\right)$

Thus, if ${\text{Cl}}_{\text{2}}$is the desired product, it is kept separately from the $\text{NaOH}$solution to avoid its disproportionation into Cl-containing species.

What determines which product forms 

(b)

Writing the ${\text{Cl}}_{\text{2}}$ reactions separately for each case, in the basic medium, the chlorine gas can form two compounds:

- for ${\text{ClO}}^{-}$the reaction is as follows:

${\text{Cl}}_{\text{2}}\text{(g) + 2NaOH(aq)}\to {\text{NaCl(aq) + NaClO(aq) +H}}_{\text{2}}\text{O(l)}$

- for ${\text{ClO}}_3^{-}$the reaction is as follows:

${\text{Cl}}_{\text{2}}\text{(g) + 6NaOH(aq)}\to {\text{5NaCl(aq) + NaClO}}_{\text{3}}{\text{(aq) + 3H}}_{\text{2}}\text{O(l)}$

Thus, the product created would be determined by the ratio of${\text{Cl}}_{\text{2}}$gas to $\text{NaOH}$available.

Find the mole ratio

(c)

Based on the reactions written in b),

- if the ratio

${\text{Cl}}_{\text{2}}\text{:NaOH = 1:2}$

It would produce sodium hypochlorite $\text{(NaOCl)}$.

- if the ratio

${\text{Cl}}_{\text{2}}\text{:NaOH = 1:6}$

Thus, the Sodium chlorate $\left({\text{NaOCl}}_3\right)$would be formed.