Question

${\mathbf{\text{F}}}_{\mathbf{\text{2}}}{\mathbf{\text{and Cl}}}_{\mathbf{\text{2}}}$are produced by electrolytic oxidation, whereas${\mathbf{\text{Br}}}_{\mathbf{\text{2}}}{\mathbf{\text{andl}}}_{\mathbf{\text{2}}}$are produced by chemical oxidation of the halide ions in a concentrated aqueous solution (brine) by a more electronegative halogen. Give two reasons why${\mathbf{\text{Cl}}}_{\mathbf{\text{2}}}$isn't prepared this way.

Short answer

1. Because${\text{Cl}}_2$ is highly electronegative, a very strong oxidizing agent such as manganese dioxide $\left({\text{MnO}}_{\text{2}}\right)$or sodium dichromate$\left({\text{Na}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\right)$ and severe conditions are required for its preparation.

2. Since the source of ${\text{- Cl}}^{-}$ions is table salt , molten${\text{NaCl/CaCl}}_{\text{2}}$ mixture is employed to obtain a pure $\text{Na}$. On a big scale, chlorine gas is obtained as a 'by-product' on the anode.

Step-by-step solution

Oxidation

Oxidation is the loss of electrons, while reduction is the gain of electrons. When neutral atoms or molecules gain or lose electrons, such as those on an electrode's surface, they become ions, which can dissolve in the electrolyte and react with other ions.

Chemical oxidation reduces residual COD, non-biodegradable chemicals, and trace organic compounds using oxidation reagents such as hydrogen peroxide, chlorine dioxide, and ozone.

Explanation

1. Chlorine is highly electronegative,it is rarely synthesized through chemical oxidation.

Halogens' electronegativity drops as they go down Group$\text{7 (17) from F to I.}$

Non-metals such as fluorine and chlorine are extremely electronegative and thus are excellent oxidizing agents to oxidize a highly powerful oxidizing substance such as manganese dioxide or sodium dichromate. As a result, the chemical oxidation of anions is not advantageous because severe conditions are required.

2. In the industry, table salt is also a common source of manufacturing. A molten mixture is utilized to obtain a pure one. On the anode, chlorine gas is produced as a 'by-product.'

Anode (oxidation):${\text{2Cl}}^{\text{-}}\text{(l)}\to {\text{Cl}}_{\text{2}}{\text{(g) + 2e}}^{\text{-}}$

Cathode (reduction): ${\text{2Na}}^{\text{+}}{\text{(l) + 2e}}^{\text{-}}\to \text{2Na(l)}$.