Question

How does chlorine gas give `Cl^(-)` and `ClO_3^(-)` upon basic hydrolysis?

Answer 1

This is a disproportionation reaction.

`Cl_2(g) + +2HO^(-) rarr Cl^(-)+ClO^(-) +H_2O`

Chlorine is simultaneously reduced and oxidized.

Explanation:

`Cl_2(g) + 2HO^(-) rarr Cl^(-) + ClO^(-) + H_2O`

Chlorine gas is OXIDIZED to hypochlorite `(Cl,+I)`, and REDUCED to chloride ion, `(Cl^-,-I)`,.

`"OXIDATION:"` `1/2Cl_2(g) + H_2O rarr ClO^(-) +2H^(+) + e^(-)`

`"REDUCTION:"` `1/2Cl_2(g) + e^(-) rarr Cl^(-)`

`"OVERALL:"` `Cl_2(g) + H_2O rarr Cl^(-)+ClO^(-) +2H^(+)`

But we note that basic conditions were specified. How do we cope? We simply add `2xxHO^-` to both sides.

`Cl_2(g) + +2HO^(-) + cancel(H_2O) rarr Cl^(-)+ClO^(-) +cancel(2)H_2O`

Here we have used the equation: `H^+ + HO^(-) rarr H_2O`, and cancelled out the waters.