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

Oct 18, 2016

This is a disproportionation reaction.

$C {l}_{2} \left(g\right) + + 2 H {O}^{-} \rightarrow C {l}^{-} + C l {O}^{-} + {H}_{2} O$

Chlorine is simultaneously reduced and oxidized.

Explanation:

$C {l}_{2} \left(g\right) + 2 H {O}^{-} \rightarrow C {l}^{-} + C l {O}^{-} + {H}_{2} O$

Chlorine gas is OXIDIZED to hypochlorite $\left(C l , + I\right)$, and REDUCED to chloride ion, $\left(C {l}^{-} , - I\right)$,.

$\text{OXIDATION:}$ $\frac{1}{2} C {l}_{2} \left(g\right) + {H}_{2} O \rightarrow C l {O}^{-} + 2 {H}^{+} + {e}^{-}$

$\text{REDUCTION:}$ $\frac{1}{2} C {l}_{2} \left(g\right) + {e}^{-} \rightarrow C {l}^{-}$

$\text{OVERALL:}$ $C {l}_{2} \left(g\right) + {H}_{2} O \rightarrow C {l}^{-} + C l {O}^{-} + 2 {H}^{+}$

But we note that basic conditions were specified. How do we cope? We simply add $2 \times H {O}^{-}$ to both sides.

$C {l}_{2} \left(g\right) + + 2 H {O}^{-} + \cancel{{H}_{2} O} \rightarrow C {l}^{-} + C l {O}^{-} + \cancel{2} {H}_{2} O$

Here we have used the equation: ${H}^{+} + H {O}^{-} \rightarrow {H}_{2} O$, and cancelled out the waters.