How does chlorine gas give $C l^{-}$ $Cl^(-)$ and $C l O_{3}^{-}$ $ClO_3^(-)$ upon basic hydrolysis?

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Answer 1 · 2016-10-18T15:09:11

This is a disproportionation reaction.

$C l_{2} (g) + + 2 H O^{-} \to C l^{-} + C l O^{-} + H_{2} O$ $Cl_2(g) + +2HO^(-) rarr Cl^(-)+ClO^(-) +H_2O$

Chlorine is simultaneously reduced and oxidized.

$C l_{2} (g) + 2 H O^{-} \to C l^{-} + C l O^{-} + H_{2} O$ $Cl_2(g) + 2HO^(-) rarr Cl^(-) + ClO^(-) + H_2O$

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

$OXIDATION:$ $"OXIDATION:"$ $\frac{1}{2} C l_{2} (g) + H_{2} O \to C l O^{-} + 2 H^{+} + e^{-}$ $1/2Cl_2(g) + H_2O rarr ClO^(-) +2H^(+) + e^(-)$

$REDUCTION:$ $"REDUCTION:"$ $\frac{1}{2} C l_{2} (g) + e^{-} \to C l^{-}$ $1/2Cl_2(g) + e^(-) rarr Cl^(-)$

$OVERALL:$ $"OVERALL:"$ $C l_{2} (g) + H_{2} O \to C l^{-} + C l O^{-} + 2 H^{+}$ $Cl_2(g) + H_2O rarr Cl^(-)+ClO^(-) +2H^(+)$

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

How does chlorine gas give Cl^(-)Cl−Cl^(-) and ClO_3^(-)ClO−3ClO_3^(-) upon basic hydrolysis?