In passing chlorine gas through a concentrated solution of alkali we get chloride and chlorate ions Obtain balanced chemical equation for this reaction.

Explanation

(ii) `Cl_(2) to 2Cl_(1)` (Reducing half)
`Cl_(2) to 2ClO_(3)^(-)` (Oxidation half)
(ii) Balancing `e^(-)`
`Cl_(2) + 2e^(-) to 2Cl^(-) `
`Cl_(2) to 2ClO_(3)^(-) + 10e^(-)`
(iv) Balancing 'O'
`Cl_(2) + 2e^(-) to 2Cl^(-)`
`Cl_(2) + 6H_(2)O to 2ClO_(3) + 10e^(-)`
(v) Balancing 'H'
`Cl_(2)+ 2e^(-) to 2Cl^(-)`
`Cl_(2)+ 6H_(2)O + 12OH^(-) to 2ClO_(3)^(-) + 10e^(-) + 12H_(2)O`
Now,
`[Cl_(2) + 2e^(-) to 2Cl^(-)]xx5`
`ul(Cl_(2)+ 12OH^(-) to 2ClO_(3)^(-) + 10e^(-)+ 6H_(2)O)`
`6Cl_(2) + 12OH^(-) to 10Cl^(-) + 2ClO_(3)^(-)+ 6H_(2)O`
or `3Cl_(2) + 6OH^(-) to 5Cl^(-) + CIO_(3)^(-) + 3H_(2)O`