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

Explanation

(i)
(ii) `Cl_(2) to 2Cl^(-) ("Reduction half"), Cl_(2) to 2ClO_(3)^(-) ("Reduction half")`
(iii) 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^(-)] x x 5`
`Cl_(2)+12OH^(-) to 2ClO_(3)^(-)+10e^(-)+6H_(2)O`
`6Cl_(2)+12OH^(-) to 10Cl^(-) +2ClO_(3)^(-) +6H_(2)O`
`3Cl_(2)+6OH^(-) to 5Cl^(-)+ClO_(3)^(-)+3H_(2)O`