Balance the following reactions by ion electron method `:` `Cl_(2) + OH^(-) rarr ClO_(3)+Cl + H_(2)O`

Explanation

Step I `:` Oxidation `: Cl_(2) rarr ClO_(3)^(-)`
Reduction`: Cl_(2) rarr Cl^(-)`
Step II `:` Oxidation `Cl_(2) rarr 2ClO_(3)^(-)`
Reduction `:` `Cl_(2) rarr2Cl^(-)`
Step III `:`
Oxidation `: Cl_(2) +12OH^(-) rarr 2ClO_(3)^(-) + 6 H_(2) O`
Reduction `:` `Cl_(2) rarr 2Cl^(-)`
Sep IV `:`
Oxidation `:` `Cl_(2) +12OH^(-) rarr 2ClO_(3) + 6 H_(2) O + 10e`
Reduction `:Cl_(2) + 2e rarr 2Cl^(-)`
Step V `:`
Oxidation `:` `Cl_(2) = 12OH^(-) rarr 2ClO_(3)^(-) + 6H_(2)O+10e `
Reduction `: Cl_(2) + 2e rarr 2Cl^(-) ] xx5`
`6Cl_(2) = 12OH^(-) rarr 2ClO_(3)^(-) + 10 Cl ^(-) + 6H_(2) O`
or `3Cl_(2) + 6 OH^(-) rarr ClO_(3) + 5Cl^(-) + 3 H_(2)O` is the balanced reaction.
Step III onwards may be replaced as `:`
Step III `:` Oxidation `: Cl_(2) +6H_(2)O rarr 2ClO_(3) + 12H^(+)`
Reduction `: Cl_(2) rarr 2Cl^(-)`
Step IV `:` Oxidation `: Cl_(2) +6 H_(2) O rarr 2ClO_(3) +12H^(+) + 10e`
Reduction `: Cl_(2) + 2e rarr 2Cl^(-)`
Step V `:` Oxidation `: Cl_(2) + 6 H_(2) rarr 2ClO_(3) ^(-) + 12H ^(+) + 10e`
Reduction `: Cl_(2) + 2e rarr 2eCl^(-) ] xx 5`
`6 Cl_(2) + 12OH^(-) rarr 2eCl^(-) + 10Cl^(-) + 6 H_(2)O`
or , `3Cl_(2) + 3H_(2) O rarr ClO_(3)^(-) + 5Cl^(-) + 6 H^(+)`
To remove `H^(+)` ion, and equal number of `OH^(-)` ions in both sides.
`3Cl_(2) + 3H_(2) O + 6 OH^(-) rarr ClO_(3) + 5Cl^(-) +6OH^(-)`
or, `3Cl_(2) + 3H_(2) O + 6 OH^(-) rarr ClO_(3) + 5Cl+ 6 H_(2) O `
or, `3Cl_(3) + 6OH^(-) rarr ClO_(3)^(-) + 5Cl^(-) +3H_(2)O` is the balanced reaction.