Balance the following equation with ion electron method:
$C{l}_2+O{H}^{-}\to C{l}^{-}+ClO{}^{-}{}_3+H_{2}O$

In Ion electron method also called half-reaction method, the reduction (R) and oxidation (O)reactions are balanced individually and thereafter added to get balanced equation. For the reaction:

${Cl}_2+H{O}^{-}\to {Cl}^{-}+Cl{O}_3^{-}+H_{2}O$

1. Determine the oxidation state of each specie and separate reduction nad oxidation reaction:

$\stackrel{0}{{Cl}_2}+H\stackrel{-2}{O^{-}}\to \stackrel{-1}{{Cl}^{-}}+\stackrel{+5}{Cl}{O}_3^{-}+H_2\stackrel{-2}{O}$

Therefore Reduction:

$Red:{Cl}_2\to {Cl}^{-}$

And oxidation: ${Cl}_2\to Cl{O}_3^{-}$

2. Balance reacting atoms both side:

${Cl}_2\to 2{Cl}^{-}$

3. Balance charge by adding electrons

${Cl}_2+2e^{-}\to 2{Cl}^{-}$

Similarly, balance oxidation process as:

$Oxd:{Cl}_2\to Cl{O}_3^{-}$

1. Balance $Cl$ atom

${Cl}_2\to 2Cl{O}_3^{-}$

2. Balance $O$ by adding water and thereafter balance $H$ by adding $H^{+}$

${Cl}_2+6H_{2}O\to 2Cl{O}_3^{-}+12H^{+}$

3.Since reaction occurs in basic medium, add equal number of $O{H}^{-}$ as that of $H^{+}$ both sides:

${Cl}_2+6H_{2}O+12H{O}^{-}\to 2Cl{O}_3^{-}+12H^{+}+12H{O}^{-}$

4.Combine $O{H}^{-}$ and $H^{+}$ to $H_{2}O$ and write net water molecules:

${Cl}_2+6H_{2}O+12H{O}^{-}\to 2Cl{O}_3^{-}+12H_{2}O{Cl}_2+12H{O}^{-}\to 2Cl{O}_3^{-}+6H_{2}O$

5. Balance charge by adding electrons:

${Cl}_2+12H{O}^{-}\to 2Cl{O}_3^{-}+6H_{2}O+10e^{-}$

Finally add the balance reduction and oxidation reactions

$5{Cl}_2+{Cl}_2+12H{O}^{-}\to 2{Cl}^{-}+2Cl{O}_3^{-}+6H_{2}O6{Cl}_2+12H{O}^{-}\to 10{Cl}^{-}+2Cl{O}_3^{-}+6H_{2}O$

Balanced redox reaction is therefore:

$3{Cl}_2+6H{O}^{-}\to 5{Cl}^{-}+Cl{O}_3^{-}+3H_{2}O$