Balancing Redox Reactions. Balance the basic solution (ClO3)- + MnO2 = Cl- + (MnO4)- using half reaction?

To start, I am not going to play the game....just balance it in ACIDIC conditions as normal....

`"Chlorate is reduced to chloride...."`
`ClO_3^(-) +6H^+ +6e^(-)rarr Cl^(-)+3H_2O(l)` `(i)`

`"Manganese dioxide is oxidized to permanganate...."`
`MnO_2(s) +2H_2O(l)rarr MnO_4^(-) + 4H^+ +3e^(-)` `(ii)`

We adds `(i)+2xx(ii)` to retire the electrons....

`2MnO_2(s) +ClO_3^(-) +6H^+ +6e^(-)+4H_2O(l)rarr 2MnO_4^(-) + 8H^+ +6e^(-)+Cl^(-)+3H_2O(l)`

..and then we cancel common reagents....

`2MnO_2(s) +ClO_3^(-) +H_2O(l)rarr 2MnO_4^(-)+2H^+ +Cl^(-)`

...the which is balanced as required. But BASIC conditions were specified, so we simply add `2xxHO^-` TO BOTH SIDES of the equation...

`2MnO_2(s) +ClO_3^(-) +H_2O(l)+2HO^(-)rarr 2MnO_4^(-)+underbrace(2H^+ +2HO^(-))_(2H_2O)+Cl^(-)`

...to give finally....

`underbrace(2MnO_2(s))_"brown solid" +ClO_3^(-) +2HO^(-)rarr underbrace(2MnO_4^(-))_"purple solution"+Cl^(-)+H_2O(l)`

The which is BALANCED with respect to mass and charge.... Whether this would work or not is another matter. I don't have a list of half equations at hand....

Consider,

`ClO_3^(-)+MnO_2 to Cl^(-) + MnO_4`, and its half reactions,

`ClO_3^(-) to Cl^(-)`, and

`MnO_2 to MnO_4^(-)`

In turn,

`6e^(-)+ 6H^(+)+ ClO_3^(-) to Cl^(-) + 3H_2O`, and

`2*[MnO_2+2H_2O to MnO_4^(-)+4H^(+)+3e^(-)]` give,

`ClO_3^(-) + 2MnO_2+H_2O to Cl^(-) + 2H^(+) +2MnO_4^(-)`

which will be neutralized in basic solution to arrive at,

`2OH^(-) + ClO_3^(-) + 2MnO_2 to Cl^(-) + H_2O + 2MnO_4^(-)`