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^(-)#