The reaction to balance is the following:
#ClO_3^(-)(aq)+SO_2(g)->SO_4^(2-)(aq)+Cl^(-)(aq)#
Step 1 : we will first split the reaction into two half equations:
Oxidation: #SO_2->SO_4^(2-)#
Reduction: #ClO_3^(-)->Cl^(-)#
Step 2 : balance the other elements other than oxygen and nitrogen.
They are balanced.
Step 3 : balance oxygen using #H_2O#:
Oxidation: #SO_2+2H_2O->SO_4^(2-)#
Reduction: #ClO_3^(-)->Cl^(-)+3H_2O#
Step 4 : balance hydrogen using #H^(+)#:
Oxidation: #SO_2+2H_2O->SO_4^(2-)+4H^(+)#
Reduction: #ClO_3^(-)+6H^(+)->Cl^(-)+3H_2O#
Step 5 : balance charges using #e^(-)#:
Oxidation: #SO_2+2H_2O->SO_4^(2-)+4H^(+)+2e^(-)#
Reduction: #ClO_3^(-)+6H^(+)+6e^(-)->Cl^(-)+3H_2O#
Step 7 : sum the two half equations while cancelling the number of electrons by multiplying each half equation by the corresponding integer:
Oxidation: #color(red)(3xx)(SO_2+2H_2O->SO_4^(2-)+4H^(+)+cancel(2e^(-)))#
Reduction: #color(red)(1xx)(ClO_3^(-)+6H^(+)+cancel(6e^(-))->Cl^(-)+3H_2O)#
RedOx: #ClO_3^(-)(aq)+color(red)(3)SO_2(g)+color(red)(3)H_2O(l)->color(red)(3)SO_4^(2-)(aq)+Cl^(-)(aq)+color(red)(6)H^(+)(aq)#
Step 8 : cancel the #H^(+)# ions by adding #OH^(-)# to each side:
RedOx: #ClO_3^(-)(aq)+color(red)(3)SO_2(g)+color(red)(3)H_2O(l)+color(blue)(6)OH^(-)(aq)->color(red)(3)SO_4^(2-)(aq)+Cl^(-)(aq)+underbrace(color(red)(6)H^(+)(aq)+color(blue)(6)OH^(-)(aq))_(6H_2O)#
Step 9 : FINALLY, cancel the #H_2O# molecules on both sides:
RedOx: #ClO_3^(-)(aq)+color(red)(3)SO_2(g)+color(blue)(6)OH^(-)(aq)->color(red)(3)SO_4^(2-)(aq)+Cl^(-)(aq)+color(red)(3)H_2O(l)#
Here is a video that explains this topic in details:
Balancing Redox Reactions | Basic Medium.
