The reaction to balance is the following:
ClO_3^(-)(aq)+SO_2(g)->SO_4^(2-)(aq)+Cl^(-)(aq)ClO−3(aq)+SO2(g)→SO2−4(aq)+Cl−(aq)
Step 1 : we will first split the reaction into two half equations:
Oxidation: SO_2->SO_4^(2-)SO2→SO2−4
Reduction: ClO_3^(-)->Cl^(-)ClO−3→Cl−
Step 2 : balance the other elements other than oxygen and nitrogen.
They are balanced.
Step 3 : balance oxygen using H_2OH2O:
Oxidation: SO_2+2H_2O->SO_4^(2-)SO2+2H2O→SO2−4
Reduction: ClO_3^(-)->Cl^(-)+3H_2OClO−3→Cl−+3H2O
Step 4 : balance hydrogen using H^(+)H+:
Oxidation: SO_2+2H_2O->SO_4^(2-)+4H^(+)SO2+2H2O→SO2−4+4H+
Reduction: ClO_3^(-)+6H^(+)->Cl^(-)+3H_2OClO−3+6H+→Cl−+3H2O
Step 5 : balance charges using e^(-)e−:
Oxidation: SO_2+2H_2O->SO_4^(2-)+4H^(+)+2e^(-)SO2+2H2O→SO2−4+4H++2e−
Reduction: ClO_3^(-)+6H^(+)+6e^(-)->Cl^(-)+3H_2OClO−3+6H++6e−→Cl−+3H2O
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^(-)))3×(SO2+2H2O→SO2−4+4H++2e−)
Reduction: color(red)(1xx)(ClO_3^(-)+6H^(+)+cancel(6e^(-))->Cl^(-)+3H_2O)1×(ClO−3+6H++6e−→Cl−+3H2O)
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)ClO−3(aq)+3SO2(g)+3H2O(l)→3SO2−4(aq)+Cl−(aq)+6H+(aq)
Step 8 : cancel the H^(+)H+ ions by adding OH^(-)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. **
