J.R. S. answered 11/21/22
Ph.D. University Professor with 10+ years Tutoring Experience
CIO3-(aq) + C2O42-(aq) → CIO2(g) + CO2(g)
ClO3- ==> ClO2 ... reduction half reaction (Cl goes from 5+ t0 4+)
ClO3- ==> ClO2 + H2O ... balanced for Cl and O
ClO3- + 2H2O ==> ClO2 + H2O + 2OH- ... balanced for Cl, O and H (using base, OH-)
ClO3- + 2H2O + e- ==> ClO2 + H2O + 2OH- ... balanced for Cl, O, H and charge = balanced reduction rxn
C2O42- ==> CO2 ... oxidation half reaction (C goes from 3+ to 4+)
C2O42- ==> 2CO2 ... balanced for C and O
C2O42- ==> 2CO2 + 2e- ... balanced for C, O and charge = balanced oxidation reaction
Multiply reduction reaction by 2 to equalize electrons and then add the two equations:
2ClO3- + 4H2O + 2e- ==> 2ClO2 + 2H2O + 4OH-
C2O42- ==> 2CO2 + 2e-
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2ClO3- + 4H2O + 2e- + C2O42- ==> 2ClO2 + 2H2O + 4OH- + 2CO2 + 2e-
Combine/cancel like terms to end up with final balanced redox equation:
2ClO3- + 2H2O + C2O42- ==> 2ClO2 + 4OH- + 2CO2 ... BALANCED REDOX EQUATION