Cr2O2−+SO2→Cr3++SO42−
(in acidic solution)
Oxidation half equation: SO2+2H2O−>SO4+4H++2e−...(i)
Reduction half equation: Cr72−O2−+14H++6e−→2Cr3++7H2O...(ii)
Multiplying eqn. (i) by 3 and adding to eqn. (ii) we get Cr2O72−+3SO2+2H+→2Cr3++3SO42−+H2O