In SO2, the charge of O is 2(-2)= -4 so the charge of S is +4. In H2SO4, the charge of O is 4(-2)= -8, the charge of H is 2(+1)= +2, so the charge of S is +6. Since S +4 becomes S +6, electrons are lost, and this is oxidation. So, SO2 is the substance oxidized, and it is also the reducing agent. The other reactant, HNO3, is the substance reduced, and it is the oxidizing agent. Solution In SO2, the charge of O is 2(-2)= -4 so the charge of S is +4. In H2SO4, the charge of O is 4(-2)= -8, the charge of H is 2(+1)= +2, so the charge of S is +6. Since S +4 becomes S +6, electrons are lost, and this is oxidation. So, SO2 is the substance oxidized, and it is also the reducing agent. The other reactant, HNO3, is the substance reduced, and it is the oxidizing agent..