The element on the reactant side which has been reduced is: HI + H2SO4 ⟶ I2+ SO2+ H2O

Explanation

Option C is correct.The element on the reactant side which has been reduced in the equation is sulfur. HI + H2SO4 → I2 + SO2 + H2OIn the reactant, sulfur has an oxidation state of +6. In the product, sulfur has an oxidation state of +4. This means that sulfur has gained 2 electrons in the reaction, which is the definition of reduction. The oxidation state of an element can be determined by following a set of rules. These rules take into account the electronegativity of the atoms involved in the bonding, as well as the number of electrons that are shared in the bond. In the reactant, sulfur is bonded to oxygen, which is more electronegative than sulfur. This means that sulfur will lose electrons to oxygen, resulting in an oxidation state of +6. In the product, sulfur is bonded to oxygen and hydrogen. Oxygen is more electronegative than sulfur, but hydrogen is less electronegative than sulfur. This means that sulfur will gain electrons from oxygen and hydrogen, resulting in an oxidation state of +4. Therefore, the element on the reactant side which has been reduced is sulfur.