Problem 14 - Entrance Test

In the chemical equation 2H₂S(g) + SO₂(g) → 3S(s) + 2H₂O(l), what happens to sulfur in H₂S and SO₂ respectively?

A. H₂S is oxidized, SO₂ is oxidizedB. H₂S is reduced, SO₂ is reducedC. H₂S is oxidized, SO₂ is reducedD. H₂S is reduced, SO₂ is oxidized

Correct: C

Let's determine the oxidation states of sulfur in each compound: - In H₂S: Hydrogen (H) is +1. Since there are two H atoms, 2(+1) + S = 0, so S = -2. - In SO₂: Oxygen (O) is -2. Since there are two O atoms, S + 2(-2) = 0, so S = +4. - In S(s) (elemental sulfur): S = 0. Comparing the changes: - For H₂S: Sulfur goes from -2 to 0. This is an increase in oxidation number, which means H₂S is oxidized. - For SO₂: Sulfur goes from +4 to 0. This is a decrease in oxidation number, which means SO₂ is reduced. Therefore, H₂S is oxidized, and SO₂ is reduced.