Question

Oxidation state of S in H2S03 and H2S08

Answer 1

In H₂SO₃ (sulfurous acid), the oxidation state of sulfur (S) is +4. This can be determined by considering the oxidation states of hydrogen (H) and oxygen (O) in the compound. Hydrogen is generally assigned an oxidation state of +1 in compounds, and oxygen is typically assigned an oxidation state of -2. Since the compound is neutral, the sum of the oxidation states must be zero. Therefore, we can calculate the oxidation state of sulfur as follows:
2(oxidation state of H) + oxidation state of S + 3(oxidation state of O) = 0 2(+1) + oxidation state of S + 3(-2) = 0 2 + oxidation state of S - 6 = 0 oxidation state of S - 4 = 0 oxidation state of S = +4
In H₂SO₈ (sulfuric acid), the oxidation state of sulfur (S) is +6. Following a similar reasoning as before, we can assign the oxidation states:
2(oxidation state of H) + oxidation state of S + 4(oxidation state of O) = 0 2(+1) + oxidation state of S + 4(-2) = 0 2 + oxidation state of S - 8 = 0 oxidation state of S - 6 = 0 oxidation state of S = +6
Therefore, the oxidation state of sulfur in H₂SO₃ (sulfurous acid) is +4, and the oxidation state of sulfur in H₂SO₈ (sulfuric acid) is +6.