About compounds of sulfur and oxygen
How sulfur is oxidized
As an example of the formation of compounds of sulfur with oxygen, we may give the reaction of the combustion of chemically pure sulfur in air:
S + O₂ → SO₂
There is also a reaction that demonstrates the transition of SO₂ to SO₃:
2SO₂ + O₂ → 2SO₃.
This reaction takes place in industry in the presence of vanadium oxide, which acts as a catalyst. The second condition is a temperature of +400-500 ⁰С. From sulfur (VI) oxide, sulfuric acid can be obtained:
SO₃ + Н₂О → Н₂SO₄
The combustion reaction of sulfur is oxidizing-reducing i.e. the elements participating in this reaction change their degree of oxidation. We can place the degrees of oxidation in an equation of reaction:
S⁰ + О₂⁰ → S⁺⁴О₂⁻²
For convenience, this reaction can be divided into two: for oxygen and for sulfur. Then it immediately becomes clear where the electrons move to:
О₂⁰ + 4е ̅ → О₂⁻² — oxygen receives electrons, so it is an oxidizer.
S⁰ - 4е ̅ → S⁺⁴ — sulfur gives up its electrons, so it acts as a reducer.
Click here to find out how to obtain sulfur dioxide at home. Obtaining sulfur (VI) oxide is also an oxidizing-reducing reaction. Its equation, taking into account degrees of oxidation, is written as follows:
2S⁺⁴О₂⁻² + О₂⁰ -> 2S⁺⁶О₃⁻².
If we show the degrees of oxidation in the equation given above for obtaining sulfuric acid, it becomes clear that this process is not an oxidizing-reducing reaction, i.e. the transfer of electrons from one atom to another does not take place, and the degrees of oxidation of elements before and after the reaction do not change:
S⁺⁶О₃⁻² + Н⁺₂О⁻² → Н₂⁺S⁺⁶O₄⁻²