About compounds of sulfur and oxygen

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