2.1.5 (d) What is Redox?

Syllabus

Redox reactions

(d) oxidation and reduction in terms of:

(i) electron transfer

(ii) changes in oxidation number

Not all reactions involve Oxidation or Reduction.

But any reaction that does involve Reduction also involves Oxidation - hence Redox.

At GCSE you almost certainly learned the acronym OILRIG.

This is about transfer of electrons.

Any substance gaining electrons is being Reduced.

But it must gain them from another substance which is clearly losing them and so being Oxidised.

If we look at some simple oxidation:

Cu_(s) + ^1/₂ O_2(g) → CuO_(s)

It is obvious that the Copper is being Oxidised because it is gaining Oxygen. But where are the electrons?

Cu_(s) + S_(l) → CuS_(s)

And in this Oxidation there isn't even any Oxygen or Hydrogen!

We could split up the equation to investigate the electron transfers.

We know that Copper tends to form Copper(II) ions.

So, Cu → Cu²⁺ + 2 e^- Where e^- represents an electron.

It is now obvious that the Copper atom has lost electrons to become a Cu²⁺ion and has been Oxidised.

Equally, we know that Sulphur tends to form Sulphide ions.

So, S + 2 e^- → S^2-

It is now obvious that the Sulphur atom has gained electrons to become an S^2-ion and has been Reduced.

Going back to our problem:

Cu_(s) + S_(l) → CuS_(s)

Since we know that this is Copper (II) Sulphide we know that the Oxidation number of Copper must have gone from 0 to +2

And the Oxidation number of Sulphur must have gone from 0 to -2

Increasing Oxidation Number is Oxidation

decreasing Oxidation Number is Reduction.

Eg. Calcium and Water

Ca_(s) + 2 H₂O_(l) → Ca(OH)_2(s)+ H_2(g)

Ox No. 0 H=+1, O=-2 Ca=+2, H=+1,O=-2 H=0

So Calcium has been Oxidised - increased Oxidation Number

Hydrogen has been reduced Oxidation number has been reduced from +1 to 0