For example: CuSO4 + 2NaOH -> Cu(OH)2 + Na2SO4
In this ion-combination reaction Copper (Cu) is in the ionic form Cu2+ on the left hand side of the equation, and as Cu2+ on the right hand side of the equation.
Cu2+ + SO42- + 2Na+ + 2OH- -> (Cu2+ + 2OH-) Solid + 2Na+ + SO42-
What we are showing here, is that the Copper (Cu) has stayed as Cu2+. It has not gained any electrons (nor lost any electrons). This is typical of ion-combination reactions.
In these reactions, the valency (oxidation number) of the reactants change.
For example: 2Fe3+ + Sn2+ -> 2Fe2+ + Sn4+ (8+ each side of the equation)
The iron (iii) + tin (ii) have reacted to give iron (ii) + tin (iv) of course, this reaction is carried out in the presence of HCl (Hydrochloric Acid), but the oxidation reduction reaction is only between the iron (iii) and tin (ii).
Now, a redox reaction is the release and uptake of electrons.
So, the Fe3+ is reduced to Fe2+, and the Sn2+ is oxidised to Sn4+.
What happened in this reaction?
Sn2+ donated electrons to the Fe3+ (an electron transfer took place).
For example: Fe (metal) + Cu2+ -> Fe2+ + Cu (metal)
Fe donates two electrons to the Cu2+ to form Cu (metal). The Fe lost 2 electrons, so is oxidised.
The Cu2+ gained 2 electrons, so is reduced (in its valency).
Confusing isnt it! - Lets try some more...
The chemical which is oxidised is the reducing agent.
For Example: Zn + 2HCl -> Zn2+ + H2 +2Cl-
In this reaction Zn + 2H+ -> Zn2+ + H2 (the chlorine is not changed in its ion state, so is not oxidised or reduced).
Zn is oxidised to Zn2+ (loses 2 electrons)
H+ is reduced to H2 (gains 2 electrons)
H+ has an oxidation number (valency) of +1, and is reduced to an oxidation number (valency) of 0.
So, reduction decreases the oxidation number (valency).
Zn has an oxidation number (valency) of 0, and is oxidised to Zn2+, an oxidation number of 2+,
So oxidation increases the oxidation number (valency).
Not only are there an exchange of electrons in these reactions, but also an exchange of protons (hydronium ions), as in any base system.
CuS + HNO3 -> Cu SO4 + NO (g) + H2O (equation not balanced).
3CuS + 8HNO3 -> 3 CuSO4 + 8NO(g) + 4H2O (equation balanced)
3CuS2+ + 3S2- + 8H+ + 8NO3- -> 3Cu2+ + 3SO42- + 8NO(g) + 4H2O (equation written in ionic nomenclature)
Copper Cu2+ does not change.
Sulphur S2- goes to S6+O4 ((S6+
Nitrogen has gone from N5+ (N5+ O32-)
to N2+ (N2+ O2-)
Sometimes it is easier to see the transfer of electrons in the system if it is split into definite steps. This will be oxidation of one substance and reduction of the other substance.
2Fe3+ + Sn2+ -> 2Fe2+ + Sn4+
Split into 2 separate steps.
2Fe3+ + 2e- -> 2Fe2+ (reduction)
(6+) + (2-) -> (4+) (balanced for charges)
Sn2+ -> Sn4+ + 2e- (oxidation)
(2+) -> (4+) + (2-)
Add the two half equations: 2Fe3+ + 2e- +
Sn2+ -> 2Fe2+ + Sn4+ + 2e-