In each of the following examples, we have to decide whether the reaction involves redox, and if so what has been oxidised and what reduced.
Example 1:
This is the reaction between magnesium and hydrochloric acid or hydrogen chloride gas:
Mg + 2HCl MgCl2 + H2
Have the oxidation states of anything changed? Yes they have - you have two elements which are in compounds on one side of the equation and as uncombined elements on the other. Check all the oxidation states to be sure:.
The magnesium's oxidation state has increased - it has been oxidised. The hydrogen's oxidation state has fallen - it has been reduced. The chlorine is in the same oxidation state on both sides of the equation - it hasn't been oxidised or reduced.
Example 2:
The reaction between sodium hydroxide and hydrochloric acid is:
NaOH + HCl NaCl + H2O
Checking all the oxidation states:
Nothing has changed. This isn't a redox reaction.
Example 3:
This is a sneaky one! The reaction between chlorine and cold dilute sodium hydroxide solution is:
2NaOH + Cl2 NaCl + NaClO + H2O
Obviously the chlorine has changed oxidation state because it has ended up in compounds starting from the original element. Checking all the oxidation states shows:
The chlorine is the only thing to have changed oxidation state. Has it been oxidised or reduced? Yes! Both! One atom has been reduced because its oxidation state has fallen. The other has been oxidised.
This is a good example of a disproportionation reaction. A disproportionation reaction is one in which a single substance is both oxidised and reduced.
Using oxidation states to identify the oxidising and reducing agent
This is just a minor addition to the last section. If you know what has been oxidised and what has been reduced, then you can easily work out what the oxidising agent and reducing agent are.
Example 1
This is the reaction between chromium(III) ions and zinc metal:
2Cr3+ + Zn 2Cr2+ + Zn2+
The chromium has gone from the +3 to the +2 oxidation state, and so has been reduced. The zinc has gone from the zero oxidation state in the element to +2. It has been oxidised.
So what is doing the reducing? It is the zinc - the zinc is giving electrons to the chromium (III) ions. So zinc is the reducing agent.
Similarly, you can work out that the oxidising agent has to be the chromium(III) ions, because they are taking electrons from the zinc.
Example 2
This is the equation for the reaction between manganate(VII) ions and iron(II) ions under acidic conditions. This is worked out further down the page.
MnO4- + 8H+ + 5Fe2+ Mn2+ + 4H2O + 5Fe3+
Looking at it quickly, it is obvious that the iron(II) ions have been oxidised to iron(III) ions. They have each lost an electron, and their oxidation state has increased from +2 to +3.
The hydrogen is still in its +1 oxidation state before and after the reaction, but the manganate(VII) ions have clearly changed. If you work out the oxidation state of the manganese, it has fallen from +7 to +2 - a reduction.
So the iron(II) ions have been oxidised, and the manganate(VII) ions reduced.
What has reduced the manganate(VII) ions - clearly it is the iron(II) ions. Iron is the only other thing that has a changed oxidation state. So the iron(II) ions are the reducing agent.
Similarly, the manganate(VII) ions must be the oxidising agent.
Using oxidation states to work out reacting proportions
This is sometimes useful where you have to work out reacting proportions for use in titration reactions where you don't have enough information to work out the complete ionic equation.
Remember that each time an oxidation state changes by one unit, one electron has been transferred. If one substance's oxidation state in a reaction falls by 2, that means that it has gained 2 electrons.
Something else in the reaction must be losing those electrons. Any oxidation state fall by one substance must be accompanied by an equal oxidation state increase by something else.
This example is based on information in an old AQA A' level question.
Ions containing cerium in the +4 oxidation state are oxidising agents. (They are more complicated than just Ce4+.) They can oxidise ions containing molybdenum from the +2 to the +6 oxidation state (from Mo2+ to MoO42-). In the process the cerium is reduced to the +3 oxidation state (Ce3+). What are the reacting proportions?
The oxidation state of the molybdenum is increasing by 4. That means that the oxidation state of the cerium must fall by 4 to compensate.
But the oxidation state of the cerium in each of its ions only falls from +4 to +3 - a fall of 1. So there must obviously be 4 cerium ions involved for each molybdenum ion.
The reacting proportions are 4 cerium-containing ions to 1 molybdenum ion.
Or to take a more common example involving iron(II) ions and manganate(VII) ions . . .
A solution of potassium manganate(VII), KMnO4, acidified with dilute sulphuric acid oxidises iron(II) ions to iron(III) ions. In the process, the manganate(VII) ions are reduced to manganese(II) ions. Use oxidation states to work out the equation for the reaction.
The oxidation state of the manganese in the manganate(VII) ion is +7. The name tells you that, but work it out again just for the practice!
In going to manganese(II) ions, the oxidation state of manganese has fallen by 5. Every iron(II) ion that reacts, increases its oxidation state by 1. That means that there must be five iron(II) ions reacting for every one manganate(VII) ion.
The left-hand side of the equation will therefore be:
MnO4- + 5Fe2+ + ?
The right-hand side will be:
Mn2+ + 5Fe3+ + ?
After that you will have to make guesses as to how to balance the remaining atoms and the charges. In this case, for example, it is quite likely that the oxygen will end up in water. That means that you need some hydrogen from somewhere.
That isn't a problem because you have the reaction in acid solution, so the hydrogens could well come from hydrogen ions.