Chemguide: Support for CIE A level Chemistry

This statement is about the changes in the relative stability of the oxidation states of the Group 4 elements as you go down the group.

Before you go on, you should find and read the statement in your copy of the syllabus.

For some simple descriptive stuff, I am going to refer you to a Chemguide page, and then look at what CIE wants in terms of explanations afterwards on this page.

Straightforward descriptive stuff

Read the beginning of the page oxidation state trends in Group 4.

This will give you virtually everything suggested in the teacher support material.

Do not read the second part of this page dealing with the explanations. These aren't the explanations which CIE want, and you will just get confused.

Explanations in terms of E° values

You can only really apply this to reactions involving tin or lead ions. The only relevant E° values quoted in the Data Booklet that you will have in an exam (and which you will find towards the end of the syllabus) are:

Remember that the E° values give a measure of the position of the various equilibria. The more positive the E° value, the further it lies to the right.

You will see that the equilibria involving lead(IV) are further to the right than the one involving tin(IV). That means that the lead(IV) ions or compound are more likely to pick up electrons and go to the the lead(II) side.

Picking up electrons happens when a compound oxidises something else. That means that lead(IV) compounds are better oxidising agents than tin(IV) compounds.

Why does this happen? Because for lead, the +2 oxidation state is more stable relative to the +4 state than tin(II) is to tin(IV).

Let's look at a couple of examples of the sort you might be asked about in an exam.

Why does lead(IV) oxide oxidise concentrated hydrochloric acid to chlorine?

You may remember that if you treat lead(IV) oxide with very cold concentrated hydrochloric acid, you get some lead(IV) chloride formed. You may also remember that if you warm this, the lead(IV) chloride decomposes to give lead(II) chloride and chlorine.

If you reacted lead(IV) oxide with hot concentrated hydrochloric acid, you could think about this in another way. The lead(IV) oxide is a strong enough oxidising agent to oxidise the hydrochloric acid to chlorine.

Think about the E° values:

Where you couple two of these equilibria together in a test tube, the more positive one moves to the right, and the less positive one moves to the left.

In this case, starting from the chloride ions in the hydrochloric acid, and the lead(IV) oxide (and the hydrogen ions from the acid), the lead equilibrium moves to the right, and the chlorine one to the left.

Chlorine gas is formed, and a lead(II) compound.

E° values apply only when the ion concentrations are 1 mol dm^-3. In the case of concentrated hydrochloric acid, they will be more like 10 mol dm^-3.

Increasing the concentration of something on one side of an equilibrium will move the position of equilibrium towards the other side.

The effect of tin or lead ions on the interconversion of iron(II) and iron(III)

Look at these E° values:

Iron(III) ions and tin(II) ions

What would be the effect of mixing solutions containing tin(II) ions and iron(III) ions? Try to work this out before you read on.

The iron has a greater tendency to move to the right than the tin one, because the E° is more positive. That means that the iron(III) ions will gain electrons (be reduced) to iron(II) ions. The tin(II) ions will lose electrons (be oxidised) to tin(IV) ions.

In other words, tin(II) ions are strong enough reducing agents to reduce iron(III) ions to iron(II) ions.

Iron(III) ions and lead(II) ions

Now suppose you tried a similar thing by mixing solutions containing lead(II) ions and iron(III) ions.

This time nothing could happen. The more positive E° value belongs to the lead equilibrium. That one is the one that would have to move to the right. But if you start from lead(II) ions, it is already as far to the right as it can go.

Any reaction would involve the equilibrium moving to the left - but that is impossible according to the E° values. The more positive one has to move to the right; the less positive one to the left.

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