THE REDOX REACTIONS BETWEEN HALIDE IONS AND CONCENTRATED SULPHURIC ACID

This page describes and explains the redox reactions involving halide ions and concentrated sulphuric acid. It uses these reactions to discuss the trend in reducing ability of the ions as you go from fluoride to chloride to bromide to iodide.

The Facts

There are two different types of reaction which might go on when concentrated sulphuric acid is added to a solid ionic halide like sodium fluoride, chloride, bromide or iodide. The concentrated sulphuric acid can act both as an acid and as an oxidising agent.

Concentrated sulphuric acid acting as an acid

The concentrated sulphuric acid gives a hydrogen ion to the halide ion to produce a hydrogen halide. Because this is a gas, it immediately escapes from the system. If the hydrogen halide is exposed to moist air, you see it as steamy fumes.

As an example, concentrated sulphuric acid reacts with solid sodium chloride in the cold to produce hydrogen chloride and sodium hydrogensulphate.

All of the halide ions (fluoride, chloride, bromide and iodide) behave similarly.

Concentrated sulphuric acid acting as an oxidising agent

With fluoride or chloride ions

Concentrated sulphuric acid isn't a strong enough oxidising agent to oxidise fluoride or chloride ions. In those cases, all you get produced are the steamy fumes of the hydrogen halide - hydrogen fluoride or hydrogen chloride.

You can look at this another way - from the point of view of the halide ions. The fluoride and chloride ions aren't strong enough reducing agents to reduce the sulphuric acid.

Whichever way you look at it, all you get is the hydrogen halide!

That isn't true, though, with bromides and iodides.

With bromide ions

The bromide ions are strong enough reducing agents to reduce the concentrated sulphuric acid. In the process the bromide ions are oxidised to bromine.

The bromide ions reduce the sulphuric acid to sulphur dioxide gas. This is a decrease of oxidation state of the sulphur from +6 in the sulphuric acid to +4 in the sulphur dioxide.

You can combine these two half-equations to give the overall ionic equation for the reaction:

What you see in this reaction are the steamy fumes of hydrogen bromide contaminated with the brown colour of bromine vapour. The sulphur dioxide is a colourless gas, so you couldn't observe its presence directly.

With iodide ions

Iodide ions are stronger reducing agents than bromide ions are. They are oxidised to iodine by the concentrated sulphuric acid.

The reduction of the sulphuric acid is more complicated than before. The iodide ions are powerful enough reducing agents to reduce it

• first to sulphur dioxide (sulphur oxidation state = +4)

• then to sulphur itself (oxidation state = 0)

• and all the way to hydrogen sulphide (sulphur oxidation state = -2).

The most important of this mixture of reduction products is probably the hydrogen sulphide. The half-equation for its formation is:

Combining these last two half-equations gives:

This time what you see is a trace of steamy fumes of hydrogen iodide, but mainly lots of iodine. The reaction is exothermic and so purple iodine vapour is formed, and probably dark grey solid iodine condensing around the top of the tube. There will also be red colours where the iodine comes into contact with the solid iodide.

The red colour is due to the I₃^- ion formed by reaction between I₂ molecules and I^- ions.

You won't see the colourless hydrogen sulphide gas, but might pick up its "bad egg" smell if you were foolish enough to smell the intensely poisonous gases evolved!

Summary of the trend in reducing ability

• Fluoride and chloride ions won't reduce concentrated sulphuric acid.

• Bromide ions reduce the sulphuric acid to sulphur dioxide. In the process, the bromide ions are oxidised to bromine.

• Iodide ions reduce the sulphuric acid to a mixture of products including hydrogen sulphide. The iodide ions are oxidised to iodine.

• Reducing ability of the halide ions increases as you go down the Group.

Explaining the trend

An over-simplified explanation

This only works (and even then, not very well!) if you ignore fluoride ions. The argument goes like this:

When a halide ion acts as a reducing agent, it gives electrons to something else. That means that the halide ion itself has to lose electrons.

The bigger the halide ion, the further the outer electrons are from the nucleus, and the more they are screened from it by inner electrons. It therefore gets easier for the halide ions to lose electrons as you go down the Group because there is less attraction between the outer electrons and the nucleus.

It sounds convincing, but it only tells part of the story. We need to look in some detail at the energetics of the change.

A more detailed explanation

Looking at how the enthalpy changes vary from halogen to halogen

We need to compare the amount of heat evolved or absorbed when you convert a solid halide (like sodium chloride) into molecules of the halogen.

Taking sodium chloride as an example:

• We need to supply the energy to break the attractions between the ions in the sodium chloride. In other words, we need to supply the lattice enthalpy.

• We need to supply the energy to remove the electron from the chloride ion. This is the reverse of the electron affinity of the chlorine. You can get this figure by looking up the electron affinity in a Data Book and giving it a positive rather than a negative sign.

• We then recover some energy when the chlorine atoms turn into chlorine molecules. Energy is released when the bonds are formed.

  Chlorine is simple because it is a gas. With bromine and iodine, heat will also be released when they condense to a liquid or solid. To take account of this, it is better to think of this in terms of atomisation energy rather than bond energy. The number we want is the reverse of atomisation energy.

  Atomisation energy is the energy needed to produce 1 mole of isolated gaseous atoms starting from an element in its standard state (gas for chlorine, and liquid for bromine, for example - both of them as X₂).

Look carefully at the diagram so that you see how this all fits together:

What we need to do is calculate the enthalpy change shown by the green arrow in the diagram for each of the halogens so that we can make a comparison. The diagram shows that the overall change involving the halide ions is endothermic - the green arrow is pointing upwards towards a higher energy.

This isn't the total enthalpy change for the whole reaction. Heat will be given out when the changes involving the sulphuric acid occur. That will be the same irrespective of which halogen you are talking about. The total enthalpy change will be the sum of the enthalpy changes for the halide ion half-reaction and the sulphuric acid half-reaction.

The table shows the energy changes which vary from halogen to halogen. We are assuming that you start from solid sodium halide. The values for the lattice enthalpies for other solid halides would be different, but the pattern will still be the same.

The overall enthalpy change for the halide half-reaction:

Look at the final column of figures.

Notice that the sum of these enthalpy changes gets less endothermic as you go down the Group. That means that the total change (including the sulphuric acid) will become easier as you go down the Group.

The amount of heat given out by the half-reaction involving the sulphuric acid must be great enough to make the reactions with the bromide or iodide feasible, but not enough to compensate for the more positive values produced by the fluoride and chloride half-reactions.

I don't know what the real value for the sulphuric acid half-reaction to produce sulphur dioxide is, but it must be something like -980 kJ mol^-1. Try the effect of combining that value with the overall values in the table to see what happens to the total enthalpy change of reaction for each halogen.

Exploring the changes in the various energy terms

Which individual energy terms in the table are most important in making the halogen half-reaction less endothermic as you go down the Group?

Chlorine to iodine

Considering the halogens from chlorine to iodine, it is the lattice enthalpy which has fallen most. It falls by 87 kJ mol^-1. By contrast, the heat needed to remove the electron has only fallen by 54 kJ mol^-1.

Both of these terms matter, but the fall in lattice enthalpy is the more important. This falls because the ions are getting bigger. That means that they aren't as close to each other, and so the attractions between positive and negative ions in the solid lattice get less.

The simplified explanation that we mentioned earlier concentrates on the less important fall in the amount of energy needed to remove the electron from the ion. That's misleading!

Fluorine

Fluoride ions are very difficult to oxidise to fluorine. The table shows that this isn't anything to do with the amount of energy needed to remove an electron from a fluoride ion. It is actually easier to remove an electron from a fluoride ion than from a chloride ion. In this case, to make the generalisation that an electron gets easier to remove as the ion gets bigger is just plain wrong!

Fluoride ions are so small that the electrons feel an abnormal amount of repulsion from each other. This outweighs the effect of their closeness to the nucleus and makes them easier to remove than you might expect.

There are two important reasons why fluoride ions are so difficult to oxidise.

The first is the comparatively very high lattice enthalpy of the solid fluoride. This is due to the small size of the fluoride ion, which means that the positive and negative ions are very close together and so strongly attracted to each other.

The other factor is the small amount of heat which is released when the fluorine atoms combine to make fluorine molecules. (Scroll back and look at the table again.)

This is because of the low bond enthalpy of the F-F bond. The reason for this low bond enthalpy is discusssed on a separate page.

What if the halide ions were in solution rather than in a solid?

We have concentrated on the energetics of the process starting from solid halide ions because that's what you use if you try to oxidise them using concentrated sulphuric acid. What about oxidising them in solution using some different oxidising agent?

The trend is exactly the same. Fluoride ions are difficult to oxidise and it gets easier as you go down the Group towards iodide ions. Looked at another way, fluoride ions aren't good reducing agents, but iodide ions are.

The explanation this time has to start from the hydrated ions in solution rather than solid ions. In a sense, this has already been done on another page.

Fluorine is a very powerful oxidising agent because it very readily forms its negative ion in solution. That means that it will be energetically difficult to reverse the process.

By contrast, for the energetic reasons you will find discussed, iodine is relatively reluctant to form its negative ion in solution. That means that it will be relatively easy to persuade it to revert to iodine molecules again.

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