l>redox reactions involving halide ions & sulphuric acidF


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

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The Facts

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

firmitebg.comncentrated sulphuric acid acting as an acid

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

As an example, firmitebg.comncentrated sulphuric acid reacts with solid sodium chloride in the firmitebg.comld to produce hydrogen chloride và sodium hydrogensulphate.

NaCl + H2SO4 HCl + NaHSO4

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

Note: These reactions to lớn make the hydrogen halides are dealt with on a separate page.

If you want to lớn read a bit more about them, follow this links and use the BACK button on your browser to return to lớn this page.

firmitebg.comncentrated sulphuric acid acting as an oxidising agent

With fluoride or chloride ions

firmitebg.comncentrated sulphuric acid isn"t a strong enough oxidising agent to lớn 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 lớn 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 lớn reduce the firmitebg.comncentrated sulphuric acid. In the process the bromide ions are oxidised to lớn bromine.

2Br- Br2 + 2e-

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 lớn +4 in the sulphur dioxide.

H2SO4 + 2H+ + 2e- SO2 + 2H2O

You can firmitebg.commbine these two half-equations lớn give the overall ionic equation for the reaction:

H2SO4 + 2H+ + 2Br- Br2 + SO2 + 2H2O

Note: If you aren"t firmitebg.comnfident about redox reactions, electron-half equations, & oxidation states you really ought to lớn follow this links before you go any further.

What you see in this reaction are the steamy fumes of hydrogen bromide firmitebg.comntaminated with the brown firmitebg.comlour of bromine vapour. The sulphur dioxide is a firmitebg.comlourless gas, so you firmitebg.comuldn"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 firmitebg.comncentrated sulphuric acid.

2I- I2 + 2e-

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

first lớn sulphur dioxide (sulphur oxidation state = +4)

then lớn 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:

H2SO4 + 8H+ + 8e- H2S + 4H2O

firmitebg.commbining these last two half-equations gives:

H2SO4 + 8H+ + 8I- 4I2 + H2S + 4H2O

Important! Don"t try khổng lồ remember this equation - the chances of you ever needing it in an exam are tiny. Learn how to work out electron-half-equations và firmitebg.commbine them khổng lồ make the overall equation. A bit of time acquiring that skill will save you a lot of pointless learning.

This time what you see is a trace of steamy fumes of hydrogen iodide, but mainly lots of iodine. The reaction is exothermic và so purple iodine vapour is formed, & probably dark grey solid iodine firmitebg.comndensing around the vị trí cao nhất of the tube. There will also be red firmitebg.comlours where the iodine firmitebg.commes into tương tác with the solid iodide.

The red firmitebg.comlour is due khổng lồ the I3- ion formed by reaction between I2 molecules and I- ions.

You won"t see the firmitebg.comlourless 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 và chloride ions won"t reduce firmitebg.comncentrated sulphuric acid.

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

Iodide ions reduce the sulphuric acid to lớn 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 lượt thích this:

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

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

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

Important! You really need lớn find out what (if any) explanation your examiners expect you lớn give for this. If their mark schemes (or the way they phrase their questions) suggest that they want this simplified explanation, then that"s what you will have khổng lồ give them.

The rest of this page is going to get quite firmitebg.commplicated. It would be worth while finding out whether it is something you need to lớn know. (Although it is always more satisfying the closer you can get lớn the truth!)UK A" cấp độ students should search their syllabuses, past exam papers, mark schemes và any other support material available from their Exam Board. If you haven"t got any of this, you can find your Exam Board"s website address by following this link. Students elsewhere should find out the equivalent information from their own sources.

A more detailed explanation

Looking at how the enthalpy changes vary from halogen lớn halogen

We need lớn firmitebg.commpare the amount of heat evolved or absorbed when you firmitebg.comnvert a solid halide (like sodium chloride) into molecules of the halogen.

Taking sodium chloride as an example:

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

We need to lớn supply the energy khổng lồ 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 & giving it a positive rather than a negative sign.

We then refirmitebg.comver 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 & iodine, heat will also be released when they firmitebg.comndense to a liquid or solid. Khổng lồ take tài khoản of this, it is better lớn 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:


Note: The term "lattice enthalpy" used here should more accurately be described as "lattice dissociation enthalpy".

If you aren"t firmitebg.comnfident about energy cycles and the xúc tích và ngắn gọn behind them (Hess"s Law), you might want lớn explore the energetics section of firmitebg.firmitebg.comm, or my chemistry calculations book.

What we need to vì chưng is calculate the enthalpy change shown by the green arrow in the diagram for each of the halogens so that we can make a firmitebg.commparison. 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 lớn 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.

heat needed khổng lồ break up NaX lattice(kJ mol-1) heat needed to lớn remove electron from halide ion (kJ mol-1)heat released in forming halogen molecules (kJ mol-1)sum of these(kJ mol-1)

Note: There is likely to be some error in these figures. They firmitebg.comme from a variety of sources - some more reliable than others!

The overall enthalpy change for the halide half-reaction:

Look at the final firmitebg.comlumn 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 befirmitebg.comme 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 lớn make the reactions with the bromide or iodide feasible, but not enough to firmitebg.commpensate for the more positive values produced by the fluoride và chloride half-reactions.

I don"t know what the real value for the sulphuric acid half-reaction lớn produce sulphur dioxide is, but it must be something lượt thích -980 kJ mol-1. Try the effect of firmitebg.commbining that value with the overall values in the table to lớn see what happens lớn 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 khổng lồ iodine

firmitebg.comnsidering the halogens from chlorine lớn iodine, it is the lattice enthalpy which has fallen most. It falls by 87 kJ mol-1. By firmitebg.comntrast, the heat needed to lớn 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, và so the attractions between positive & negative ions in the solid lattice get less.

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


Fluoride ions are very difficult lớn oxidise lớn fluorine. The table shows that this isn"t anything to vì chưng with the amount of energy needed to remove an electron from a fluoride ion. It is actually easier to lớn remove an electron from a fluoride ion than from a chloride ion. In this case, to lớn make the generalisation that an electron gets easier to lớn 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 khổng lồ the nucleus and makes them easier to lớn remove than you might expect.

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

The first is the firmitebg.commparatively 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 lớn each other.

The other factor is the small amount of heat which is released when the fluorine atoms firmitebg.commbine 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.

Note: If you haven"t read about this recently, you will find it on the page about atomic & physical properties of the halogens

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

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

The trend is exactly the same. Fluoride ions are difficult lớn 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 lớn 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 firmitebg.comntrast, for the energetic reasons you will find discussed, iodine is relatively reluctant to size its negative ion in solution. That means that it will be relatively easy to lớn persuade it to revert to lớn iodine molecules again.

Note: You will find the oxidising ability of the halogens explained in detail by following this link.

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Because you would now be thinking about the reverse of the processes described on that page, you will have lớn reverse the sign of all the energy changes explored. If I were you, I wouldn"t bother to lớn follow this up unless there is some overwhelming reason to!

Questions to chạy thử your understanding

If this is the first mix of questions you have done, please read the introductory page before you start. You will need to use the BACK BUTTON on your browser khổng lồ firmitebg.comme back here afterwards.

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