INORGANIC CHEMISTRY
DEFINITIONS OF OXIDATION AND REDUCTION (REDOX)
In this section various definitions of oxidation and reduction (redox) in terms of the transfer of oxygen, hydrogen and electrons will be looked at. It will also explain the terms oxidising agent and reducing agent.
Oxidation and reduction in terms of oxygen transfer
Definitions
- Oxidation is gain of oxygen.
- Reduction is loss of oxygen.
For example, in the extraction of iron from its ore:
Because both reduction and oxi
dation are going on side-by-side, this is known as a redox reaction.
Oxidising and reducing agents
An oxidising agent is a substance which oxidises something else. In the above example, the iron(III) oxide is the oxidising agent.
A reducing agent reduces something else. In the equation, the carbon monoxide is the reducing agent.
· Oxidising agents give oxygen to another substance.
· Reducing agents remove oxygen from another substance.
Oxidation and reduction in terms of hydrogen transfer
These are old definitions which aren't used very much nowadays. The most likely place you will come across them is in organic chemistry.
Definitions
· Oxidation is loss of hydrogen.
· Reduction is gain of hydrogen.
Notice that these are exactly the opposite of the oxygen definitions.
For example, ethanol can be oxidised to ethanal:
You would need to use an oxidising agent to remove the hydrogen from the ethanol. A commonly used oxidising agent is potassium dichromate(VI) solution acidified with dilute sulphuric acid.
Ethanal can also be reduced back to ethanol again by adding hydrogen to it. A possible reducing agent is sodium tetrahydridoborate, NaBH4. Again the equation is too complicated to be worth bothering about at this point.
An update on oxidising and reducing agents
· Oxidising agents give oxygen to another substance or remove hydrogen from it.
· Reducing agents remove oxygen from another substance or give hydrogen to it.
Oxidation and reduction in terms of electron transfer
This is easily the most important use of the terms oxidation and reduction at A' level.
Definitions
- Oxidation is loss of electrons.
- Reduction is gain of electrons.
It is essential that you remember these definitions. There is a very easy way to do this. As long as you remember that you are talking about electron transfer:
A simple example
The equation shows a simple redox reaction which can obviously be described in terms of oxygen transfer.
Copper(II) oxide and magnesium oxide are both ionic. The metals obviously aren't. If you rewrite this as an ionic equation, it turns out that the oxide ions are spectator ions and you are left with:
A last comment on oxidising and reducing agents
If you look at the equation above, the magnesium is reducing the copper(II) ions by giving them electrons to neutralise the charge. Magnesium is a reducing agent.
Looking at it the other way round, the copper(II) ions are removing electrons from the magnesium to create the magnesium ions. The copper(II) ions are acting as an oxidising agent.
WRITING IONIC EQUATIONS FOR REDOX REACTIONS
This page explains how to work out electron-half-reactions for oxidation and reduction processes, and then how to combine them to give the overall ionic equation for a redox reaction. This is an important skill in inorganic chemistry.
Don't worry if it seems to take you a long time in the early stages. It is a fairly slow process even with experience. Take your time and practice as much as you can.
Electron-half-equations
What is an electron-half-equation?
When magnesium reduces hot copper(II) oxide to copper, the ionic equation for the reaction is:
You can split the ionic equation into two parts, and look at it from the point of view of the magnesium and of the copper(II) ions separately. This shows clearly that the magnesium has lost two electrons, and the copper(II) ions have gained them.
These two equations are described as "electron-half-equations" or "half-equations" or "ionic-half-equations" or "half-reactions" - lots of variations all meaning exactly the same thing!
Any redox reaction is made up of two half-reactions: in one of them electrons are being lost (an oxidation process) and in the other one those electrons are being gained (a reduction process).
Working out electron-half-equations and using them to build ionic equations
In the example above, we've got at the electron-half-equations by starting from the ionic equation and extracting the individual half-reactions from it. That's doing everything entirely the wrong way round!
In reality, you almost always start from the electron-half-equations and use them to build the ionic equation.
Example 1: The reaction between chlorine and iron(II) ions
Chlorine gas oxidises iron(II) ions to iron(III) ions. In the process, the chlorine is reduced to chloride ions.
You would have to know this, or be told it by an examiner. In building equations, there is quite a lot that you can work out as you go along, but you have to have somewhere to start from!
You start by writing down what you know for each of the half-reactions. In the chlorine case, you know that chlorine (as molecules) turns into chloride ions:
The first thing to do is to balance the atoms that you have got as far as you possibly can:
ALWAYS check that you have the existing atoms balanced before you do anything else. If you forget to do this, everything else that you do afterwards is a complete waste of time!
Now you have to add things to the half-equation in order to make it balance completely.
All you are allowed to add are:
- electrons
- water
- hydrogen ions (unless the reaction is being done under alkaline conditions - in which case, you can add hydroxide ions instead)
In the chlorine case, all that is wrong with the existing equation that we've produced so far is that the charges don't balance. The left-hand side of the equation has no charge, but the right-hand side carries 2 negative charges.
That's easily put right by adding two electrons to the left-hand side. The final version of the half-reaction is:
Now you repeat this for the iron(II) ions. You know (or are told) that they are oxidised to iron(III) ions. Write this down:
The atoms balance, but the charges don't. There are 3 positive charges on the right-hand side, but only 2 on the left.
You need to reduce the number of positive charges on the right-hand side. That's easily done by adding an electron to that side:
Combining the half-reactions to make the ionic equation for the reaction
What we've got at the moment is this:
It is obvious that the iron reaction will have to happen twice for every chlorine molecule that reacts. Allow for that, and then add the two half-equations together.
But don't stop there!! Check that everything balances - atoms and charges. It is very easy to make small mistakes, especially if you are trying to multiply and add up more complicated equations.
You will notice that I haven't bothered to include the electrons in the added-up version. If you think about it, there are bound to be the same number on each side of the final equation, and so they will cancel out. If you aren't happy with this, write them down and then cross them out afterwards!
Example 2: The reaction between hydrogen peroxide and manganate(VII) ions
The first example was a simple bit of chemistry which you may well have come across. The technique works just as well for more complicated (and perhaps unfamiliar) chemistry.
Manganate(VII) ions, MnO4-, oxidise hydrogen peroxide, H2O2, to oxygen gas. The reaction is done with potassium manganate(VII) solution and hydrogen peroxide solution acidified with dilute sulphuric acid.
During the reaction, the manganate(VII) ions are reduced to manganese(II) ions.
Let's start with the hydrogen peroxide half-equation. What we know is:
The oxygen is already balanced. What about the hydrogen?
All you are allowed to add to this equation are water, hydrogen ions and electrons. If you add water to supply the extra hydrogen atoms needed on the right-hand side, you will mess up the oxygens again - that's obviously wrong!
Add two hydrogen ions to the right-hand side.
Now all you need to do is balance the charges. You would have to add 2 electrons to the right-hand side to make the overall charge on both sides zero.
Now for the manganate(VII) half-equation:
You know (or are told) that the manganate(VII) ions turn into manganese(II) ions. Write that down.
The manganese balances, but you need four oxygens on the right-hand side. These can only come from water - that's the only oxygen-containing thing you are allowed to write into one of these equations in acid conditions.
By doing this, we've introduced some hydrogens. To balance these, you will need 8 hydrogen ions on the left-hand side.
Now that all the atoms are balanced, all you need to do is balance the charges. At the moment there are a net 7+ charges on the left-hand side (1- and 8+), but only 2+ on the right. Add 5 electrons to the left-hand side to reduce the 7+ to 2+.
This is the typical sort of half-equation which you will have to be able to work out. The sequence is usually:
· Balance the atoms apart from oxygen and hydrogen.
· Balance the oxygens by adding water molecules.
· Balance the hydrogens by adding hydrogen ions.
· Balance the charges by adding electrons.
Combining the half-reactions to make the ionic equation for the reaction
The two half-equations we've produced are:
You have to multiply the equations so that the same number of electrons are involved in both. In this case, everything would work out well if you transferred 10 electrons.
But this time, you haven't quite finished. During the checking of the balancing, you should notice that there are hydrogen ions on both sides of the equation:
You can simplify this down by subtracting 10 hydrogen ions from both sides to leave the final version of the ionic equation - but don't forget to check the balancing of the atoms and charges!
You will often find that hydrogen ions or water molecules appear on both sides of the ionic equation in complicated cases built up in this way. Always check, and then simplify where possible.
Example 3: The oxidation of ethanol by acidified potassium dichromate(VI)
This technique can be used just as well in examples involving organic chemicals. Potassium dichromate(VI) solution acidified with dilute sulphuric acid is used to oxidise ethanol, CH3CH2OH, to ethanoic acid, CH3COOH.
The oxidising agent is the dichromate(VI) ion, Cr2O72-. This is reduced to chromium(III) ions, Cr3+.
We'll do the ethanol to ethanoic acid half-equation first. Using the same stages as before, start by writing down what you know:
Balance the oxygens by adding a water molecule to the left-hand side:
Add hydrogen ions to the right-hand side to balance the hydrogens:
And finally balance the charges by adding 4 electrons to the right-hand side to give an overall zero charge on each side:
The dichromate(VI) half-equation contains a trap which lots of people fall into!
Start by writing down what you know:
What people often forget to do at this stage is to balance the chromiums. If you don't do that, you are doomed to getting the wrong answer at the end of the process! When you come to balance the charges you will have to write in the wrong number of electrons - which means that your multiplying factors will be wrong when you come to add the half-equations . . . A complete waste of time!
Now balance the oxygens by adding water molecules . . .
. . . and the hydrogens by adding hydrogen ions:
Now all that needs balancing is the charges. Add 6 electrons to the left-hand side to give a net 6+ on each side.
Combining the half-reactions to make the ionic equation for the reaction
What we have so far is:
What are the multiplying factors for the equations this time? The simplest way of working this out is to find the smallest number of electrons which both 4 and 6 will divide into - in this case, 12. That means that you can multiply one equation by 3 and the other by 2.
The multiplication and addition looks like this:
Now you will find that there are water molecules and hydrogen ions occurring on both sides of the ionic equation. You can simplify this to give the final equation:
Reactions done under alkaline conditions
Working out half-equations for reactions in alkaline solution is decidedly trickier than those above.
WRITING IONIC EQUATIONS FOR REDOX REACTIONS DONE UNDER ALKALINE CONDITIONS
This section explains how to work out electron-half-reactions for oxidation and reduction processes which are carried out under alkaline conditions, and then how to combine them to give the overall ionic equation for the redox reaction. Combining them is easy; working them out may be more difficult than under acidic conditions.
Why is it more difficult to write electron-half-equations for these reactions?
What you already know
When you are trying to balance electron-half-equations, you are only allowed to add:
- electrons
- water
- hydrogen ions (unless the reaction is being done under alkaline conditions - in which case, you can add hydroxide ions instead)
If you are working under acidic or neutral conditions, the sequence is usually:
- Balance the atoms apart from oxygen and hydrogen.
- Balance the oxygens by adding water molecules.
- Balance the hydrogens by adding hydrogen ions.
- Balance the charges by adding electrons.
The whole process is fairly automatic, and provided you take care, there isn't much to go wrong.
How is this different under alkaline conditions?
The problem is that the water and the hydroxide ions that you add to balance the equations under alkaline conditions contain both hydrogen and oxygen.
To balance the oxygens, you could in principle add either H2O or OH- to the equation. The same thing is true for balancing the hydrogens. How do you know what to start with?
How to tackle the problem
In some cases, it is obvious how to build up the half-equation using hydroxide ions. Always check this before you get involved in anything more difficult. You will see what I mean shortly.
If it isn't immediately obvious, work out the electron-half equation as if it were being done under acidic conditions just as you have learnt to do on the previous page - in other words by writing in water molecules, hydrogen ions and electrons.
Once you have got a balanced half-equation, you then convert it to alkaline conditions. You will see how to do that in the following examples.
Four examples
Don't worry if the chemistry in these examples is unfamiliar to you. It doesn't matter in the slightest. All that matters is how you work out the equations.
The oxidation of cobalt(II) to cobalt(III) by hydrogen peroxide
If you add an excess of ammonia solution to a solution containing cobalt(II) ions, you get a complex ion formed called the hexaamminecobalt(II) ion, Co(NH3)62+. This is oxidised rapidly by hydrogen peroxide solution to the hexaamminecobalt(III) ion, Co(NH3)63+.
Ammonia solution is, of course, alkaline.
The half-equation for the cobalt reaction is easy. Start by writing down what you know (or are told):
Everything balances apart from the charges. Add an electron to the right-hand side to give both sides an overall charge of 2+.
The hydrogen peroxide half-equation isn't very difficult either, except that you aren't told what is formed and so have to make a guess. It would balance very nicely if you ended up with 2 hydroxide ions on the right-hand side.
This is a good example of a case where it is fairly obvious where to put hydroxide ions.
You would then just have to add 2 electrons to the left-hand side to balance the charges.
Combining the half-reactions to make the ionic equation for the reaction
What we have so far is:
The multiplication and addition looks like this:
And that's it - an easy example!
The oxidation of iron(II) hydroxide by the air
If you add sodium hydroxide solution to a solution of an iron(II) compound you get a green precipitate of iron(II) hydroxide. This is quite quickly oxidised by oxygen in the air to an orange-brown precipitate of iron(III) hydroxide.
The half-equation for the iron(II) hydroxide is straightforward. Start with what you know:
You obviously need another hydroxide ion on the left-hand side. This is even more straightforward than the previous example.
To balance the charges, add an electron to the right-hand side.
The half-equation for the oxygen isn't so easy. You don't know what is being formed.
It isn't at all obvious whether you need to balance the oxygens by adding water molecules or hydroxide ions to the right-hand side. OK - treat it as if it were being done under acidic conditions, and the problem disappears!
In this case, you can only balance the oxygens by adding water molecules to the right-hand side.
Balance the hydrogens by adding hydrogen ions to the left-hand side.
And then balance the charges by adding 4 electrons:
Now you have got a perfectly balanced half-equation. The problem is, of course, that it only applies under acidic conditions. We should have alkaline conditions - with hydroxide ions present not hydrogen ions.
So . . . get rid of the hydrogen ions! Add enough hydroxide ions to both sides of the equation so that you can neutralise all the hydrogen ions. Because it is now a balanced equation, you must add the same number to both sides to maintain the balance.
The hydrogen ions and hydroxide ions on the left-hand side would turn into 4 water molecules:
Finally, there are water molecules on both sides of the equation. Cancel out any which aren't changed.
This has all been a bit tedious - although you haven't actually had to think very much! Don't forget to re-check that everything balances now that you have finished.
Combining the half-reactions to make the ionic equation for the reaction
From here on it's all back to the usual routine. We've worked out the two half-equations:
The iron equation will have to happen 4 times to supply enough electrons to the oxygen.
Notice that the hydroxide ions on each side cancel out. (Perhaps to your surprise - certainly to mine when I worked this out!)
The reduction of manganate(VII) ions to manganate(VI) ions by hydroxide ions
This is a fairly obscure reaction but it is not too difficult to work out and balance. It is unusual in that hydroxide ions are acting as reducing agents.
Dark purple potassium manganate(VII) solution is slowly reduced to dark green potassium manganate(VI) solution by concentrated potassium hydroxide solution. Bubbles of oxygen gas are also given off.
Note: Unless the potassium manganate(VII) solution is very dilute, its strong purple colour tends to mask the green of the potassium manganate(VI) in the short term.
The half-equation for the conversion of manganate(VII) ions to manganate(VI) ions is easy (provided, of course, that you know their formulae!).
So what is happening to the hydroxide ions to turn them into oxygen gas?
It isn't actually very difficult to work out the half-equation directly, but suppose you want to avoid thinking and go through a standard routine:
Write down what you know, balancing the oxygens in the process:
Balance the hydrogens by adding hydrogen ions:
. . . and now balance the charges:
Now get rid of the hydrogen ions by adding enough hydroxide ions to both sides of the equation:
. . . and tidy it all up!
Combining the half-reactions to make the ionic equation for the reaction
What we have so far is:
The manganese reaction will have to happen four times in order to use up the four electrons produced by the hydroxide equation. Putting this together, you get:
The chemistry may be unfamiliar, but working out the equation isn't too hard!
The oxidation of chromium(III) to chromium(VI)
If you add an excess of sodium hydroxide solution to a solution containing chromium(III) ions, you get a dark green solution containing the complex ion hexahydroxochromate(III), Cr(OH)63-.
This can be oxidised to bright yellow chromate(VI) ions, CrO42- by warming it with hydrogen peroxide solution.
We've already worked out the hydrogen peroxide half-equation where it is acting as an oxidising agent under alkaline conditions:
So this time we just need to put together the half-equation for the chromium ions. What we know is:
It isn't the least bit obvious where to put hydroxide ions or water molecules, so treat it as if it were being done under acidic conditions. That way, you start by balancing the oxygens by adding water molecules.
To get 6 oxygens on each side, you need two waters on the right-hand side:
Now balance the hydrogens by adding hydrogen ions:
. . . and then balance the charges by adding electrons:
Finally, convert from acidic to alkaline conditions by adding enough hydroxide ions to both sides to turn the hydrogen ions into water:
. . . and tidy it all up:
Combining the half-reactions to make the ionic equation for the reaction
The two half-equations are:
If you multiply one equation by 3 and the other by 2, that transfers a total of 6 electrons.
Finally, tidy up the hydroxide ions that occur on both sides to leave the overall ionic equation:
OXIDATION STATES (OXIDATION NUMBERS)
This section explains what oxidation states (oxidation numbers) are and how to calculate them and make use of them.
Oxidation states are straightforward to work out and to use, but it is quite difficult to define what they are in any quick way.
Explaining what oxidation states (oxidation numbers) are
Oxidation states simplify the whole process of working out what is being oxidised and what is being reduced in redox reactions. However, for the purposes of this introduction, it would be helpful if you knew about:
- oxidation and reduction in terms of electron transfer
- electron-half-equations
We are going to look at some examples from vanadium chemistry. If you don't know anything about vanadium, it doesn't matter in the slightest.
Vanadium forms a number of different ions - for example, V2+ and V3+. If you think about how these might be produced from vanadium metal, the 2+ ion will be formed by oxidising the metal by removing two electrons:
The vanadium is now said to be in an oxidation state of +2.
Removal of another electron gives the V3+ ion:
The vanadium now has an oxidation state of +3.
Removal of another electron gives a more unusual looking ion, VO2+.
The vanadium is now in an oxidation state of +4. Notice that the oxidation state isn't simply counting the charge on the ion (that was true for the first two cases but not for this one).
The positive oxidation state is counting the total number of electrons which have had to be removed - starting from the element.
It is also possible to remove a fifth electron to give another ion (easily confused with the one before!). The oxidation state of the vanadium is now +5.
Every time you oxidise the vanadium by removing another electron from it, its oxidation state increases by 1.
Fairly obviously, if you start adding electrons again the oxidation state will fall. You could eventually get back to the element vanadium which would have an oxidation state of zero.
What if you kept on adding electrons to the element? You can't actually do that with vanadium, but you can with an element like sulphur.
The sulphur has an oxidation state of -2.
Summary
Oxidation state shows the total number of electrons which have been removed from an element (a positive oxidation state) or added to an element (a negative oxidation state) to get to its present state.
Oxidation involves an increase in oxidation state
Reduction involves a decrease in oxidation state
Recognising this simple pattern is the single most important thing about the concept of oxidation states. If you know how the oxidation state of an element changes during a reaction, you can instantly tell whether it is being oxidised or reduced without having to work in terms of electron-half-equations and electron transfers.
Working out oxidation states
You don't work out oxidation states by counting the numbers of electrons transferred. It would take far too long. Instead you learn some simple rules, and do some very simple sums!
- The oxidation state of an uncombined element is zero. That's obviously so, because it hasn't been either oxidised or reduced yet! This applies whatever the structure of the element - whether it is, for example, Xe or Cl2 or S8, or whether it has a giant structure like carbon or silicon.
- The sum of the oxidation states of all the atoms or ions in a neutral compound is zero.
- The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion.
- The more electronegative element in a substance is given a negative oxidation state. The less electronegative one is given a positive oxidation state. Remember that fluorine is the most electronegative element with oxygen second.
- Some elements almost always have the same oxidation states in their compounds:
| element | usual oxidation state | exceptions |
| Group 1 metals | always +1 |
|
| Group 2 metals | always +2 |
|
| Oxygen | usually -2 | except in peroxides and F2O (see below) |
| Hydrogen | usually +1 | except in metal hydrides where it is -1 (see below) |
| Fluorine | always -1 |
|
| Chlorine | usually -1 | except in compounds with O or F (see below) |
The reasons for the exceptions
Hydrogen in the metal hydrides
Metal hydrides include compounds like sodium hydride, NaH. In this, the hydrogen is present as a hydride ion, H-. The oxidation state of a simple ion like hydride is equal to the charge on the ion - in this case, -1.
Alternatively, you can think of it that the sum of the oxidation states in a neutral compound is zero. Since Group 1 metals always have an oxidation state of +1 in their compounds, it follows that the hydrogen must have an oxidation state of -1 (+1 -1 = 0).
Oxygen in peroxides
Peroxides include hydrogen peroxide, H2O2. This is an electrically neutral compound and so the sum of the oxidation states of the hydrogen and oxygen must be zero.
Since each hydrogen has an oxidation state of +1, each oxygen must have an oxidation state of -1 to balance it.
Oxygen in F2O
The problem here is that oxygen isn't the most electronegative element. The fluorine is more electronegative and has an oxidation state of -1. In this case, the oxygen has an oxidation state of +2.
Chlorine in compounds with fluorine or oxygen
There are so many different oxidation states that chlorine can have in these, that it is safer to simply remember that the chlorine doesn't have an oxidation state of -1 in them, and work out its actual oxidation state when you need it. You will find an example of this below.
Warning!
Don't get too bogged down in these exceptions. In most of the cases you will come across, they don't apply!
Examples of working out oxidation states
What is the oxidation state of chromium in Cr2+?
That's easy! For a simple ion like this, the oxidation state is the charge on the ion - in other words: +2 (Don't forget the + sign.)
What is the oxidation state of chromium in CrCl3?
This is a neutral compound so the sum of the oxidation states is zero. Chlorine has an oxidation state of -1. If the oxidation state of chromium is n:
n + 3(-1) = 0
n = +3 (Again, don't forget the + sign!)
What is the oxidation state of chromium in Cr(H2O)63+?
This is an ion and so the sum of the oxidation states is equal to the charge on the ion. There is a short-cut for working out oxidation states in complex ions like this where the metal atom is surrounded by electrically neutral molecules like water or ammonia.
The sum of the oxidation states in the attached neutral molecule must be zero. That means that you can ignore them when you do the sum. This would be essentially the same as an unattached chromium ion, Cr3+. The oxidation state is +3.
What is the oxidation state of chromium in the dichromate ion, Cr2O72-?
The oxidation state of the oxygen is -2, and the sum of the oxidation states is equal to the charge on the ion. Don't forget that there are 2 chromium atoms present.
2n + 7(-2) = -2
n = +6
What is the oxidation state of copper in CuSO4?
Unfortunately, it isn't always possible to work out oxidation states by a simple use of the rules above. The problem in this case is that the compound contains two elements (the copper and the sulphur) whose oxidation states can both change.
The only way around this is to know some simple chemistry! There are two ways you might approach it. (There might be others as well, but I can't think of them at the moment!)
- You might recognise this as an ionic compound containing copper ions and sulphate ions, SO42-. To make an electrically neutral compound, the copper must be present as a 2+ ion. The oxidation state is therefore +2.
- You might recognise the formula as being copper(II) sulphate. The "(II)" in the name tells you that the oxidation state is 2 (see below).
You will know that it is +2 because you know that metals form positive ions, and the oxidation state will simply be the charge on the ion.
Using oxidation states
In naming compounds
You will have come across names like iron(II) sulphate and iron(III) chloride. The (II) and (III) are the oxidation states of the iron in the two compounds: +2 and +3 respectively. That tells you that they contain Fe2+ and Fe3+ ions.
This can also be extended to the negative ion. Iron(II) sulphate is FeSO4. There is also a compound FeSO3 with the old name of iron(II) sulphite. The modern names reflect the oxidation states of the sulphur in the two compounds.
The sulphate ion is SO42-. The oxidation state of the sulphur is +6 (work it out!). The ion is more properly called the sulphate(VI) ion.
The sulphite ion is SO32-. The oxidation state of the sulphur is +4 (work that out as well!). This ion is more properly called the sulphate(IV) ion. The ate ending simply shows that the sulphur is in a negative ion.
So FeSO4 is properly called iron(II) sulphate(VI), and FeSO3 is iron(II) sulphate(IV). In fact, because of the easy confusion between these names, the old names sulphate and sulphite are normally still used in introductory chemistry courses.
Using oxidation states to identify what's been oxidised and what's been reduced
This is easily the most common use of oxidation states.
Remember:
Oxidation involves an increase in oxidation state
Reduction involves a decrease in oxidation state
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:
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:
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:
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 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.
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.
PERIOD 3 ELEMENTS
Atomic and Physical Properties of the Period 3 Elements
This page describes and explains the trends in atomic and physical properties of the Period 3 elements from sodium to argon. It covers ionisation energy, atomic radius, electronegativity, electrical conductivity, melting point and boiling point.
These topics are covered in various places elsewhere on the site and this page simply brings everything together - with links to the original pages if you need more information about particular points.
Atomic Properties
Electronic structures
In Period 3 of the Periodic Table, the 3s and 3p orbitals are filling with electrons. Just as a reminder, the shortened versions of the electronic structures for the eight elements are:
| Na | [Ne] 3s1 |
| Mg | [Ne] 3s2 |
| Al | [Ne] 3s2 3px1 |
| Si | [Ne] 3s2 3px1 3py1 |
| P | [Ne] 3s2 3px1 3py1 3pz1 |
| S | [Ne] 3s2 3px2 3py1 3pz1 |
| Cl | [Ne] 3s2 3px2 3py2 3pz1 |
| Ar | [Ne] 3s2 3px2 3py2 3pz2 |
In each case, [Ne] represents the complete electronic structure of a neon atom.
First ionisation energy
The first ionisation energy is the energy required to remove the most loosely held electron from one mole of gaseous atoms to produce 1 mole of gaseous ions each with a charge of 1+.
It is the energy needed to carry out this change per mole of X.
The pattern of first ionisation energies across Period 3
Notice that the general trend is upwards, but this is broken by falls between magnesium and aluminium, and between phosphorus and sulphur.
Explaining the pattern
First ionisation energy is governed by:
- the charge on the nucleus;
- the distance of the outer electron from the nucleus;
- the amount of screening by inner electrons;
- whether the electron is alone in an orbital or one of a pair.
The upward trend
In the whole of period 3, the outer electrons are in 3-level orbitals. These are all the same sort of distances from the nucleus, and are screened by the same electrons in the first and second levels.
The major difference is the increasing number of protons in the nucleus as you go from sodium across to argon. That causes greater attraction between the nucleus and the electrons and so increases the ionisation energies.
In fact the increasing nuclear charge also drags the outer electrons in closer to the nucleus. That increases ionisation energies still more as you go across the period.
The fall at aluminium
You might expect the aluminium value to be more than the magnesium value because of the extra proton. Offsetting that is the fact that aluminium's outer electron is in a 3p orbital rather than a 3s.
The 3p electron is slightly more distant from the nucleus than the 3s, and partially screened by the 3s electrons as well as the inner electrons. Both of these factors offset the effect of the extra proton.
The fall at sulphur
As you go from phosphorus to sulphur, something extra must be offsetting the effect of the extra proton
The screening is identical in phosphorus and sulphur (from the inner electrons and, to some extent, from the 3s electrons), and the electron is being removed from an identical orbital.
The difference is that in the sulphur case the electron being removed is one of the 3px2 pair. The repulsion between the two electrons in the same orbital means that the electron is easier to remove than it would otherwise be.
Atomic radius
The trend
The diagram shows how the atomic radius changes as you go across Period 3.
The figures used to construct this diagram are based on:
- metallic radii for Na, Mg and Al;
- covalent radii for Si, P, S and Cl;
- the van der Waals radius for Ar because it doesn't form any strong bonds.
It is fair to compare metallic and covalent radii because they are both being measured in tightly bonded circumstances. It isn't fair to compare these with a van der Waals radius, though.
The general trend towards smaller atoms across the period is NOT broken at argon. You aren't comparing like with like. The only safe thing to do is to ignore argon in the discussion which follows.
Explaining the trend
A metallic or covalent radius is going to be a measure of the distance from the nucleus to the bonding pair of electrons. If you aren't sure about that, go back and follow the last link.
From sodium to chlorine, the bonding electrons are all in the 3-level, being screened by the electrons in the first and second levels. The increasing number of protons in the nucleus as you go across the period pulls the bonding electrons more tightly to it. The amount of screening is constant for all of these elements.
Electronegativity
Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons.
The Pauling scale is the most commonly used. Fluorine (the most electronegative element) is assigned a value of 4.0, and values range down to caesium and francium which are the least electronegative at 0.7.
The trend
The trend across Period 3 looks like this:
Notice that argon isn't included. Electronegativity is about the tendency of an atom to attract a bonding pair of electrons. Since argon doesn't form covalent bonds, you obviously can't assign it an electronegativity.
Explaining the trend
The trend is explained in exactly the same way as the trend in atomic radii.
As you go across the period, the bonding electrons are always in the same level - the 3-level. They are always being screened by the same inner electrons.
All that differs is the number of protons in the nucleus. As you go from sodium to chlorine, the number of protons steadily increases and so attracts the bonding pair more closely.
Physical Properties
This section is going to look at the electrical conductivity and the melting and boiling points of the elements. To understand these, you first have to understand the structure of each of the elements.
Structures of the elements
The structures of the elements change as you go across the period. The first three are metallic, silicon is giant covalent, and the rest are simple molecules.
Three metallic structures
Sodium, magnesium and aluminium all have metallic structures.
In sodium, only one electron per atom is involved in the metallic bond - the single 3s electron. In magnesium, both of its outer electrons are involved, and in aluminium all three.
The other difference you need to be aware of is the way the atoms are packed in the metal crystal.
Sodium is 8-co-ordinated - each sodium atom is touched by only 8 other atoms.
Both magnesium and aluminium are 12-co-ordinated (although in slightly different ways). This is a more efficient way to pack atoms, leading to less wasted space in the metal structures and to stronger bonding in the metal.
A giant covalent structure
Silicon has a giant covalent structure just like diamond. A tiny part of the structure looks like this:
The structure is held together by strong covalent bonds in all three dimensions.
Four simple molecular structures
The structures of phosphorus and sulphur vary depending on the type of phosphorus or sulphur you are talking about. For phosphorus, I am assuming the common white phosphorus. For sulphur, I am assuming one of the crystalline forms - rhombic or monoclinic sulphur.
The atoms in each of these molecules are held together by covalent bonds (apart, of course, from argon).
In the liquid or solid state, the molecules are held close to each other by van der Waals dispersion forces.
Electrical conductivity
- Sodium, magnesium and aluminium are all good conductors of electricity. Conductivity increases as you go from sodium to magnesium to aluminium.
- Silicon is a semiconductor.
- None of the rest conducts electricity.
The three metals, of course, conduct electricity because the delocalised electrons (the "sea of electrons") are free to move throughout the solid or the liquid metal.
In the silicon case, explaining how semiconductors conduct electricity is beyond the scope of A level chemistry courses. With a diamond structure, you mightn't expect it to conduct electricity, but it does!
The rest don't conduct electricity because they are simple molecular substances. There are no electrons free to move around.
Melting and boiling points
The chart shows how the melting and boiling points of the elements change as you go across the period. The figures are plotted in kelvin rather than °C to avoid having negative values.
It is best to think of these changes in terms of the types of structure that we have talked about further up the page.
The metallic structures
Melting and boiling points rise across the three metals because of the increasing strength of the metallic bonds.
The number of electrons which each atom can contribute to the delocalised "sea of electrons" increases. The atoms also get smaller and have more protons as you go from sodium to magnesium to aluminium.
The attractions and therefore the melting and boiling points increase because:
- The nuclei of the atoms are getting more positively charged.
- The "sea" is getting more negatively charged.
- The "sea" is getting progressively nearer to the nuclei and so more strongly attracted.
Note: Boiling point is a better guide to the strength of the metallic bonds than melting point. Metallic bonds still exist in the liquid metals and aren't completely broken until the metal boils.
I don't know why there is such a small increase in melting point as you go from magnesium to aluminium. The boiling point of aluminium is much higher than magnesium's - as you would expect.
Silicon
Silicon has high melting and boiling points because it is a giant covalent structure. You have to break strong covalent bonds before it will melt or boil.
Because you are talking about a different type of bond, it isn't profitable to try to directly compare silicon's melting and boiling points with aluminium's.
The four molecular elements
Phosphorus, sulphur, chlorine and argon are simple molecular substances with only van der Waals attractions between the molecules. Their melting or boiling points will be lower than those of the first four members of the period which have giant structures.
The sizes of the melting and boiling points are governed entirely by the sizes of the molecules. Remember the structures of the molecules:
Phosphorus
Phosphorus contains P4 molecules. To melt phosphorus you don't have to break any covalent bonds - just the much weaker van der Waals forces between the molecules.
Sulphur
Sulphur consists of S8 rings of atoms. The molecules are bigger than phosphorus molecules, and so the van der Waals attractions will be stronger, leading to a higher melting and boiling point.
Chlorine
Chlorine, Cl2, is a much smaller molecule with comparatively weak van der Waals attractions, and so chlorine will have a lower melting and boiling point than sulphur or phosphorus.
Argon
Argon molecules are just single argon atoms, Ar. The scope for van der Waals attractions between these is very limited and so the melting and boiling points of argon are lower again.
CHEMICAL REACTIONS OF THE PERIOD 3 ELEMENTS
This page describes the reactions of the Period 3 elements from sodium to argon with water, oxygen and chlorine.
Reactions with water
Sodium
Sodium has a very exothermic reaction with cold water producing hydrogen and a colourless solution of sodium hydroxide.
Magnesium
Magnesium has a very slight reaction with cold water, but burns in steam.
A very clean coil of magnesium dropped into cold water eventually gets covered in small bubbles of hydrogen which float it to the surface. Magnesium hydroxide is formed as a very thin layer on the magnesium and this tends to stop the reaction.
Magnesium burns in steam with its typical white flame to produce white magnesium oxide and hydrogen.
Aluminium
Aluminium powder heated in steam produces hydrogen and aluminium oxide. The reaction is relatively slow because of the existing strong aluminium oxide layer on the metal, and the build-up of even more oxide during the reaction.
Silicon
There is a fair amount of disagreement in the books and on the web about what silicon does with water or steam. The truth seems to depend on the precise form of silicon you are using.
The common shiny grey lumps of silicon with a rather metal-like appearance are fairly unreactive. Most sources suggest that this form of silicon will react with steam at red heat to produce silicon dioxide and hydrogen.
But it is also possible to make much more reactive forms of silicon which will react with cold water to give the same products.
Phosphorus and sulphur
These have no reaction with water.
Chlorine
Chlorine dissolves in water to some extent to give a green solution. A reversible reaction takes place to produce a mixture of hydrochloric acid and chloric(I) acid (hypochlorous acid).
In the presence of sunlight, the chloric(I) acid slowly decomposes to produce more hydrochloric acid, releasing oxygen gas, and you may come across an equation showing the overall change:
Argon
There is no reaction between argon and water.
Reactions with oxygen
Sodium
Sodium burns in oxygen with an orange flame to produce a white solid mixture of sodium oxide and sodium peroxide.
For the simple oxide:
For the peroxide:
Magnesium
Magnesium burns in oxygen with an intense white flame to give white solid magnesium oxide.
Note: If magnesium is burns in air rather than in pure oxygen, it also reacts with the nitrogen in the air. You get a mixture of magnesium oxide and magnesium nitride formed.
Aluminium
Aluminium will burn in oxygen if it is powdered; otherwise the strong oxide layer on the aluminium tends to inhibit the reaction. If you sprinkle aluminium powder into a Bunsen flame, you get white sparkles. White aluminium oxide is formed.
Silicon
Silicon will burn in oxygen if heated strongly enough. Silicon dioxide is produced.
Phosphorus
White phosphorus catches fire spontaneously in air, burning with a white flame and producing clouds of white smoke - a mixture of phosphorus(III) oxide and phosphorus(V) oxide.
The proportions of these depend on the amount of oxygen available. In an excess of oxygen, the product will be almost entirely phosphorus(V) oxide.
For the phosphorus(III) oxide:
For the phosphorus(V) oxide:
Note: You may come across these oxides written as P2O3 and P2O5. Don't use these forms! They are as logical as writing, say, ethene as CH2 and ethane as CH3.
Sulphur
Sulphur burns in air or oxygen on gentle heating with a pale blue flame. It produces colourless sulphur dioxide gas.
Note: Sulphur dioxide can, of course, be converted further into sulphur trioxide in the presence of oxygen, but it needs the presence of a catalyst and fairly carefully controlled conditions.
Chlorine and argon
Despite having several oxides, chlorine won't react directly with oxygen. Argon doesn't react either.
Reactions with chlorine
Sodium
Sodium burns in chlorine with a bright orange flame. White solid sodium chloride is produced.
Magnesium
Magnesium burns with its usual intense white flame to give white magnesium chloride.
Aluminium
Aluminium is often reacted with chlorine by passing dry chlorine over aluminium foil heated in a long tube. The aluminium burns in the stream of chlorine to produce very pale yellow aluminium chloride. This sublimes (turns straight from solid to vapour and back again) and collects further down the tube where it is cooler.
Note: You may find versions of this equation showing the aluminium chloride as Al2Cl6. In fact, this exists in the vapour at temperatures not too far above the sublimation temperature - not in the solid.
Silicon
If chlorine is passed over silicon powder heated in a tube, it reacts to produce silicon tetrachloride. This is a colourless liquid which vaporises and can be condensed further along the apparatus.
Phosphorus
White phosphorus burns in chlorine to produce a mixture of two chlorides, phosphorus(III) chloride and phosphorus(V) chloride (phosphorus trichloride and phosphorus pentachloride).
Phosphorus(III) chloride is a colourless fuming liquid.
Phosphorus(V) chloride is an off-white (going towards yellow) solid.
Note: These equations are often given starting from P rather than P4. It depends which form of phosphorus you are talking about.
If you are talking about white phosphorus (as I am here), P4 is the correct version. If you are talking about red phosphorus, then P is correct. Red phosphorus has a different (polymeric) structure, and P4 would be wrong for it.
In my experience, red phosphorus is less commonly used in labs at this level (it isn't as excitingly reactive as white phosphorus!) - which is why I am concentrating on the white form.
Sulphur
If a stream of chlorine is passed over some heated sulphur, it reacts to form an orange, evil-smelling liquid, disulphur dichloride, S2Cl2.
Chlorine and argon
It obviously doesn't make sense to talk about chlorine reacting with itself, and argon doesn't react with chlorine.
PERIODIC TABLE GROUP 1
Atomic and Physical Properties of the Group 1 Elements
This page explores the trends in some atomic and physical properties of the Group 1 elements - lithium, sodium, potassium, rubidium and caesium. You will find separate sections below covering the trends in atomic radius, first ionisation energy, electronegativity, melting and boiling points, and density.
Even if you aren't currently interested in all these things, it would probably pay you to read the whole page. The same ideas tend to recur throughout the atomic properties, and you may find that earlier explanations help to you understand later ones.
Trends in Atomic Radius
You can see that the atomic radius increases as you go down the Group.
Explaining the increase in atomic radius
The radius of an atom is governed by
- the number of layers of electrons around the nucleus
- the pull the outer electrons feel from the nucleus.
Compare lithium and sodium:
| Li |
| 1s22s1 |
| Na |
| 1s22s22p63s1 |
In each case, the outer electron feels a net pull of 1+ from the nucleus. The positive charge on the nucleus is cut down by the negativeness of the inner electrons.
This is equally true for all the other atoms in Group 1. Work it out for potassium if you aren't convinced.
The only factor which is going to affect the size of the atom is therefore the number of layers of inner electrons which have to be fitted in around the atom. Obviously, the more layers of electrons you have, the more space they will take up - electrons repel each other. That means that the atoms are bound to get bigger as you go down the Group.
Trends in First Ionisation Energy
First ionisation energy is the energy needed to remove the most loosely held electron from each of one mole of gaseous atoms to make one mole of singly charged gaseous ions - in other words, for 1 mole of this process:
Notice that first ionisation energy falls as you go down the group.
Explaining the decrease in first ionisation energy
Ionisation energy is governed by
- the charge on the nucleus,
- the amount of screening by the inner electrons,
- the distance between the outer electrons and the nucleus.
As you go down the Group, the increase in nuclear charge is exactly offset by the increase in the number of inner electrons. Just as when we were talking about atomic radius further up this page, in each of the elements in this Group, the outer electrons feel a net attraction of 1+ from the centre.
However, as you go down the Group, the distance between the nucleus and the outer electrons increases and so they become easier to remove - the ionisation energy falls.
Trends in Electronegativity
Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. It is usually measured on the Pauling scale, on which the most electronegative element (fluorine) is given an electronegativity of 4.0.
All of these elements have a very low electronegativity. (Remember that the most electronegative element, fluorine, has an electronegativity of 4.0.) Notice that electronegativity falls as you go down the Group. The atoms become less and less good at attracting bonding pairs of electrons.
Explaining the decrease in electronegativity
Imagine a bond between a sodium atom and a chlorine atom. Think of it to start with as a covalent bond - a pair of shared electrons. The electron pair will be dragged towards the chlorine because there is a much greater net pull from the chlorine nucleus than from the sodium one.
The electron pair ends up so close to the chlorine that there is essentially a transfer of an electron to the chlorine - ions are formed. The large pull from the chlorine nucleus is why chlorine is much more electronegative than sodium is.
Now compare this with the lithium-chlorine bond.
The net pull from each end of the bond is the same as before, but you have to remember that the lithium atom is smaller than a sodium atom. That means that the electron pair is going to be closer to the net 1+ charge from the lithium end, and so more strongly attracted to it.
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