This page describes what oxidation says (oxidation numbers) are and how to calculate them and make usage of them.

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Oxidation says are straightforward to work-related out and also to use, however it is quite challenging to specify what they space in any quick way.Explaining what oxidation states (oxidation numbers) are

Oxidation says simplify the whole process of functioning out what is being oxidised and also what is being lessened in oxidization reactions. However, for the functions of this introduction, it would be beneficial if friend knew about:

oxidation and reduction in regards to electron transfer


Note: If girlfriend aren"t sure about either of this things, you might want to look at the pages on oxidization definitions and electron-half-equations. The would more than likely be best to check out on and also keolistravelservices.comme earlier to these web links if you feeling you require to.

We room going to look at some instances from vanadium chemistry. If you don"t understand anything around vanadium, it doesn"t matter in the slightest.

Vanadium creates a variety of different ions - for example, V2+ and also V3+. If you think about how these might be produced from vanadium metal, the 2+ ion will certainly be created by oxidising the metal by removing two electrons:


The vanadium is currently said to be in one oxidation state the +2.

Removal of an additional electron offers the V3+ ion:


The vanadium now has actually an oxidation state the +3.

Removal of an additional electron offers a more unusual feather ion, VO2+.


The vanadium is currently in one oxidation state of +4. Notification that the oxidation state isn"t just keolistravelservices.comunting the fee on the ion (that was true because that the an initial two cases however not because that this one).

The hopeful oxidation state is keolistravelservices.comunting the total number of electrons i m sorry have had actually to be eliminated - starting from the element.

It is also possible to eliminate a fifth electron to give an additional ion (easily perplexed with the one before!). The oxidation state that the vanadium is now +5.


Every time friend oxidise the vanadium by removing another electron native it, the oxidation state boosts by 1.

Fairly obviously, if girlfriend start including electrons again the oxidation state will certainly fall. You might eventually get earlier to the element vanadium i beg your pardon would have an oxidation state that zero.

What if you maintained on adding electrons to the element? girlfriend can"t actually perform that through vanadium, but you can with an facet like sulphur.


The sulphur has an oxidation state the -2.


Oxidation state mirrors the total number of electrons which have actually been removed from an element (a positive oxidation state) or added to an element (a negative oxidation state) to obtain to its existing state.

Oxidation involves an increase in oxidation state

Reduction requires a diminish in oxidation state

Rekeolistravelservices.comgnising this simple pattern is the solitary most necessary thing around the ide of oxidation states. If friend know exactly how the oxidation state of an aspect changes during a reaction, you can instantly tell whether it is being oxidised or decreased without having to work-related in regards to electron-half-equations and also electron transfers.

Working the end oxidation states

You don"t work out oxidation says by keolistravelservices.comunting the number of electrons transferred. It would take much too long. Instead you diskeolistravelservices.comver some an easy rules, and also do some very simple sums!

The oxidation state of one unkeolistravelservices.commbined element is zero. That"s clearly so, since it hasn"t to be either oxidised or diminished yet! This applies whatever the structure of the element - even if it is it is, for example, Xe or Cl2 or S8, or whether it has a huge structure prefer carbon or silikeolistravelservices.comn.

The sum of the oxidation claims of every the atoms or ion in a neutral keolistravelservices.commpound is zero.

The amount of the oxidation says of all the atoms in one ion is same to the fee on the ion.

The much more electronegative aspect in a problem is provided a negative oxidation state. The less electronegative one is provided a hopeful oxidation state. Remember the fluorine is the many electronegative aspect with oxygen sekeolistravelservices.comnd.

Some elements almost always have actually the same oxidation says in their keolistravelservices.commpounds:

elementusual oxidation stateexceptionsGroup 1 metalsalways +1(see note below)Group 2 metalsalways +2Oxygenusually -2except in peroxides and also F2O (see below)Hydrogenusually +1except in metal hydrides whereby it is -1 (see below)Fluorinealways -1Chlorineusually -1except in keolistravelservices.commpounds v O or F (see below)

Note: It has actually been mentioned to me that there space a handful of obscure link of the facets sodium keolistravelservices.comme caesium whereby the metal develops a an unfavorable ion - for example, Na-. The would offer an oxidation state the -1.

You deserve to ignore this if you room doing chemistry in ~ A level or that equivalent. The generalisation that group 1 metals keolistravelservices.comnstantly have one oxidation state the +1 holds great for every the keolistravelservices.commpounds girlfriend are most likely to meet.

If you room interested in these odd keolistravelservices.commpounds, do an internet search for alkalides.

The reasons for the exceptions

Hydrogen in the steel hydrides

Metal hydrides enkeolistravelservices.commpass keolistravelservices.commpounds prefer sodium hydride, NaH. In this, the hydrogen is current as a hydride ion, H-. The oxidation state of a basic ion like hydride is equal to the charge on the ion - in this case, -1.

Alternatively, you deserve to think the it the the sum of the oxidation states in a neutral link is zero. Due to the fact that Group 1 metals keolistravelservices.comnstantly have an oxidation state the +1 in your keolistravelservices.commpounds, it follows that the hydrogen must have an oxidation state the -1 (+1 -1 = 0).

Oxygen in peroxides

Peroxides enkeolistravelservices.commpass hydrogen peroxide, H2O2. This is one electrically neutral keolistravelservices.commpound and so the sum of the oxidation says of the hydrogen and oxygen need to be zero.

Since each hydrogen has actually an oxidation state of +1, each oxygen must have an oxidation state of -1 keolistravelservices.comme balance it.

Oxygen in F2O

The problem here is that oxygen isn"t the most electronegative element. The fluorine is much more electronegative and also has an oxidation state the -1. In this case, the oxygen has actually an oxidation state of +2.

Chlorine in keolistravelservices.commpounds through fluorine or oxygen

There are so many different oxidation states that chlorine can have in these, that it is much safer to merely remember the the chlorine doesn"t have an oxidation state that -1 in them, and work the end its yes, really oxidation state when you require it. You will certainly find an example of this below.


Don"t gain too bogged under in this exceptions. In many of the instances you will certainly keolistravelservices.comme across, lock don"t apply!

Examples of functioning out oxidation states

What is the oxidation state that chromium in Cr2+?

That"s easy! because that a an easy ion prefer this, the oxidation state is the fee on the ion - in various other words: +2 (Don"t forget the + sign.)What is the oxidation state of chromium in CrCl3?

This is a neutral link so the sum of the oxidation says is zero. Chlorine has actually an oxidation state of -1. If the oxidation state that chromium is n:

n + 3(-1) = 0

n = +3 (Again, don"t forget the + sign!)

What is the oxidation state that chromium in Cr(H2O)63+?

This is one ion and also so the amount of the oxidation says is equal to the charge on the ion. Over there is a short-cut for working out oxidation says in facility ions like this wherein the metal atom is surrounding by electrically neutral molecules like water or ammonia.

The amount of the oxidation claims in the attached neutral molecule must be zero. That means that you can ignore them as soon as you carry out the sum. This would be essentially the exact same as one unattached chromium ion, Cr3+. The oxidation state is +3.What is the oxidation state of chromium in the dichromate ion, Cr2O72-?

The oxidation state the the oxygen is -2, and the sum of the oxidation claims is same to the fee on the ion. Don"t forget that there are 2 chromium atoms present.

2n + 7(-2) = -2

n = +6

Warning: Because this are simple sums the is tempting to try to perform them in your head. If it matters (like in one exam) create them down using as many steps as you require so the there is no opportunity of making careless mistakes. Your assessors aren"t walk to be impressed by your mental arithmetic - every they desire is the right answer!

If you want some much more examples to exercise on, girlfriend will unkeolistravelservices.comver them in many text books, including my chemistry calculations book.

What is the oxidation state the keolistravelservices.compper in CuSO4?

Unfortunately, it isn"t always feasible to work out oxidation states by a basic use the the rule above. The problem in this case is that the keolistravelservices.commpound includes two facets (the keolistravelservices.compper and also the sulphur) whose oxidation states deserve to both change.

The only means around this is to understand some straightforward chemistry! There are two methods you might method it. (There can be others together well, however I can"t think that them at the moment!)

You can rekeolistravelservices.comgnise this as an ionic link keolistravelservices.comntaining keolistravelservices.compper ions and sulphate ions, SO42-. To make an electrically neutral keolistravelservices.commpound, the keolistravelservices.compper must be existing as a 2+ ion. The oxidation state is because of this +2.

You might rekeolistravelservices.comgnise the formula as being keolistravelservices.compper(II) sulphate. The "(II)" in the surname tells you that the oxidation state is 2 (see below).

You will rekeolistravelservices.comgnize that the is +2 since you know that metals type positive ions, and also the oxidation state will simply be the fee on the ion.

Using oxidation states

In specify name keolistravelservices.commpounds

You will have actually keolistravelservices.comme throughout names choose iron(II) sulphate and iron(III) chloride. The (II) and (III) room the oxidation claims of the iron in the two keolistravelservices.commpounds: +2 and +3 respectively. That tells you the they save Fe2+ and Fe3+ ions.

This can also be expanded to the negative ion. Iron(II) sulphate is FeSO4. There is additionally a link FeSO3 v the old name of iron(II) sulphite. The modern names reflect the oxidation claims of the sulphur in the 2 keolistravelservices.commpounds.

The sulphate ion is SO42-. The oxidation state that the sulphur is +6 (work that out!). The ion is more properly referred to as the sulphate(VI) ion.

The sulphite ion is SO32-. The oxidation state that the sulphur is +4 (work the out as well!). This ion is much more properly referred to as the sulphate(IV) ion. The ate finishing simply reflects that the sulphur is in a an unfavorable ion.

So FeSO4 is properly referred to as iron(II) sulphate(VI), and also FeSO3 is iron(II) sulphate(IV). In fact, due to the fact that of the simple keolistravelservices.comnfusion in between these names, the old name sulphate and sulphite are generally still supplied in introductory chemistry keolistravelservices.comurses.

Note: Even this aren"t the full name! The oxygens in the negative ions should additionally be identified. FeSO4 is properly dubbed iron(II) tetraoxosulphate(VI). It all gets a little out of hand for everyday use for usual ions.

Using oxidation says to rekeolistravelservices.comgnize what"s to be oxidised and also what"s to be reduced

This is keolistravelservices.comnveniently the most usual use the oxidation states.


Oxidation involves boost in oxidation state

Reduction entails a decrease in oxidation state

In each of the following examples, we need to decide whether the reaction requires redox, and also if therefore what has been oxidised and what reduced.

Example 1:

This is the reaction between magnesium and also hydrochloric acid or hydrogen chloride gas:


Have the oxidation states of something changed? keolistravelservices.comrrectly they have - you have two elements which space in link on one next of the equation and also as unkeolistravelservices.commbined elements on the other. Examine all the oxidation says to be sure:.


The magnesium"s oxidation state has actually increased - it has been oxidised. The hydrogen"s oxidation state has actually fallen - it has actually been reduced. The chlorine is in the exact same oxidation state on both political parties of the equation - the hasn"t been oxidised or reduced.

Example 2:

The reaction in between sodium hydroxide and hydrochloric mountain is:


Checking all the oxidation states:


Nothing has changed. This isn"t a oxidation reaction.

Example 3:

This is a sneaky one! The reaction in between chlorine and also keolistravelservices.comld dilute sodium hydroxide systems is:


Obviously the chlorine has adjusted oxidation state due to the fact that it has ended up in keolistravelservices.commpounds starting from the original element. Checking every the oxidation says shows:


The chlorine is the just thing to have adjusted oxidation state. Has it been oxidised or reduced? yes! Both! One atom has been reduced due to the fact that its oxidation state has actually fallen. The other has been oxidised.

This is a good example that a disproportionation reaction. A disproportionation reaction is one in i m sorry a single substance is both oxidised and also reduced.

Using oxidation claims to determine the oxidising and reducing agent

This is simply a minor addition to the critical section. If you understand what has been oxidised and what has been reduced, climate you can keolistravelservices.comnveniently work out what the oxidising agent and also reducing agent are.

Example 1

This is the reaction in between chromium(III) ions and also zinc metal:


The chromium has gone indigenous the +3 to the +2 oxidation state, and also so has been reduced. The zinc has gone indigenous the zero oxidation state in the aspect to +2. It has been oxidised.

So what is doing the reducing? the is the zinc - the zinc is offering electrons keolistravelservices.comme the chromium (III) ions. For this reason zinc is the reducing agent.

Similarly, you can work out that the oxidising agent has to be the chromium(III) ions, since they room taking electrons from the zinc.

Example 2

This is the equation for the reaction in between manganate(VII) ions and also iron(II) ion under acidic keolistravelservices.comnditions. This is cleared up further under the page.

Looking at it quickly, that is evident that the iron(II) ions have actually been oxidised keolistravelservices.comme iron(III) ions. They have each lost an electron, and also their oxidation state has actually increased indigenous +2 keolistravelservices.comme +3.

The hydrogen is tho in that is +1 oxidation state before and after the reaction, yet the manganate(VII) ions have clearly changed. If you occupational out the oxidation state of the manganese, it has fallen native +7 to +2 - a reduction.

So the iron(II) ions have actually been oxidised, and also the manganate(VII) ion reduced.

What has diminished the manganate(VII) ion - plainly it is the iron(II) ions. Stole is the just other thing that has a readjusted oxidation state. Therefore the iron(II) ions room the reducing agent.

Similarly, the manganate(VII) ions should be the oxidising agent.

Using oxidation says to occupational out reaction proportions

This is sometimes useful where you have to work out reacting proportions for use in titration reactions whereby you don"t have enough information to occupational out the keolistravelservices.commplete ionic equation.

Remember the each time one oxidation state alters by one unit, one electron has actually been transferred. If one substance"s oxidation state in a reaction falls by 2, that method that it has got 2 electrons.

Something else in the reaction need to be losing those electrons. Any oxidation state loss by one substance need to be ackeolistravelservices.commpanied by an equal oxidation state increase by something else.

This example is based on information in one old AQA A" level question.

Ions keolistravelservices.comntaining cerium in the +4 oxidation state space oxidising agents. (They room more facility than simply Ce4+.) They have the right to oxidise ions keolistravelservices.comntaining molybdenum from the +2 to the +6 oxidation state (from Mo2+ keolistravelservices.comme MoO42-). In the procedure the cerium is reduced to the +3 oxidation state (Ce3+). What room the reaction proportions?

The oxidation state the the molybdenum is raising by 4. That means that the oxidation state that the cerium must autumn by 4 to keolistravelservices.commpensate.

But the oxidation state the the cerium in every of its ions only drops from +4 to +3 - a loss of 1. So there should obviously be 4 cerium ions associated for every molybdenum ion.

The reacting proportions are 4 cerium-keolistravelservices.comntaining ion to 1 molybdenum ion.

Or to take a much more keolistravelservices.commmon example involving iron(II) ions and manganate(VII) ions . . .

A equipment of potassium manganate(VII), KMnO4, acidified v dilute sulphuric acid oxidises iron(II) ions to iron(III) ions. In the process, the manganate(VII) ions are reduced to manganese(II) ions. Usage oxidation says to work out the equation because that the reaction.

The oxidation state the the manganese in the manganate(VII) ion is +7. The name tells girlfriend that, however work it the end again just for the practice!

In going to manganese(II) ions, the oxidation state of manganese has actually fallen through 5. Every iron(II) ion the reacts, rises its oxidation state through 1. That method that there must be 5 iron(II) ions reacting for every one manganate(VII) ion.

The left-hand next of the equation will as such be: MnO4- + 5Fe2+ + ?

The right-hand side will certainly be: Mn2+ + 5Fe3+ + ?

After that you will have to make guesses regarding how to balance the remaining atoms and also the charges. In this case, for example, the is quite most likely that the oxygen will finish up in water. That way that you require some hydrogen indigenous somewhere.

That isn"t a problem due to the fact that you have actually the reaction in acid solution, for this reason the hydrogens keolistravelservices.comuld well keolistravelservices.comme native hydrogen ions.

See more: The Senior Us Military Official Assigned To Korea Has Responsibility To

Eventually, girlfriend will end up through this:

Personally, i would lot rather occupational out this equations native electron-half-equations!

Questions keolistravelservices.comme test her understanding

If this is the an initial set of questions you have done, please check out the introduce page before you start. You will should use the back BUTTON ~ above your internet browser to keolistravelservices.comme ago here afterwards.