Redox Reactions and Oxidation Numbers, Grade 11
Oxidation is loss of electrons. Reduction is gain of electrons. If the oxidation number of any atom changes, the reaction is redox. That third sentence is the one that earns marks, because it is the only test you can apply to an equation you have never seen before.
This page covers the Grade 11 Term 4 work on redox reactions and oxidation numbers: the definitions, the full rules for assigning oxidation numbers with the two exceptions that catch people out, how to tell a redox reaction from one that only looks like one, oxidising and reducing agents, half-reactions, and the practicals that are worth running.
The five definitions, and nothing more
Six marks in a Grade 11 paper go to definitions, and they are the cheapest marks in the topic. Learn them in this form.
| Term | Definition |
|---|---|
| Oxidation | The loss of electrons |
| Reduction | The gain of electrons |
| Redox reaction | A reaction in which electrons are transferred |
| Oxidising agent | A substance that gains electrons and is therefore reduced |
| Reducing agent | A substance that loses electrons and is therefore oxidised |
| Half-reaction | The part of a redox reaction that shows either the oxidation or the reduction on its own |
OIL RIG still works: Oxidation Is Loss, Reduction Is Gain. It is about electrons, not oxygen, and that matters because plenty of redox reactions have no oxygen anywhere in them.
The agent is always the opposite of what happens to it
This single confusion costs more marks in this topic than anything else.
A reducing agent does the reducing to something else. To reduce something else it has to hand over electrons. Handing over electrons is oxidation. So the reducing agent is the thing that gets oxidised.
Read the two rows in the table above again with that in mind. If you write "the reducing agent is reduced" you have lost the mark, and it is a mark almost every paper offers.
One more habit worth building: name the substance, not the atom. In a reaction between potassium permanganate and iron(II) sulfate, the oxidising agent is potassium permanganate, not manganese. The agent is a reactant you could weigh out of a bottle.
How to work out an oxidation number
An oxidation number is the charge an atom would have if every bond in the compound were completely ionic. It is a bookkeeping device, not a real charge, and its whole job is to let you see at a glance whether electrons have moved.
There is a slower method that works from electronegativity differences and drawn-out bonds, and it is genuinely useful for understanding why the numbers come out as they do. For getting answers in an exam, use the rules. Work down this list and stop as soon as the atom you want is settled.
| # | Rule | Examples |
|---|---|---|
| 1 | An uncombined element is 0, whatever shape it comes in | Cu, Fe, O2, Cl2, S8, P4 are all 0 |
| 2 | A single-atom ion equals its charge | Na+ is +1, Cl− is −1, Fe3+ is +3, S2− is −2 |
| 3 | Fluorine in a compound is always −1. No exceptions at all | HF, CaF2, SF6 |
| 4 | Group 1 is +1, Group 2 is +2 in any compound | Na in NaCl is +1, Ca in CaCO3 is +2 |
| 5 | Hydrogen is +1, except when bonded to a metal, where it is −1 | +1 in H2O and HCl. −1 in NaH and CaH2 |
| 6 | Oxygen is −2, except in peroxides where it is −1 | −2 in H2O and CO2. −1 in H2O2 and Na2O2 |
| 7 | The other halogens are −1 unless they are bonded to oxygen or to fluorine | Cl is −1 in NaCl, but +1 in HOCl and +5 in ClO3− |
| 8 | The numbers must add up to the overall charge. Zero for a neutral molecule, the ion charge for an ion | Use this last, to solve for whatever is left |
The two exceptions that catch people out
Rules 5 and 6 are the only ones with exceptions, and both exceptions are examinable.
Take hydrogen peroxide, H2O2. Apply oxygen equals −2 without thinking and you get 2(+1) + 2(−2) = −2 for a molecule with no charge. That is not a hard question, it is an impossible one, and the impossibility is the signal. Oxygen is −1 in a peroxide, which gives 2(+1) + 2(−1) = 0.
The same logic runs the other way for sodium hydride, NaH. Sodium is Group 1, so it is +1, and the total must be zero, so hydrogen has to be −1. Hydrogen bonded to a metal is the one place it goes negative.
If your numbers do not add up, you have hit an exception rather than made an arithmetic mistake. Check hydrogen and oxygen first.
Solving for the one you do not know
Once the easy atoms are assigned, put x on the unknown and let the arithmetic finish it.
Worked example: the manganese in potassium permanganate, KMnO4
Potassium is Group 1, so it is +1. Each oxygen is −2, and there are four of them, so oxygen contributes −8. The molecule is neutral, so everything adds to zero.
(+1) + x + (−8) = 0, so x = +7.
Worked example: the sulfur in sodium thiosulfate, Na2S2O3
Two sodiums at +1 give +2. Three oxygens at −2 give −6. The total is zero, and there are two sulfur atoms sharing what is left.
(+2) + 2x + (−6) = 0, so 2x = +4 and each sulfur is +2.
Worked example: an ion rather than a molecule, the chromium in Cr2O72−
The only change is what the numbers add up to. For an ion they add to the charge on the ion, here −2, not to zero.
2x + 7(−2) = −2, so 2x = +12 and each chromium is +6.
Worked example: two different answers in one compound, ammonium nitrate NH4NO3
This one is worth doing because it looks like a trick and is not. Ammonium nitrate is built from two ions, NH4+ and NO3−, so treat each ion separately.
In NH4+: x + 4(+1) = +1, so nitrogen is −3.
In NO3−: x + 3(−2) = −1, so nitrogen is +5.
The same element, in the same compound, eight apart. Nothing is wrong. Oxidation numbers belong to atoms in a setting, not to elements, and this is the clearest demonstration of that you will find. If a question hands you a compound built from two polyatomic ions, split it before you start.
Common oxidation numbers, for checking your answer
Do not memorise this table. Work the numbers out, then use it to check.
| Element | Where it is | Oxidation number |
|---|---|---|
| Oxygen | O2 | 0 |
| H2O, CO2, most compounds | −2 | |
| H2O2, Na2O2 (peroxides) | −1 | |
| OF2 | +2 | |
| Hydrogen | H2 | 0 |
| H2O, HCl, NH3 | +1 | |
| NaH, CaH2 (metal hydrides) | −1 | |
| Manganese | KMnO4 | +7 |
| MnO2 | +4 | |
| Mn2+, MnSO4 | +2 | |
| Chromium | Cr2O72−, CrO42− | +6 |
| Cr3+ | +3 | |
| Sulfur | S8 | 0 |
| H2SO4, SO42− | +6 | |
| SO2, SO32− | +4 | |
| H2S, S2− | −2 | |
| Nitrogen | N2 | 0 |
| NH3, NH4+ | −3 | |
| NO2 | +4 | |
| NO3−, HNO3 | +5 | |
| Carbon | CH4 | −4 |
| CO | +2 | |
| CO2, CO32− | +4 | |
| C2O42− (oxalate) | +3 | |
| Iron | Fe | 0 |
| FeO, FeCl2, Fe2+ | +2 | |
| Fe2O3, FeCl3, Fe3+ | +3 |
Is this reaction actually redox?
This is the question Grade 11 papers ask most often, usually as a multiple choice, and it is answerable in about fifteen seconds once you know the routine.
Assign oxidation numbers to every atom on both sides. If even one atom changes, it is redox. If nothing changes, it is not. That is the whole test.
Two shortcuts save you most of the arithmetic.
| What you see | What it means |
|---|---|
| An uncombined element on one side and combined on the other | Redox, guaranteed. Something went from 0 to not-zero, and that is a change. Look for Mg, Zn, Cu, H2, O2, Cl2 |
| Only ions swapping partners, with no element on its own anywhere | Almost certainly not redox. Check it, but expect nothing to have moved |
Three whole families of reaction are never redox, and recognising them by name is faster than working through them.
| Family | Example | Why not |
|---|---|---|
| Neutralisation, acid plus base | H2SO4 + 2NaOH → Na2SO4 + 2H2O | A proton moves, not an electron. H stays +1, O stays −2, S stays +6, Na stays +1 |
| Precipitation | Ag+(aq) + Cl−(aq) → AgCl(s) | Silver was +1 and still is. Chloride was −1 and still is. They have only stopped being dissolved |
| Dissolving a salt | NaCl(s) → Na+(aq) + Cl−(aq) | Same numbers, different phase |
Compare that with an acid attacking a metal, which looks superficially similar and is redox:
Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g)
Magnesium goes 0 to +2 and hydrogen goes +1 to 0. Two changes, so it is redox. Magnesium is oxidised and is therefore the reducing agent. Sulfuric acid is reduced and is therefore the oxidising agent. An uncombined element appears on each side, which is the shortcut telling you before you have written anything down.
Naming the oxidising and reducing agent
Once the numbers are on the page, the rest is mechanical.
| Step | What to do |
|---|---|
| 1 | Find the atom whose oxidation number went up. That atom was oxidised |
| 2 | The reactant containing it is the reducing agent |
| 3 | Find the atom whose oxidation number went down. That atom was reduced |
| 4 | The reactant containing it is the oxidising agent |
Worked example: iron(III) oxide reacting with carbon monoxide, the reaction inside a blast furnace
Fe2O3 + 3CO → 2Fe + 3CO2
Iron goes from +3 in Fe2O3 down to 0 in the metal, so iron is reduced and iron(III) oxide is the oxidising agent. Carbon goes from +2 in CO up to +4 in CO2, so carbon is oxidised and carbon monoxide is the reducing agent. Oxygen sits at −2 throughout and takes no part.
Note that iron ends up as the metal we want, and it got there by being reduced. That is what smelting is. Every tonne of iron made at Newcastle or Vanderbijlpark comes out of a reaction that is doing exactly this.
Half-reactions
A half-reaction shows one side of the exchange with the electrons written in.
For zinc dropped into copper(II) sulfate solution:
| Half-reaction | |
|---|---|
| Oxidation | Zn → Zn2+ + 2e− |
| Reduction | Cu2+ + 2e− → Cu |
| Net | Zn + Cu2+ → Zn2+ + Cu |
Electrons are written on the left for a reduction and on the right for an oxidation, because that is the side they arrive at or leave from. Get that the wrong way round and you have written the reverse reaction.
The electrons must cancel exactly when you add the two halves. If they do not, multiply one or both half-reactions until they do, and only then add.
What Grade 11 actually requires here
The teaching plan changed, and a lot of Grade 11 material still in circulation has not caught up.
Balancing redox equations by tracking oxidation-number changes, sometimes called the ion-electron method, is no longer the required Grade 11 approach. The teaching plan replaces it with balancing from half-reactions taken off the Table of Standard Reduction Potentials, and increases the time allowed for the topic to fit that in.
Which means Grade 11 learners now need the Table of Standard Reduction Potentials, and most Grade 11 textbooks do not print one. It is the same table used in Grade 12 electrochemistry and it is on the official data sheet. If your Grade 11 class does not have it in front of them, they cannot do the balancing the plan asks for. Print it or borrow it from the Grade 12 book before you start this topic.
Oxidation numbers are still very much in. What changed is their job. They are there to let you recognise a redox reaction and to identify what was oxidised and what was reduced. They are not the balancing tool any more.
Balancing from half-reactions, worked through
Worked example: permanganate against oxalate in acid
This is the reaction behind a standard permanganate titration, and it is a good one to practise on because the numbers are awkward enough to be honest.
The two half-reactions, in acid:
MnO4− + 8H+ + 5e− → Mn2+ + 4H2O
C2O42− → 2CO2 + 2e−
Five electrons on one side, two on the other. The lowest common multiple is ten, so double the first and multiply the second by five:
2MnO4− + 16H+ + 10e− → 2Mn2+ + 8H2O
5C2O42− → 10CO2 + 10e−
Add them and the ten electrons cancel:
2MnO4− + 5C2O42− + 16H+ → 2Mn2+ + 10CO2 + 8H2O
Always check the charge as well as the atoms. On the left, 2(−1) + 5(−2) + 16(+1) = +4. On the right, 2(+2) + 0 + 0 = +4. Balanced. Learners check the atoms and forget the charge, and an equation that balances by atoms and not by charge is still wrong.
The oxidation numbers confirm it: manganese drops from +7 to +2, five electrons each, ten in total across two atoms. Carbon rises from +3 to +4, one electron each, ten in total across ten atoms. Ten out and ten in.
The three kinds of redox reaction
Grade 11 sorts redox reactions into three shapes.
| Kind | What happens | Example |
|---|---|---|
| Synthesis | Substances combine into one product | 2Mg(s) + O2(g) → 2MgO(s) |
| Decomposition | One substance breaks into two or more | 2H2O(l) → 2H2(g) + O2(g) |
| Displacement | A more reactive element takes the place of a less reactive one | Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s) |
The trap in this section
Every displacement reaction is redox. Not every synthesis or decomposition reaction is.
Plenty of summaries state the classification the other way round, as though the three names were subdivisions of redox. They are not. They are shapes an equation can take, and each shape has redox and non-redox members.
| Reaction | Shape | Redox? |
|---|---|---|
| 2Mg + O2 → 2MgO | Synthesis | Yes. Mg 0 to +2, O 0 to −2 |
| CaO + H2O → Ca(OH)2 | Synthesis | No. Ca stays +2, O stays −2, H stays +1 |
| 2H2O → 2H2 + O2 | Decomposition | Yes. H +1 to 0, O −2 to 0 |
| CaCO3 → CaO + CO2 | Decomposition | No. Ca +2, C +4, O −2, all the way through |
| Zn + CuSO4 → ZnSO4 + Cu | Displacement | Yes, and displacement always is |
If a question gives you a synthesis or a decomposition and asks whether it is redox, it is testing exactly this. Check the numbers. Do not answer from the shape.
Disproportionation, briefly
Occasionally one substance is both oxidised and reduced in the same reaction. Hydrogen peroxide breaking down is the everyday example:
2H2O2(l) → 2H2O(l) + O2(g)
Oxygen starts at −1 in the peroxide. Some of it ends at −2 in the water, reduced, and some ends at 0 in the oxygen gas, oxidised. The same element, from the same bottle, going both ways.
It is beyond what Grade 11 is examined on, but it is worth thirty seconds of a lesson because it is the cleanest possible proof that oxidation numbers belong to atoms in a setting rather than to elements.
The practicals worth running
All three of these work, cost almost nothing and take a single period.
| Practical | What it shows | Time |
|---|---|---|
| Iron nail in copper(II) sulfate | Displacement. The nail goes copper-coloured and the blue fades | Set up in five minutes, read the next day |
| Zinc granules in dilute hydrochloric acid | Metal plus acid, and hydrogen collected and tested | 15 minutes |
| Electrolysis of acidified water | Decomposition, driven electrically, with two gases in a 2:1 ratio | 20 minutes |
Iron nail in copper(II) sulfate, in full
Apparatus: a test tube, a rack, a clean iron nail, and copper(II) sulfate solution of roughly 0,5 mol·dm−3. Any concentration that is clearly blue will do.
Method: clean the nail until it is bright, stand it in the test tube, pour in enough solution to cover it, and leave it somewhere it will not be knocked. Look at it the next day.
What you should see: a dull orange-brown coating of copper on the part of the nail that was submerged, and a solution that has faded from blue towards pale green. The green is iron(II) in solution, and pointing that out is worth doing because learners assume the colour has simply drained away.
Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s)
Iron goes 0 to +2 and is oxidised, so iron is the reducing agent. Copper goes +2 to 0 and is reduced, so copper(II) sulfate is the oxidising agent. The conclusion learners should write is that iron is more reactive than copper, and that they can see it because the reaction ran in that direction and not the other.
Worth doing as a control: stand a piece of copper wire in iron(II) sulfate solution alongside it. Nothing happens. One tube that reacts and one that does not is a far better lesson about reactivity than one tube on its own.
Zinc and hydrochloric acid, with the pop test
Quarter-fill a test tube with about 2 mol·dm−3 hydrochloric acid, add two or three clean zinc granules, and invert a second test tube over the mouth of the first to collect the hydrogen. The teacher does the match, not the class.
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
Zinc goes 0 to +2, hydrogen goes +1 to 0. The squeaky pop is the hydrogen burning in air, which is a second redox reaction happening on top of the first: 2H2 + O2 → 2H2O. Two reactions for the price of one, and it is worth saying so out loud.
If nothing fizzes, the granules are oxidised. Zinc that has sat in a cupboard grows a dull coating. Rub it bright first.
Electrolysis of acidified water
Two inverted test tubes full of water, standing over carbon electrodes in a beaker of water with a few drops of sulfuric acid in it, and a battery of at least 4,5 V.
What you should see: bubbles at both electrodes, and roughly twice as much gas over the electrode joined to the negative terminal. That larger volume is hydrogen. Oxygen collects at the positive side.
2H2O(l) → 2H2(g) + O2(g)
The acid is there only to make the water conduct, and learners regularly write it into the equation as a reactant. It is not one. It is a spectator, and the two-to-one ratio in the test tubes is the reason the equation has to have those coefficients.
Two demonstrations we will not recommend
Both of these appear in Grade 11 material and neither belongs in a school. We would rather say so plainly than have a teacher find out the hard way.
| The demonstration | Why not | Run this instead |
|---|---|---|
| Heating mercury(II) oxide to show decomposition | It produces mercury vapour, which is a cumulative neurotoxin with no safe classroom exposure. A test tube held in a flame has no containment at all, and disposing of the residue properly is beyond most schools | Hydrogen peroxide with a spatula of manganese(IV) oxide. Vigorous, visible, glowing-splint positive for oxygen, and everything on the bench is safe |
| Generating hydrogen sulfide to show a reducing agent | Hydrogen sulfide is lethal, and it deadens your sense of smell as the concentration rises, so the warning smell disappears exactly when the danger arrives. Instructions that say to take the class outside and stand well back are not a control measure | Vitamin C or sodium sulfite decolourising iodine solution. Same lesson, an obvious colour change, and nothing that can hurt anyone |
Neither substitution costs you any curriculum content. One is a decomposition, the other is a reducing agent in action, and the substitutes demonstrate both more clearly than the originals.
Where this fits in the curriculum
| Grade | 11, Physical Sciences |
| Term | Term 4 |
| Theme | Chemical change |
| Topic | Types of reactions |
| Time allocated | 4 hours for redox reactions and 2 hours for oxidation numbers |
| Assessment | Final examination, Paper 2. There is no prescribed formal practical for this topic |
| Changed by the current teaching plan | Balancing by oxidation number is replaced by balancing from the Table of Standard Reduction Potentials |
Six hours is not much for a topic this size, and the definitions and the oxidation-number rules are where the hours pay back. A learner who can assign oxidation numbers reliably can answer most of what a Paper 2 question will ask, even the parts they have not seen before.
It also sets up all of Grade 12 electrochemistry. Everything on this page reappears there with a voltmeter attached.
The mistakes that cost the marks
| The mistake | What to do instead |
|---|---|
| Saying the oxidising agent is oxidised | It gains electrons, so it is reduced. The agent is always the opposite of what happens to it |
| Applying oxygen equals −2 to a peroxide | In H2O2 and Na2O2 oxygen is −1. If the sum will not come out, suspect this first |
| Naming the atom as the agent | Name the substance. Not "manganese" but "potassium permanganate" |
| Making a polyatomic ion add up to zero | It adds up to the charge on the ion. Only a neutral molecule adds to zero |
| Calling a precipitation or a neutralisation redox | Check the numbers. Nothing changes in either, so neither is |
| Assuming every synthesis and decomposition is redox | Many are not. CaCO3 → CaO + CO2 changes nothing |
| Writing "clear" for "colourless" | They are different words. A permanganate solution that has been decolourised is colourless, and it was clear the whole time |
| Balancing the atoms and not the charge | Both sides must match on both. Add the charges up every time |
| Putting electrons on the wrong side | Left for a reduction, right for an oxidation |
How the 60 marks are made up
| Section | Marks |
|---|---|
| Multiple choice | 6 |
| Definitions and terminology | 10 |
| Assigning oxidation numbers | 14 |
| Identifying redox reactions | 8 |
| Oxidising and reducing agents | 8 |
| Half-reactions and balancing | 6 |
| The practical and its observations | 8 |
No mark allocation is prescribed for this topic. The split above is ours, weighted towards oxidation numbers and identification because that is what the curriculum keeps and what the examination keeps asking.
Free worksheet and marking memo
Both free, no sign up, straight to the PDF.
- Learner worksheet, 60 marks, with multiple choice, the definitions, fourteen marks of oxidation numbers to assign, redox identification, half-reactions and the nail practical
- Marking memorandum, with the working shown for every oxidation number and a note on the nine places learners drop marks
- Teacher guide, with how to spend the six hours, the three practicals and the two demonstrations we will not recommend
Related pages
- Electrochemical cells, Grade 12. Everything on this page again, with a voltmeter across it. This is where the Table of Standard Reduction Potentials properly belongs, and Grade 11 now borrows it
- Titration, Grade 12. The permanganate example above is a redox titration
- Endothermic and exothermic reactions, Grade 11. The other half of Chemical Change in the same year
Apparatus
This topic is unusually cheap to equip. Nothing here needs specialist glassware.
Borosilicate test tubes from R4 each are the only vessel any of the three practicals needs, and a wire test tube holder keeps fingers away from anything being warmed. Disposable pipettes in packs of 100 are what you want for adding solution drop by drop, and at that price a class can have one each.
For the electrolysis, carbon rod electrodes are the right choice because carbon is inert and will not join in the reaction. Electrodes for student cells cover the metals if you want to run the displacement work as a set of half-cells instead.
Goggles for the acid work are in the safety range.
One thing we cannot sell you is a bright zinc granule. Whatever you buy will have dulled by the time it reaches a classroom, so keep a sheet of fine abrasive paper with the zinc and rub a few bright at the start of the lesson. It is the difference between a reaction that fizzes and one that does not.