IB Chemistry SLTopic 6 — Electron TransferPaper 1 & 2Core idea~12 min read
Oxidation and Reduction
Proton transfer had one moving part. So does this one — except now it is electrons that move, and the bookkeeping device that tracks them is the oxidation number. Get comfortable with that and redox stops being guesswork.
📚 What you need to know
Oxidation is loss of electrons; reduction is gain of electrons. OIL RIG.
The two always happen together — that is why we call it a redox reaction.
Oxidation number increases when a species is oxidised and decreases when it is reduced.
An oxidising agent accepts electrons and is itself reduced.
A reducing agent donates electrons and is itself oxidised.
Quicker checks: oxidation is also the gain of oxygen or the loss of hydrogen.
Roman numerals in names give the oxidation state: iron(II) is Fe2+, manganate(VII) contains Mn at +7.
Following the electrons
Drop a strip of magnesium into blue copper(II) sulfate solution and within minutes the blue fades and the strip is coated in brown copper. Two electrons have moved from each magnesium atom to a copper ion.
This is the single most-confused point in the topic. The reducing agent does the reducing, so it must be the one giving electrons away — which means it is the one being oxidised.
The definitionsOxidation Is Loss · Reduction Is Gain — of electrons
Oxidation numbers
Electrons are not always transferred cleanly. In a covalent molecule they are shared, unevenly, and nothing has actually gained or lost a whole electron. The oxidation number is an accounting fiction that lets us track redox anyway: pretend every bond is fully ionic, and count what each atom would have.
Rule
Value
Watch out for
An uncombined element
0
includes O2, Cl2, S8 and every metal
A simple monatomic ion
equal to its charge
Fe3+ is +3, S2– is –2
Group 1 / group 2 / aluminium
+1 / +2 / +3
always, in any compound
Fluorine
–1
no exceptions at all
Hydrogen
+1
–1 in metal hydrides such as NaH
Oxygen
–2
–1 in peroxides such as H2O2
Sum over a neutral compound
0
this is what lets you find an unknown
Sum over a polyatomic ion
the charge on the ion
–2 for SO42–, –1 for NO3–
Notice that +7 is not a charge on the manganese — there is no Mn7+ sitting in the ion. It is a bookkeeping number, and that is all it needs to be.
Reading the change
Once every atom has a number, redox becomes something you can see at a glance. Compare each element on the left with the same element on the right: up is oxidation, down is reduction.
Neutralisation, precipitation and most acid–base chemistry sit still on this scale. That is precisely why they are not redox reactions.
Agents
Oxidising agent
Reducing agent
Does what to the other species
oxidises it
reduces it
Electrons
accepts them
donates them
What happens to itself
is reduced
is oxidised
Its own oxidation number
decreases
increases
Typical examples
O2, Cl2, MnO4–, Cr2O72–
metals, H2, CO, I–
Some species appear in both columns depending on the company they keep. Hydrogen peroxide is the classic: against Fe2+ it takes electrons and acts as an oxidising agent, while against Fe3+ it gives them up and acts as a reducing agent. Its oxygen sits at –1, halfway between the –2 of water and the 0 of O2, so it has room to move in either direction.
WORKED EXAMPLE
Deduce the oxidation state of the underlined element in each species. (a) Cr2O72– (b) H2SO4 (c) NaH (d) H2O2
(a) dichromate2Cr + 7(−2) = −2, so 2Cr = +12Cr = +6(b) sulfuric acid2(+1) + S + 4(−2) = 0S = +6(c) sodium hydrideSodium is group 1, so it must be +1, and the compound is neutral.H = −1(d) hydrogen peroxide2(+1) + 2O = 0O = −1The last two are the standard exceptions. If your answer for H or O comes out unusual, check whether you are looking at a hydride or a peroxide before assuming you are wrong.
WORKED EXAMPLE
For the extraction of iron in the blast furnace, identify what is oxidised, what is reduced, and name the oxidising and reducing agents. Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)
Step 1 — number every elementFe: +3 → 0 C: +2 → +4 O: −2 throughoutStep 2 — read the direction of travelIron falls from +3 to 0, so it is reduced. Carbon climbs from +2 to +4, so it is oxidised.Fe₂O₃ is the oxidising agentCO is the reducing agentName the whole species, not just the atom. The reducing agent is carbon monoxide, not “carbon”.
WORKED EXAMPLE
In which of these is the species in bold acting as an oxidising agent? A Cl2 + 2Br– → 2Cl– + Br2 B Zn + Cu2+ → Zn2+ + Cu C H2 + CuO → Cu + H2O
A — chlorineCl: 0 → −1, so it gained electronsA is the oxidising agentB — zincZn: 0 → +2, so it lost electronsZinc is oxidised, which makes it the reducing agent.C — hydrogenH: 0 → +1, again a lossHydrogen is oxidised, so it too is a reducing agent. Only A gains electrons.
💡 Exam tip
Assign oxidation states to every element on both sides. It takes fifteen seconds and removes all the guesswork.
Define oxidation and reduction in terms of electrons, not oxygen, unless the question is about combustion.
Name the whole species as the agent, with its formula.
Remember the agent is the opposite of what happens to it. Say it out loud if you have to.
Watch for H in hydrides (–1) and O in peroxides (–1). Examiners include them deliberately.
Write oxidation states as +2, with the sign first, to distinguish them from the charge 2+.
⚠️ Common mix-up
Swapping the agents. The reducing agent is the one that gets oxidised.
Assuming oxygen is always –2 and hydrogen always +1.
Treating the oxidation number as a real charge. The +7 in MnO4– is a bookkeeping figure.
Calling every reaction redox. If no oxidation number changes, it is not.
Forgetting to multiply by the number of atoms when solving for an unknown, e.g. the two chromiums in Cr2O72–.
Up next: Writing Half-Equations — you can now say which way the electrons went. The next step is writing them into the equation explicitly, which is how every serious redox calculation begins.
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