IB Chemistry SLTopic 6 — Electron TransferPaper 1 & 2Core skill~13 min read
Writing Half-Equations
A full redox equation hides the interesting part. Split it in two and the electrons appear — how many moved, and in which direction. That is the number every redox calculation is built on.
📚 What you need to know
A half-equation shows one half of a redox reaction, with the electrons written in.
Electrons appear on the left for reduction and on the right for oxidation.
Build one in order: balance the main atom, then O with H2O, then H with H+, then charge with e–.
To combine two halves, multiply each so the electron numbers match, then add and cancel.
The electrons must cancel completely — none may appear in the overall equation.
Manganate(VII) titrations are self-indicating; iodine–thiosulfate titrations use starch.
Why bother splitting it up
Take the reaction between acidified manganate(VII) ions and iron(II) ions. Written in full it is a formidable-looking equation with a 5 and an 8 in it, and there is no obvious way to arrive at those numbers by trial and error. Written as two halves, each one is straightforward, and the coefficients fall out of the electron count.
The two halves
oxidation: Fe2+ → Fe3+ + e– reduction: MnO4– + 8H+ + 5e– → Mn2+ + 4H2O
Building a half-equation
The order matters. Balance the element that is actually changing first, and leave the charge until last — the electrons are what fix it.
Step 4 is the only one that requires thought, and even then it is arithmetic: work out the total charge on each side and add electrons until they match.
🧩 The method, every time
Balance the element being oxidised or reduced.
Balance oxygen by adding H2O to the side that needs it.
Balance hydrogen by adding H+.
Add e– to whichever side makes the total charges equal.
Check: atoms balanced, charges balanced. Electrons on the left means reduction.
These rules assume acidic conditions, which is why H+ is available. That is not a technicality — manganate(VII) titrations are acidified with dilute sulfuric acid for exactly this reason. Hydrochloric acid is avoided because the chloride ions would themselves be oxidised.
Putting the halves together
One iron(II) ion releases one electron. One manganate(VII) ion consumes five. So five iron(II) ions are needed for every manganate(VII) ion — and there is the 5 that looked so arbitrary in the full equation.
Check the overall equation both ways: the atoms must balance, and so must the total charge. Here it is +17 on each side.
Notice where the 1 : 5 ratio came from. It is not in either formula and you cannot see it in the full equation — it comes purely from the electron counts. That ratio is what you will use in every redox titration calculation, so find it from the half-equations rather than trying to remember it.
Redox titrations
A redox titration works exactly like an acid–base one, except an oxidising agent is titrated against a reducing agent. The awkward part is spotting the end point, and there are two standard solutions to that problem.
Manganate(VII) is deep purple and its product Mn2+ is almost colourless, so the solution itself does the indicating. Iodine needs help, and starch provides it.
The two standard systems
MnO4– + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O 2S2O32– + I2 → 2I– + S4O62–
WORKED EXAMPLE
Construct the half-equation for the reduction of dichromate(VI) ions, Cr2O72–, to Cr3+ ions in acidic solution.
Step 1 — balance the chromiumCr₂O₇²⁻ → 2Cr³⁺Step 2 — seven oxygens need seven watersCr₂O₇²⁻ → 2Cr³⁺ + 7H₂OStep 3 — fourteen hydrogensCr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂OStep 4 — balance the chargeLeft is −2 + 14 = +12. Right is 2 × (+3) = +6. Six electrons are needed on the left.Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂OSanity check with oxidation numbers: Cr goes from +6 to +3, a drop of 3, and there are two of them — six electrons.
WORKED EXAMPLE
Combine the half-equations below into a balanced overall equation. Zn(s) → Zn2+(aq) + 2e– Ag+(aq) + e– → Ag(s)
Step 1 — match the electronsThe zinc half releases two; the silver half uses one. Multiply the silver half by 2.2Ag⁺ + 2e⁻ → 2AgStep 2 — add and cancelZn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s)Step 3 — checkAtoms balance, and the charge is +2 on each side. No electrons remain, which is the sign you have done it correctly.
WORKED EXAMPLE
A 1.50 g iron supplement tablet containing iron(II) sulfate was dissolved in dilute sulfuric acid and titrated against 0.0200 mol dm–3 potassium manganate(VII). The titre was 24.60 cm3. Calculate the percentage by mass of iron in the tablet. (Ar Fe = 55.85)
Step 1 — moles of manganate(VII)n = 0.02460 × 0.0200 = 4.92 × 10⁻⁴ molStep 2 — use the 1 : 5 ratioOne MnO₄⁻ takes five electrons, and each Fe²⁺ gives one.n(Fe²⁺) = 5 × 4.92 × 10⁻⁴ = 2.46 × 10⁻³ molStep 3 — convert to massm = 2.46 × 10⁻³ × 55.85 = 0.1374 gStep 4 — as a percentage(0.1374 ÷ 1.50) × 1009.16 % iron by massThe 1 : 5 ratio is the whole calculation. Get it from the half-equations, not from memory, and the rest is ordinary mole arithmetic.
💡 Exam tip
Follow the order: main atom, then oxygen, then hydrogen, then charge. Out of order it becomes guesswork.
Add H2O and H+, never oxygen atoms or hydrogen molecules.
Check the total charge on both sides before you write the final answer.
In the overall equation, no electrons should appear. If any survive, your multipliers are wrong.
For titration calculations, state the ratio you are using and where it came from.
Say that manganate(VII) titrations are self-indicating, and that starch is added late in an iodine titration.
⚠️ Common mix-up
Putting the electrons on the wrong side. Reduction takes them in, so they sit with the reactants.
Balancing oxygen with O atoms instead of water molecules.
Forgetting to multiply the whole half-equation, only scaling part of it.
Using a 1 : 1 ratio in a manganate titration out of habit.
Acidifying with hydrochloric acid, which would itself be oxidised.
Adding starch at the start of an iodine–thiosulfate titration.
Up next: Relative Ease of Redox — you can write any half-equation now, but not every pair of them actually happens. The next question is which species wins the tug of war for the electrons.
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