Redox Chemistry & Acid-Base Titrations
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Redox Chemistry & Acid–Base Titrations
The Big Idea: Every chemical reaction where electrons move (redox) can be tracked using oxidation numbers like a scorecard — and every acid-base titration is just a carefully measured "electron-free" reaction used to find an unknown concentration.
Quick Summary — Everything in this Chapter
- Oxidation numbers are a "bookkeeping charge" — the charge an atom would have if all bonding were 100% ionic.
- OIL RIG: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).
- Six simple rules let you calculate the oxidation number of any atom in any compound or ion.
- Roman numerals (Stock notation) show the oxidation state of a variable element, e.g. iron(III) oxide.
- An oxidising agent gets reduced itself (it "steals" electrons); a reducing agent gets oxidised itself (it "gives away" electrons).
- Disproportionation = the SAME element is oxidised AND reduced in one reaction.
- Half-equations show electron gain/loss separately; combine two half-equations (matching electrons) to build a full ionic equation.
- Acid-base indicators are weak acids/bases that change colour sharply at the equivalence point of a titration.
- Concentration = moles ÷ volume (in dm³). Titration calculations combine this with reaction stoichiometry.
- Uncertainties from multiple pieces of equipment (balance, pipette, burette, flask) are added together to get total experimental uncertainty.
1. Oxidation Numbers — Introduction
Think of oxidation number as a label tag you stick on every atom in a compound, telling you "how it would behave" if the compound were fully ionic. It's not a real charge in covalent molecules — it's a bookkeeping tool that lets chemists track electron movement without drawing every single bond.
Oxidation vs Reduction — Three Definitions
Depending on the type of reaction, oxidation and reduction can be defined in three equivalent ways:
| Oxidation | Reduction |
|---|---|
Addition of oxygen2Mg + O₂ → 2MgO | Loss of oxygen2CuO + C → 2Cu + CO₂ |
Loss of hydrogenCH₃OH → CH₂O + H₂O | Addition of hydrogenC₂H₄ + H₂ → C₂H₆ |
Loss of electronsAl → Al³⁺ + 3e⁻ | Gain of electronsF₂ + 2e⁻ → 2F⁻ |
Oxidation Numbers of Simple Ions
For a simple (monatomic) ion, the oxidation number is literally just the charge on the ion. No calculation needed.
| Type | Example | Why |
|---|---|---|
| Atoms | Na in Na = 0 | Neutral already — no electrons need adding or removing |
| Cations | Na in Na⁺ = +1 | Need to add 1 electron to make Na⁺ neutral |
| Anions | Cl in Cl⁻ = −1 | Need to take 1 electron away to make Cl⁻ neutral |
Q: What are the oxidation numbers of the elements in: (a) C (b) Fe³⁺ (c) Fe²⁺ (d) O²⁻ (e) He (f) Al³⁺?
A: a) 0 b) +3 c) +2 d) −2 e) 0 f) +3
The Six Oxidation Number Rules
| # | Rule | Example |
|---|---|---|
| 1 | An uncombined element's oxidation number is zero | H₂, Zn, O₂ all = 0 |
| 2 | Many atoms have fixed oxidation numbers in compounds | Group 1 = +1, Group 2 = +2, F = −1 always. H = +1 (except metal hydrides, −1). O = −2 (except peroxides −1, and OF₂ where it's +2) |
| 3 | A mono-atomic ion's oxidation number = its charge | Zn²⁺ = +2, Fe³⁺ = +3, Cl⁻ = −1 |
| 4 | The sum of oxidation numbers in a neutral compound = 0 | NaCl: (+1) + (−1) = 0 |
| 5 | The sum of oxidation numbers in an ion = the charge on that ion | SO₄²⁻: S(+6) + 4×O(−2) = −2 |
| 6 | The more electronegative element gets the negative oxidation number | F₂O: F = −1 (×2), O = +2 |
Roman Numerals (Stock Notation)
Transition metals have variable oxidation numbers, so we write the oxidation state in Roman numerals after the name — this is called Stock Notation (after chemist Alfred Stock). It's mainly used for metals; non-metals like SO₂ are usually just called "sulfur dioxide" rather than "sulfur(IV) oxide."
- Fe²⁺ in FeO → iron(II) oxide
- Fe³⁺ in Fe₂O₃ → iron(III) oxide
- Potassium manganate(VII), KMnO₄ → Mn is +7 (K⁺ = +1, four O = −8, so Mn must balance to give overall −1 charge on MnO₄⁻)
Q: Deduce the oxidation number of the underlined element: (1) NO₂⁻ (2) NO₃⁻ (3) S₂O₃²⁻ (4) S₄O₆²⁻
- 2 oxygens = 2×(−2) = −4. Overall charge is −1, so N + (−4) = −1 → N = +3
- 3 oxygens = 3×(−2) = −6. Overall charge is −1, so N + (−6) = −1 → N = +5
- 3 oxygens = −6. Overall charge is −2, so 2S + (−6) = −2 → 2S = +4 → S = +2
- 6 oxygens = −12. Overall charge is −2, so 4S + (−12) = −2 → 4S = +10 → S = +2.5 (a fractional average — the 4 sulfur atoms sit in two different chemical environments, so this is the average, not a "real" number for any single atom)
What is the oxidation number of chromium in the dichromate ion, Cr₂O₇²⁻?
Name Na₂Cr₂O₇ using Stock notation, and explain why the prefix "di" is used before "chromate."
2. Oxidation & Reduction — Electron Transfer
Once you can calculate oxidation numbers, you can instantly spot oxidation and reduction in any equation — just track how the numbers change.
2NH₃ + 3Br₂ → N₂ + 6HBr
- N in NH₃ changes from −3 to 0 → oxidation number increased → nitrogen is oxidised
- Br changes from 0 to −1 → oxidation number decreased → bromine is reduced
Oxidising Agents vs Reducing Agents
This is the part students most often get backwards, so let's be really precise about it:
2Fe²⁺ + H₂O₂ + 2H⁺ → 2Fe³⁺ + 2H₂O
- Fe: +2 → +3 (increased → oxidised)
- O in H₂O₂: −1 → −2 (decreased → reduced)
2Fe³⁺ + H₂O₂ + 2OH⁻ → 2Fe²⁺ + 2H₂O + O₂
- Fe: +3 → +2 (decreased → reduced)
- O in H₂O₂: −1 → 0 (increased → oxidised)
In the reaction Mg + Fe²⁺ → Mg²⁺ + Fe, which species is acting as the oxidising agent, and why?
3. Disproportionation Reactions
Cl₂ + H₂O → HCl + HClO
- Cl₂ starts at oxidation number 0
- One Cl atom ends up at −1 in HCl (reduced)
- The other Cl atom ends up at +1 in HClO (oxidised)
Balancing a Disproportionation Equation — Step by Step
Balance: Cl₂ + OH⁻ → Cl⁻ + ClO₃⁻ + H₂O
- Write the unbalanced equation and identify atoms that change oxidation number: Cl goes from 0 → −1 (in Cl⁻) and 0 → +5 (in ClO₃⁻)
- Deduce the oxidation number changes: one Cl changes by −1, the other by +5
- Balance the oxidation number changes — the total increase must equal the total decrease. Change of −1 (×5) balances change of +5 (×1):
Cl₂ + OH⁻ → 5Cl⁻ + ClO₃⁻ + H₂O - Balance the charges: Reactant side charge = OH⁻ contributes 1− per ion. Product side = 5(1−) + 1(1−) = 6−. So we need 6 OH⁻ ions to balance:
Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + H₂O - Balance the atoms (Cl and H/O last): final answer needs 3Cl₂ to give 6 Cl atoms total (5 as Cl⁻ + 1 as ClO₃⁻), and 3H₂O to balance the 6 H atoms from 6OH⁻
In the reaction 3ClO⁻ → 2Cl⁻ + ClO₃⁻, is this a disproportionation reaction? Justify your answer using oxidation numbers.
4. Ionic Equations & Half-Equations
Naming Compounds with Oxidation Numbers
For simple two-element compounds, naming is straightforward (PCl₃ = phosphorus(III) chloride, PCl₅ = phosphorus(V) chloride). For more complex compounds involving elements with variable oxidation states, Roman numerals pin down exactly which oxidation state is present — e.g. K₂CrO₄ is potassium chromate(VI).
- Cu₂O → O is −2, compound neutral, so 2Cu = +2 → Cu = +1 → copper(I) oxide
- MnSO₄ → sulfate is −2, so Mn = +2 → manganese(II) sulfate
- Na₂CrO₄ → 2Na = +2, CrO₄ must be −2 overall → Cr = +6 → sodium chromate(VI)
- KMnO₄ → K = +1, MnO₄ must be −1 overall → Mn = +7 → potassium manganate(VII)
- Na₂Cr₂O₇ → 2Na = +2, Cr₂O₇ must be −2 overall → Cr = +6 → sodium dichromate(VI)
Metals vs Non-Metals Forming Ions
| Behaviour | Oxidation number change | Example | |
|---|---|---|---|
| Metals | Form positive ions by losing electrons | Increases (oxidised) | 2Na + 2H₂O → 2NaOH + H₂ (Na: 0→+1) |
| Non-metals | Form negative ions by gaining electrons | Decreases (reduced) | 4Na + O₂ → 2Na₂O (O: 0→−2) |
Constructing Half-Equations
A half-equation shows only "half" of a redox reaction — just the electron gain or loss for one species. Some are simple (like Pb²⁺ + 2e⁻ → Pb), but others need water and H⁺ added to balance oxygen and hydrogen atoms first.
Step 2: Add H₂O to balance O atoms
Step 3: Add H⁺ to balance H atoms
Step 4: Add e⁻ to balance the charge
Dichromate ions (Cr₂O₇²⁻) react with Fe²⁺ in acid, forming Cr³⁺ and Fe³⁺.
Half-equation 1 (oxidation): Fe²⁺ → Fe³⁺ + e⁻
Half-equation 2 (reduction) — built step by step:
- Balance Cr:
Cr₂O₇²⁻ → 2Cr³⁺ - Add H₂O for O:
Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O - Add H⁺ for H:
Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O - Add e⁻ for charge:
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Combine: multiply half-equation 1 by 6 so electrons match (6e⁻ on each side), then add together and cancel the electrons:
Balancing Full Ionic Equations Directly (Without Half-Equations)
MnO₄⁻ reacts with Fe²⁺ in acid to form Mn²⁺, Fe³⁺ and water.
- Identify oxidation number changes: Mn: +7 → +2 (change of −5). Fe: +2 → +3 (change of +1)
- Balance the changes: 1×(−5) must equal 5×(+1), so we need 5 Fe²⁺ for every 1 MnO₄⁻:
MnO₄⁻ + 5Fe²⁺ + H⁺ → Mn²⁺ + 5Fe³⁺ + H₂O - Balance charges (ignoring H⁺ for now): reactant charge = (5×2+)+(1−) = 9+. Product charge = (2+)+5×(3+) = 17+. Difference = 8, so we need 8H⁺:
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + ... - Balance atoms (O and H last): 8H⁺ gives us 4H₂O on the product side
Construct the half-equation for the reduction of MnO₄⁻ to Mn²⁺ in acidic solution using the 4-step method.
5. Acid–Base Titrations with Indicators
A titration is essentially a very precise way of asking "how much of solution A exactly cancels out solution B?" You use an indicator to spot the exact moment the reaction is complete — the equivalence point.
Acid-Base Indicators
Indicators are themselves weak acids or bases, and their conjugate acid/base forms have different colours. As pH changes, the indicator flips from one form to the other, giving a visible colour change.
| Indicator | Colour in Acid | Colour in Alkali |
|---|---|---|
| Litmus | Red | Blue |
| Methyl Orange | Red | Yellow |
| Phenolphthalein | Colourless | Pink |
Concentration Calculations
25.0 cm³ of 0.050 mol dm⁻³ sodium carbonate was completely neutralised by 20.00 cm³ of dilute hydrochloric acid. Find the concentration of the HCl.
- Write the balanced equation:
Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂ - Moles of Na₂CO₃: 0.025 dm³ × 0.050 mol dm⁻³ = 0.00125 mol
- Use stoichiometry (1:2 ratio): 0.00125 mol Na₂CO₃ reacts with 2 × 0.00125 = 0.00250 mol HCl
- Find concentration of HCl: 0.00250 mol ÷ 0.0200 dm³
A student dissolved 10 g of NaOH in 2 dm³ of distilled water. Find the concentration.
30.0 cm³ of 0.100 mol dm⁻³ NaOH exactly neutralises 25.0 cm³ of hydrochloric acid (1:1 ratio, NaOH + HCl → NaCl + H₂O). Find the concentration of the HCl.
Uncertainty Calculations
- Adding/subtracting readings: when you read an instrument twice (e.g. initial and final burette readings), you add the absolute uncertainties together, since each reading could be "out" by the stated amount.
- Multiple pieces of equipment: add up each instrument's individual percentage uncertainty to get the total experimental uncertainty.
A titration uses: balance (±0.5%), volumetric flask (±0.1%), volumetric pipette (±0.2%), burette (±0.4%).
What to Memorise
Concepts Checklist
Exam Tips & Common Mistakes
- 2. Oxidation & Reduction — Electron Transfer
- 4. Ionic Equations & Half-Equations
- Exam Tips & Common Mistakes
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