2.3Unit 2. Energetics, Group Chemistry, Halogenoalkanes and Alcohols
Redox Chemistry & Acid-Base Titrations — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions
5 min read 13 parts Lesson 3 of 53 1 quiz
Curriculum Equation Key pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: pH + pOH = 14
Lesson summary
2.3 Redox Chemistry & Acid-Base Titrations — oxidation numbers, ionic equations and titrations — IAL EDEXCEL LINEAR (CHEMISTRY) notes: Assign oxidation numbers, identify oxidation, reduction and disproportionation, construct half-equations and ionic equations, and perform and int…
- Oxidising agent
- Reducing agent
- Fractional values
- Equivalence point
- Concordant titres
- Standard solution
Part 1 of 11Oxidation and reduction
OIL RIG
- OIL RIG oxidation is loss of electrons and reduction is gain of electrons.
- Oxidation is also an increase in oxidation number and reduction a decrease.
In organic chemistry oxidation is often gain of oxygen or loss of hydrogen, and reduction is the reverse.
Oxidising agent
- Oxidising agent oxidises another species by taking its electrons and is itself reduced, so its oxidation number decreases.
Reducing agent
- Reducing agent reduces another species by giving it electrons and is itself oxidised, so its oxidation number increases.
A redox reaction needs both an oxidising agent and a reducing agent; some substances can act as either depending on what they react with.
2NH3 + 3Br2 → N2 + 6HBr
- Nitrogen goes from −3 to 0 and is oxidised.
- Bromine goes from 0 to −1 and is reduced.
Part 2 of 11Rules for oxidation numbers
Oxidation number
- Oxidation number the charge an atom would have if the bonding were completely ionic, used to track electron transfer.
It refers to a single atom, and in a simple ion it equals the ion charge.
| Species | Oxidation number | Exception |
|---|---|---|
| Uncombined element | 0 | None |
| Oxygen in compounds | -2 | -1 in peroxides; +2 in OF2 |
| Hydrogen in compounds | +1 | -1 in metal hydrides, e.g. NaH |
| Group 1 / Group 2 metals | +1 / +2 | None |
| Fluorine | -1 | None |
| Halogens in halides | -1 | Positive when bonded to O or a more electronegative halogen |
Rules for oxidation numbers
Oxidation numbers sum to 0 in a neutral compound and to the charge in an ion.
The more electronegative atom takes the negative value, e.g. oxygen in CO2.
Part 3 of 11Calculating oxidation numbers
Ions
- In SO42−: S + 4(−2) = −2, so S is +6.
- In Cr2O72−: 2Cr = +12, so Cr is +6.
- In NO3− N is +5 and in NO2− N is +3.
- In S2O32− S is +2.
Hydrides and peroxides
- In NaH hydrogen is −1 because sodium is +1.
- In H2O2 oxygen is −1 because hydrogen is +1.
Fractional values
- S4O62− gives S = +2.5.
- This is an average because the sulfur atoms are in two different environments.
Part 4 of 11Naming with Roman numerals (Stock notation)
1
Roman numerals in a name give the oxidation number of an element with variable oxidation states.
2
Fe2+ in FeO is iron(II) oxide and Fe3+ in Fe2O3 is iron(III) oxide.
3
Potassium manganate(VII), KMnO4: four O at −2 gives −8, the ion is −1, so Mn is +7.
4
Other examples: Cu2O is copper(I) oxide, MnSO4 is manganese(II) sulfate, Na2CrO4 is sodium chromate(VI) and Na2Cr2O7 is sodium dichromate(VI).
5
PCl3 is phosphorus(III) chloride and PCl5 is phosphorus(V) chloride.
Part 5 of 11Metals, non-metals and disproportionation
Metals form positive ions by losing electrons, so they are oxidised, e.g. Na goes 0 → +1 and Mg goes 0 → +2.
Non-metals form negative ions by gaining electrons, so they are reduced, e.g. oxygen goes 0 → −2 in Na2O.
Disproportionation
- Disproportionation the same species is simultaneously oxidised and reduced.
- Cl2 + 2NaOH → NaCl + NaClO + H2O, with chlorine going 0 → −1 and 0 → +1 (cold dilute alkali).
- Hot concentrated alkali3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O, with chlorine going to −1 and +5.
- Cu+ in water: 2Cu+ → Cu + Cu2+.
To balance a redox equation by oxidation numbers
- Write the unbalanced equation and find the atoms that change.
- Work out the oxidation number changes.
- Balance the oxidation number changes.
- Balance the charges.
- Balance the atoms.
Part 6 of 11Half-equations and ionic equations
- Half-equation
- shows only the gain or loss of electrons, e.g. Pb2+ + 2e− → Pb and 2Br− → Br2 + 2e−.
1
Write the unbalanced half-equation and balance the atom being oxidised or reduced.
2
Add H2O to balance O atoms.
3
Add H+ to balance H atoms.
4
Add e− to balance charge.
Example
- Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O.
Combine half-equations by multiplying so that the electrons are equal, adding and cancelling.
- 6Fe2+ + Cr2O72− + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O.
- MnO4− + 8H+ + 5e− → Mn2+ + 4H2O.
Ionic equation
- Ionic equation shows only the species that react.
- Write the full equation and split soluble ionic compounds into ions;
- keep solids, gases, water and weak acids whole.
- Cancel spectator ions; atoms and charges must both balance, e.g. Ag+(aq) + Cl−(aq) → AgCl(s).
Neutralisation ionic equation: H+(aq) + OH−(aq) → H2O(l).
Part 7 of 11Indicators
- Indicators are weak acids or bases whose conjugate forms have different colours;
- the colour changes reversibly with H+ concentration.
| Indicator | Colour in acid | Colour in alkali |
|---|---|---|
| Litmus | red | blue |
| Methyl orange | red | yellow |
| Phenolphthalein | colourless | pink |
Indicators
A good indicator changes colour sharply at the equivalence point, so its range must lie within the vertical section of the titration curve.
Litmus and universal indicator change gradually and are unsuitable.
With phenolphthalein, put the base in the burette so the first permanent pink is easy to see.
| Titration | pH at equivalence | Suitable indicator |
|---|---|---|
| Strong acid + strong base | about 7 | Methyl orange or phenolphthalein |
| Weak acid + strong base | about 9 | Phenolphthalein |
| Strong acid + weak base | about 5 | Methyl orange |
| Weak acid + weak base | no sharp change | None suitable |
Part 8 of 11Acid-base titration procedure
1
Rinse the burette with the titrant and the pipette with the solution to be analysed.
2
Pipette 25.0 cm3 of solution into a conical flask and add 2–3 drops of indicator.
3
Fill the burette, remove the funnel and record the initial reading.
4
Run in titrant while swirling, adding dropwise near the end point until a permanent colour change.
5
Record the final reading and repeat until two titres are concordant (within 0.10 cm3).
6
Calculate the mean titre from the concordant results.
Do a rough titration first, use a white tile and read the bottom of the meniscus at eye level.
Equivalence point
- Equivalence point the point where the reactants have reacted in the stoichiometric ratio;
- the end point is where the indicator changes colour.
Part 9 of 11Concentration calculations
Concentration calculations
Concentration (mol dm−3) = moles of solute ÷ volume of solution (dm3).
Moles = concentration × volume (dm3), so convert cm3 to dm3 by dividing by 1000.
Mass of solute (g) = moles × molar mass; concentration in g dm−3 = mass ÷ volume.
1
Write the balanced equation.
2
Calculate the moles of the known solution.
3
Use the mole ratio to find the moles of the other reagent.
4
Divide by its volume (dm3) to find the concentration.
Part 10 of 11Uncertainty in titrations
1
Percentage uncertainty (%) = absolute uncertainty ÷ measured value × 100.
2
Readings taken twice (burette initial and final, thermometer, balance before and after) double the absolute uncertainty.
3
A burette reading is ±0.05 cm3, so a titre is ±0.10 cm3.
4
Total experimental uncertainty is the sum of the percentage uncertainties of each piece of equipment.
- Balance, volumetric flask, pipette and burette add together, e.g. 0.5 + 0.1 + 0.2 + 0.4 = ±1.2%.
Part 11 of 11More from the specification
- Standard solution
- A solution whose concentration is known accurately.
Key terms
| Term | Meaning |
|---|---|
| Oxidation number | The charge an atom would have if all its bonds were ionic. |
| Oxidation | Loss of electrons, or an increase in oxidation number. |
| Reduction | Gain of electrons, or a decrease in oxidation number. |
| Oxidising agent | A species that accepts electrons and is itself reduced. |
| Disproportionation | A reaction in which the same species is simultaneously oxidised and reduced. |
| Half-equation | An equation showing only the oxidation or only the reduction with electrons. |
| Equivalence point | The point where reactants have reacted in the exact stoichiometric ratio. |
| Concordant titres | Titres that agree within 0.10 cm3. |
Avoid these mistakes
1
Forgetting that oxygen is −1 in peroxides — H2O2 has H +1 and O −1.
2
Treating all oxidation numbers as whole numbers — averages such as +2.5 occur.
3
Cancelling the wrong ions in an ionic equation — only spectator ions cancel and charge must balance.
4
Using universal indicator in a titration — it changes gradually.
5
Counting uncertainty once for a burette titre — two readings double it.
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