2.5Unit 2. Energetics, Group Chemistry, Halogenoalkanes and Alcohols

Redox Chemistry & Acid-Base Titrations — IAL Edexcel Modular (Chemistry) Revision Notes & Topical Questions

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2.5 Redox Chemistry & Acid-Base Titrations — IAL EDEXCEL MODULAR (CHEMISTRY) notes: Assign oxidation numbers, identify oxidation, reduction and disproportionation, combine half-equations into ionic equations, and carry out and calculate acid–base titrations.Oxidation = loss of el…

Non-metals
Exceptions
Fractional values
Stock notation
Oxidising agent
Reducing agent

Section 1 of 12Oxidation and reduction

Oxidation

Oxidation loss of electrons (OIL RIG), an increase in oxidation number, or gain of oxygen or loss of hydrogen

Reduction

Reduction gain of electrons, a decrease in oxidation number, or loss of oxygen or gain of hydrogen

Metals

Metals form positive ions by losing electrons, so they are oxidised and their oxidation number increases
Na goes from 0 to +1 in NaOH and Mg goes from 0 to +2 in MgCl₂.

Non-metals

Non-metals form negative ions by gaining electrons, so they are reduced and their oxidation number decreases
Oxygen goes from 0 to −2 in Na₂O.

Section 2 of 12Oxidation number rules

Oxidation number the charge an atom would have if its bonding were completely ionic

The oxidation number of an uncombined element is 0.

In a simple ion, the oxidation number equals the charge on the ion.

The oxidation numbers in a neutral compound add up to 0, and in an ion they add up to the overall charge.

Fluorine is always −1, oxygen is usually −2 and hydrogen is usually +1.

The more electronegative element takes the negative value.

Exceptions

Exceptions hydrogen is −1 in metal hydrides such as NaH, and oxygen is −1 in peroxides such as H₂O₂

Fractional values

Fractional values can occur when atoms of one element are in different environments
Sulfur in S₄O₆²⁻ has an average oxidation number of +2.5.

Section 3 of 12Naming with oxidation numbers

Stock notation

Stock notation the oxidation number is written in Roman numerals after elements with variable oxidation numbers
Fe₂O₃ is iron(III) oxide and PCl₅ is phosphorus(V) chloride.

Oxyanions

Oxyanions the Roman numeral gives the oxidation number of the central atom
Potassium manganate(VII) contains MnO₄⁻, in which manganese is +7.

Others

K₂CrO₄ is potassium chromate(VI), Na₂Cr₂O₇ is sodium dichromate(VI) and Cu₂O is copper(I) oxide

Section 4 of 12Oxidising and reducing agents

Oxidising agent

Oxidising agent a substance that oxidises another species by taking electrons from it

It is itself reduced and its oxidation number decreases.

Reducing agent

Reducing agent a substance that reduces another species by giving it electrons

It is itself oxidised and its oxidation number increases.

Redox reaction

Redox reaction needs both an oxidising agent and a reducing agent, so electrons are transferred from the reducing agent to the oxidising agent

Example

Example in Zn + Cu²⁺ → Zn²⁺ + Cu, Zn is oxidised (0 to +2) and Cu²⁺ is reduced (+2 to 0)

Section 5 of 12Disproportionation

1Disproportionation

Disproportionation a reaction in which the same species is simultaneously oxidised and reduced

2Chlorine with cold dilute sodium hydroxide

Chlorine with cold dilute sodium hydroxide gives chloride (−1) and chlorate(I) (+1)
Cl₂ + 2NaOH → NaCl + NaClO + H₂O.

3Chlorine with hot concentrated sodium hydroxide

Chlorine with hot concentrated sodium hydroxide gives chloride and chlorate(V)
3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O.

4Copper(I) in water

2Cu⁺ → Cu + Cu²⁺

Section 6 of 12Half-equations

Half-equation
- shows only the species gaining or losing electrons
- Examples are Pb²⁺ + 2e^- → Pb and 2Br^- → Br₂ + 2e^-.

Write the species that is oxidised or reduced and balance the atom that changes.

Add H₂O to balance oxygen atoms.

Add H⁺ to balance hydrogen atoms.

Add e^- to balance the charge.

Example MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O, and Cr₂O₇²⁻ + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O

Section 7 of 12Constructing ionic equations

Combining half-equations
- multiply so that the electrons lost equal the electrons gained, add them, then cancel the electrons and species on both sides

Write the unbalanced equation and identify the atoms that change oxidation number.

Deduce the oxidation number changes.

Balance the oxidation number changes.

Balance the charges.

Balance the atoms.

Ionic equations
- split soluble ionic compounds into ions, but not solids, gases, water or weak acids, and cancel spectator ions
- Both atoms and charges must balance, as in Ag⁺(aq) + Cl^-(aq) → AgCl(s).

Section 8 of 12Acid–base titration method

Purpose
- finds the unknown concentration of an acid or base using a standard solution of known concentration

Rinse the burette with the titrant and the pipette with the solution being analysed.

Pipette 25.0 cm³ of solution into a conical flask and add 2–3 drops of indicator.

Fill the burette, remove the funnel and record the initial reading.

Run in titrant while swirling, adding it dropwise near the end point until the colour change is permanent.

Record the final reading, reading the bottom of the meniscus at eye level.

Do a rough titre first, then repeat accurately until two titres are concordant (within 0.10 cm³) and take their mean.

Equivalence point
- where the reactants have reacted in the ratio of the balanced equation, and a white tile makes the colour change easier to see

Section 9 of 12Indicators

Indicator	Colour in acid	Colour in alkali
Litmus	Red	Blue
Methyl orange	Red	Yellow
Phenolphthalein	Colourless	Pink

Acid–base indicator

Acid–base indicator a weak acid or base whose conjugate forms have different colours
The colour changes reversibly with the H⁺ concentration.

Choice

Choice the colour change range must lie within the vertical section of the titration curve,
so litmus is unsuitable and universal indicator changes too gradually

Phenolphthalein

Phenolphthalein better with the base in the burette, because the appearance of pink is easier to see than the loss of colour

Section 10 of 12Titration calculations and uncertainty

Write the balanced equation.

2Calculate moles of the solution of known concentration from n

Calculate moles of the solution of known concentration from n c × V, converting cm³ to dm³ by dividing by 1000.

Use the mole ratio to find moles of the other reagent.

Divide by the volume in dm³ to find its concentration.

Concentration in g dm⁻³

Concentration in g dm⁻³ mass in g ÷ volume in dm³, and mass of solute = moles × molar mass

Percentage uncertainty

(absolute uncertainty ÷ measured value) × 100

Uncertainties are added when two readings are subtracted, so a burette reading with ±0.05 cm³ each gives ±0.10 cm³ for a titre.
The total experimental uncertainty is the sum of the individual percentage uncertainties.

Section 11 of 12Key terms

Term	Meaning
Oxidation number	The charge an atom would have if its bonding were completely ionic.
Oxidising agent	A substance that oxidises another species and is itself reduced.
Reducing agent	A substance that reduces another species and is itself oxidised.
Disproportionation	A reaction in which the same species is simultaneously oxidised and reduced.
Half-equation	An equation showing only the loss or gain of electrons by one species.
Equivalence point	The point at which reactants have reacted in the ratio of the balanced equation.
Concordant titres	Titres that agree within 0.10 cm³.
Indicator	A weak acid or base that changes colour with pH.

Section 12 of 12Exam technique

Confusing oxidation with gain of electrons — oxidation is loss (OIL RIG)

Using an oxidation number for a molecule rather than for a single atom

Leaving electrons in the final ionic equation — they must cancel

Using unconcordant titres in the mean — only use titres within 0.10 cm³

Forgetting to convert cm³ to dm³ in n = c × V

Last updated: 18 October 2026