These notes align with SEAB GCE O-Level Chemistry (6092) content used in IP programmes (exams from 2026).
Status: SEAB O-Level Chemistry 6092 syllabus (exams from 2026) checked 2026-06-02 - scope unchanged; remains the reference for this note.
6092 Route Map
Use this page for Topic 7 Redox Chemistry and the electrochemistry part of 6092: oxidation numbers, selective discharge, copper purification, electroplating, simple cells, and hydrogen fuel cells.
The core idea is simple: Redox tracks electron loss and electron gain.
Use it as a working check: Oxidation is loss of electrons or increase in oxidation state. Reduction is gain of electrons or decrease in oxidation state. The oxidising agent is reduced.
Then go one layer deeper: Example: in electrolysis, reduction happens at the cathode and oxidation happens at the anode. Keep that rule even when the electrode signs feel confusing.
What you must know
Redox definitions: oxidation = loss of electrons/oxygen gain/hydrogen loss/increase in oxidation state; reduction = gain of electrons/oxygen loss. Oxidising agents are reduced; reducing agents are oxidised.
Redox tests: acidified KMnOX4 turns purple → colourless when reduced; acidified KX2CrX2OX7 orange → green; IX− turns brown on oxidation; starch test gives blue-black with IX2.
Electrolysis: molten compounds give metal at cathode, non-metal at anode. Aqueous electrolysis depends on cation reactivity (Cu²⁺/Ag⁺ discharged before H⁺) and anion rules (concentrated chloride → ClX2; dilute → OX2 from OHX−; SOX4X2− typically not discharged).
Half-equations and selective discharge must consider electrode material (inert vs copper). Products may change with concentration (e.g., NaCl dilute vs concentrated).
Applications: copper purification (impure anode dissolves; pure cathode gains Cu), electroplating with object as cathode, simple cells produce current when two metals with different reactivity are connected via electrolyte; hydrogen fuel cell oxidises HX2, reduces OX2 to water.
Oxidation number assignment checkpoint
Before deciding what is oxidised or reduced, assign oxidation numbers with a charge-balance routine. Work from fixed rules first, then solve for the unknown atom.
Species type
First fixed values
Charge-balance move
Common trap
Element alone, such as Fe or OX2
Elemental atoms have oxidation number 0.
No solving is needed.
Giving OX2 oxygen a value of -2 just because oxygen often has -2 in compounds.
Simple ion, such as FeX3+
The oxidation number equals the ion charge.
FeX3+ has Fe at +3.
Neutral compound, such as FeX2OX3
Polyatomic ion, such as CrX2OX7X2−
Worked check: in FeX2OX3+3CO2Fe+3COX2, Fe changes from +3 in FeX2OX3 to 0 in Fe, so Fe is reduced. Carbon changes from +2 in CO to +4 in COX2, so carbon is oxidised.
Misconception check: oxidation numbers are bookkeeping values, not physical charges on every atom. Their job is to show whether electron ownership has increased or decreased across the reaction.
Redox role checkpoint
When a question asks for the oxidised species, reduced species, oxidising agent, or reducing agent, work in this order. It keeps the role names tied to evidence instead of memory.
Step
What to check
What it tells you
1
Did the oxidation number increase?
That species is oxidised.
2
Did the oxidation number decrease?
That species is reduced.
3
Which species caused another species to be oxidised?
That species is the oxidising agent and is itself reduced.
4
Which species caused another species to be reduced?
That species is the reducing agent and is itself oxidised.
Common trap: the oxidising agent is not the substance that gets oxidised. The oxidising agent causes oxidation in another species, so it gains electrons or has a lower oxidation number after the reaction.
Aqueous electrolysis checkpoint
For aqueous electrolysis, decide each electrode separately. The solution contains water, so HX+ or OHX− can compete with dissolved ions.
Electrode
First question
Usual product with inert electrodes
Key exception
Cathode
Is the metal ion less reactive than hydrogen?
Less reactive metal ions such as CuX2+ discharge as metal; otherwise HX2 forms.
Very reactive cations such as NaX+ remain in solution.
Anode
Is there concentrated halide ion?
Concentrated ClX−, BrX−, or IX−
Worked check: aqueous CuSOX4 with carbon electrodes gives Cu at the cathode because CuX2+ is less reactive than hydrogen, and OX2 at the anode because sulfate ions are not discharged. With copper electrodes, the copper anode dissolves to replace CuX2+ ions removed at the cathode.
Common trap: do not copy molten-salt products into aqueous electrolysis. Water adds competing ions, so aqueous NaCl does not produce sodium metal at the cathode.
Simple Cell Direction Checkpoint
When a simple cell uses two different metals, start with the reactivity difference before drawing arrows. The more reactive metal is oxidised and supplies electrons to the external circuit.
What to decide
Rule to use
Common trap
Which metal is oxidised
The more reactive metal loses electrons and becomes ions.
Calling the less reactive metal the anode just because it is the positive terminal.
Electron flow in the wire
Electrons flow from the more reactive metal to the less reactive metal.
Drawing electron flow from positive to negative.
Conventional current direction
Conventional current is opposite to electron flow.
Mixing the chemistry arrow with the circuit-current arrow.
Cell voltage size
Bigger reactivity difference usually gives a larger potential difference.
Saying bigger electrodes always give a larger voltage.
Worked check: in a zinc-copper cell, zinc is more reactive, so zinc atoms are oxidised:
ZnZnX2++2eX−
Electrons leave the zinc electrode and travel through the wire to the copper electrode. The voltmeter's positive terminal is connected to copper, but the electrons still flow from zinc to copper.
Misconception check: a simple cell and electrolysis use different energy routes. A simple cell produces electrical energy from a spontaneous redox reaction; electrolysis uses electrical energy to force a reaction.
Detailed notes
Half-equations: balance atoms, then charge with e⁻; in alkaline medium, add OHX−/HX2O as needed. Combine by equalising electrons.
Oxidation number changes: use to identify oxidised vs reduced species and disproportionation (e.g., ClX2+2OHX−ClX−+ClOX−+HX2O
For halogen displacement questions, rank the halogen molecule first, then identify which halide ion can be oxidised. A more reactive halogen displaces a less reactive halide from solution.
Added halogen
Halide solution
Reaction outcome
Electron-transfer explanation
Common trap
ClX2
BrX−
BrX2 forms.
Chlorine gains electrons and oxidises bromide ions to bromine.
Saying chlorine is oxidised because it is the stronger halogen.
ClX2
IX−
BrX2
ClX−
IX2
BrX−
Worked check: chlorine water added to potassium iodide solution produces iodine because ClX2+2IX−2ClX−+IX2. Chlorine is reduced from 0 to -1, while iodide is oxidised from -1 to 0.
Misconception check: the more reactive halogen is the oxidising agent, so it is reduced. The halide ion that is displaced is oxidised to the halogen molecule.
Half-equation balancing checkpoint
Before combining half-equations, make each half-equation balance atoms and charge on its own. The electron side tells you whether oxidation or reduction is happening.
Half-equation clue
Where electrons go
What it means
Common trap
Species loses electrons
Electrons appear on the product side.
Oxidation occurs.
Writing electrons on both sides and cancelling too early.
Species gains electrons
Electrons appear on the reactant side.
Reduction occurs.
Calling it oxidation because the charge becomes more positive or less negative.
Charges do not match after atoms are balanced
Add electrons to the more positive side until total charge matches.
Charge balance is the final check.
Balancing atoms only and leaving unequal charges.
Two half-equations are added
Multiply one or both equations so electrons lost equals electrons gained.
Electrons must cancel completely in the final ionic equation.
Adding half-equations with unequal electron counts.
Worked check: FeX2+FeX3++eX− is oxidation because the electron is produced. CuX2++2eX−Cu is reduction because electrons are used up. If these are combined, multiply the iron half-equation by 2 so the two electrons cancel.
Misconception check: electrons are not optional decoration in half-equations. They are the charge-balance evidence that proves which species is oxidised and which species is reduced.
Electroplating setup checkpoint
For electroplating questions, decide what must gain metal first. The object to be plated must be the cathode because metal ions gain electrons there and form metal atoms on its surface.
Setup part
What to choose
Why
Common trap
Object to be plated
Cathode, connected to the negative terminal
Metal cations are reduced and deposit on this object.
Making the object the anode, so it would lose metal instead.
Plating metal
Anode, connected to the positive terminal
It can dissolve to replace metal ions used up at the cathode.
Using an inert anode when the question expects the metal ion concentration to stay steady.
Electrolyte
Solution containing ions of the plating metal
The solution supplies the cations that are reduced at the cathode.
Choosing a salt solution with the wrong metal ion.
Surface preparation
Clean the object before plating
Grease or oxide layers stop an even coating from forming.
Blaming poor plating only on current when the surface was dirty.
Worked check: to copper-plate a steel key, make the key the cathode, use copper as the anode, and use a copper(II) salt solution such as CuSOX4. At the key, CuX2++2eX−Cu, so copper atoms coat the surface.
Misconception check: electroplating is not just dipping an object into a salt solution. The circuit must force reduction of metal ions onto the object.
Worked walkthroughs
JC preview - The permanganate titration below previews content from H2 Chemistry. It is included because IP students benefit from early exposure, but it is not assessed in the O-Level 6092 examination.
Half-equation combo: MnOX4X−+8HX++5eX−MnX2++4HX2O with FeX2+FeX3++eX−; multiply iron equation by 5, add to show MnOX4X− oxidises FeX2+.
Not cancelling electrons when adding half-equations.
Using KMnOX4/KX2CrX2OX7 without acid-state acidified to avoid brown MnOX2 or incomplete reduction.
Forgetting electrode roles: oxidation at anode, reduction at cathode, regardless of sign.
Ignoring concentration/electrode effects on selective discharge.
Practice drills
Assign oxidation numbers in CrX2OX7X2−, NOX3X−, ClOX− and identify oxidised/reduced species in a given reaction.
Write the ionic equation for acidified KX2CrX2OX7
Predict products at each electrode for electrolysis of CuSOX4 with carbon vs copper electrodes; explain colour changes.
Describe observations when chlorine water is added to BrX− and IX− solutions; link to redox displacement.
Quick applications
Assign oxidation states in FeX2OX3+3CO2Fe+3COX2: Fe +3 → 0 (reduced), C +2 → +4 (oxidised); CO is reducing agent.
Electrolysis: molten NaCl → Na at cathode (Na⁺ + e⁻ → Na), Cl₂ at anode (2Cl⁻ → Cl₂ + 2e⁻). Aqueous CuSOX4 with carbon electrodes → Cu at cathode, O₂ at anode (because SOX4X2−
Simple cell: zinc strip + copper strip in electrolyte; electrons flow from Zn (more reactive, oxidised) to Cu; voltmeter positive at copper.
Hydrogen fuel cell alkaline version: at anode 2HX2+4OHX−4HX2O+4eX−
Exam cues
Balance half-equations for both atoms and charge; ensure electrons added to the correct side; include state symbols.
Selective discharge: for aqueous solutions with inert electrodes, compare cation reactivity and anion rules; remember concentrated ClX− or BrX− gives halogen, dilute gives oxygen.
Colour cues: KMnOX4 test for reducing agents; iodine formed from iodide gives brown solution/blue-black with starch.
In electroplating, object to be plated is the cathode so metal cations reduce onto it.
