IP Chemistry Upper Sec 07: Redox Chemistry

Study guide26 Nov 2025, 00:00 ZUpdated 17 Jul 2026

Reviewed by

Azmi·Senior Chemistry Specialist

Oxidation-reduction definitions, electrochemistry, and cells for IP Sec 3-4 Chemistry.

For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.

How this chapter applies

Eclat blended core: oxidation and reduction, oxidation states, redox tests, electrolysis, simple cells, electroplating, copper purification, and fuel cells are taught together because assessments often combine them.
School-sensitive extension: acidic-medium half-equation methods, dichromate chemistry, disproportionation, and detailed fuel-cell half-equations should be used only where the current school teaches them.
2027 national comparison: K324 Topic C7 covers the full redox and electrochemistry core, including selective discharge, concentration effects, copper electrodes, simple cells, and the hydrogen fuel cell.
Check your school: electrode conventions, permitted half-equation methods, and the depth of cell chemistry can differ. Match the current school materials and question data.
Exam-track route: use the separate O-Level and SEC G3 Chemistry notes for K324 topic ownership.

Choose the right Chemistry resource

Use this IP page for redox and electrochemistry: oxidation numbers, selective discharge, copper purification, electroplating, simple cells, and hydrogen fuel cells.

Need	Best Eclat page
Official 6092 paper format and topic order	O-Level Chemistry 6092 syllabus guide
Bonding foundation before electrolysis	Chemical Bonding and Structure notes
Electrolysis as a Paper 3 investigation	O-Level Chemistry Electrolysis and Redox Practical Guide
Paper 2 answer layout and timing	O-Level Chemistry Paper 2 structured question strategy

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 $\ce{KMnO4}$ turns purple → colourless when reduced; acidified $\ce{K2Cr2O7}$ orange → green; $\ce{I-}$ turns brown on oxidation; starch test gives blue-black with $\ce{I2}$ .
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 → $\ce{Cl2}$ ; dilute → $\ce{O2}$ from $\ce{OH-}$ ; $\ce{SO4^{2-}}$ typically not discharged).
Half-equations and selective discharge must consider electrode material (inert vs copper). Products may change with concentration (e.g., $\ce{NaCl}$ 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 $\ce{H2}$ , reduces $\ce{O2}$ 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 $\ce{Fe}$ or $\ce{O2}$	Elemental atoms have oxidation number 0.	No solving is needed.	Giving $\ce{O2}$ oxygen a value of -2 just because oxygen often has -2 in compounds.
Simple ion, such as $\ce{Fe^{3+}}$	The oxidation number equals the ion charge.	$\ce{Fe^{3+}}$ has Fe at +3.	Dropping the sign and writing only 3.
Neutral compound, such as $\ce{Fe2O3}$	Oxygen is usually -2.	Let Fe be $x$ : $2x + 3(-2) = 0$ , so $x = +3$ .	Forgetting to multiply by the subscript 2 on Fe.
Polyatomic ion, such as $\ce{Cr2O7^{2-}}$	Oxygen is usually -2.	Let Cr be $x$ : $2x + 7(-2) = -2$ , so $x = +6$ .	Setting the total to 0 when the species has an overall ion charge.

Worked check: in $\ce{Fe2O3 + 3CO -> 2Fe + 3CO2}$ , Fe changes from +3 in $\ce{Fe2O3}$ to 0 in $\ce{Fe}$ , so Fe is reduced. Carbon changes from +2 in $\ce{CO}$ to +4 in $\ce{CO2}$ , 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 $\ce{H+}$ or $\ce{OH-}$ 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 $\ce{Cu^{2+}}$ discharge as metal; otherwise $\ce{H2}$ forms.	Very reactive cations such as $\ce{Na+}$ remain in solution.
Anode	Is there concentrated halide ion?	Concentrated $\ce{Cl-}$ , $\ce{Br-}$ , or $\ce{I-}$ gives the halogen. Otherwise $\ce{O2}$ forms from water or hydroxide ions.	Copper anode can dissolve instead of oxygen forming.

Worked check: aqueous $\ce{CuSO4}$ with carbon electrodes gives $\ce{Cu}$ at the cathode because $\ce{Cu^{2+}}$ is less reactive than hydrogen, and $\ce{O2}$ at the anode because sulfate ions are not discharged. With copper electrodes, the copper anode dissolves to replace $\ce{Cu^{2+}}$ ions removed at the cathode.

Common trap: do not copy molten-salt products into aqueous electrolysis. Water adds competing ions, so aqueous $\ce{NaCl}$ 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:

$\ce{Zn -> Zn^{2+} + 2e-}$

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 $\ce{OH-}$ / $\ce{H2O}$ as needed. Combine by equalising electrons.
Oxidation number changes: use to identify oxidised vs reduced species and disproportionation (e.g., $\ce{Cl2 + 2OH- -> Cl- + ClO- + H2O}$ ).
Colour cues: acidified $\ce{KMnO4}$ purple → colourless; acidified $\ce{K2Cr2O7}$ orange → green; $\ce{I2}$ brown → blue-black with starch; $\ce{SO2}$ bleaches purple $\ce{KMnO4}$ .
Electrolysis (aqueous, inert electrodes): cathode deposits less reactive cation (Cu²⁺/Ag⁺) else $\ce{H2}$ ; anode gives $\ce{O2}$ unless concentrated $\ce{Cl-}$ / $\ce{Br-}$ / $\ce{I-}$ present. Copper electrodes shift products (anode dissolves).

Halogen displacement checkpoint

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
$\ce{Cl2}$	$\ce{Br-}$	$\ce{Br2}$ forms.	Chlorine gains electrons and oxidises bromide ions to bromine.	Saying chlorine is oxidised because it is the stronger halogen.
$\ce{Cl2}$	$\ce{I-}$	$\ce{I2}$ forms.	Chlorine gains electrons and oxidises iodide ions to iodine.	Forgetting iodine can give a blue-black colour with starch.
$\ce{Br2}$	$\ce{Cl-}$	No displacement.	Bromine is less reactive than chlorine, so it cannot oxidise chloride ions.	Assuming any halogen will displace any halide.
$\ce{I2}$	$\ce{Br-}$	No displacement.	Iodine is less reactive than bromine, so bromide remains unchanged.	Predicting a reaction just because both species contain halogens.

Worked check: chlorine water added to potassium iodide solution produces iodine because $\ce{Cl2 + 2I- -> 2Cl- + I2}$ . 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: $\ce{Fe^{2+} -> Fe^{3+} + e-}$ is oxidation because the electron is produced. $\ce{Cu^{2+} + 2e- -> Cu}$ 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 $\ce{CuSO4}$ . At the key, $\ce{Cu^{2+} + 2e- -> Cu}$ , 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

School-sensitive extension: The acidic permanganate half-equation below previews later Chemistry work. Use it only where the current IP school teaches it; it is not part of the K324 comparison scope.

Half-equation combo: $\ce{MnO4- + 8H+ + 5e- -> Mn^{2+} + 4H2O}$ with $\ce{Fe^{2+} -> Fe^{3+} + e-}$ ; multiply iron equation by 5, add to show $\ce{MnO4-}$ oxidises $\ce{Fe^{2+}}$ .
Oxidation number check: $\ce{Fe2O3 + 3CO -> 2Fe + 3CO2}$ : Fe +3 → 0 (reduced), C +2 → +4 (oxidised); CO is reducing agent.
Electrolysis prediction: Molten $\ce{PbBr2}$ : cathode $\ce{Pb}$ , anode $\ce{Br2}$ . Aqueous concentrated $\ce{NaCl}$ : cathode $\ce{H2}$ , anode $\ce{Cl2}$ ; dilute $\ce{NaCl}$ : cathode $\ce{H2}$ , anode $\ce{O2}$ .

Pitfalls and fixes

Not cancelling electrons when adding half-equations.
Using $\ce{KMnO4}$ / $\ce{K2Cr2O7}$ without acid-state acidified to avoid brown $\ce{MnO2}$ 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 $\ce{Cr2O7^{2-}}$ , $\ce{NO3-}$ , $\ce{ClO-}$ and identify oxidised/reduced species in a given reaction.
Write the ionic equation for acidified $\ce{K2Cr2O7}$ oxidising $\ce{I-}$ to $\ce{I2}$ .
Predict products at each electrode for electrolysis of $\ce{CuSO4}$ with carbon vs copper electrodes; explain colour changes.
Describe observations when chlorine water is added to $\ce{Br-}$ and $\ce{I-}$ solutions; link to redox displacement.

Quick applications

Assign oxidation states in $\ce{Fe2O3 + 3CO -> 2Fe + 3CO2}$ : Fe +3 → 0 (reduced), C +2 → +4 (oxidised); CO is reducing agent.
Electrolysis: molten $\ce{NaCl}$ → Na at cathode (Na⁺ + e⁻ → Na), Cl₂ at anode (2Cl⁻ → Cl₂ + 2e⁻). Aqueous $\ce{CuSO4}$ with carbon electrodes → Cu at cathode, O₂ at anode (because $\ce{SO4^{2-}}$ not discharged); with copper electrodes, anode dissolves $\ce{Cu -> Cu^{2+} + 2e-}$ .
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 $\ce{2H2 + 4OH- -> 4H2O + 4e-}$ ; at cathode $\ce{O2 + 2H2O + 4e- -> 4OH-}$ ; overall $\ce{2H2 + O2 -> 2H2O}$ .

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 $\ce{Cl-}$ or $\ce{Br-}$ gives halogen, dilute gives oxygen.
Colour cues: $\ce{KMnO4}$ 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.

Next steps

Revise the full hub: https://eclatinstitute.sg/blog/ip-chemistry-upper-sec-notes

New to the Integrated Programme? Start with What is IP? | Browse all free IP notes.

Sources

Add Eclat Institute as a Preferred Source on Google.