Full chapter
Redox Chemistry
Track electrons through reactions, electrolysis and electricity-producing cells.
O-Level 6092 (2026) / SEC G3 K324 (2027)
Recognise oxidation and reduction
Use the model that makes the electron change easiest to see.
| Model | Oxidation | Reduction |
|---|---|---|
| Oxygen transfer | Gain of oxygen | Loss of oxygen |
| Hydrogen transfer | Loss of hydrogen | Gain of hydrogen |
| Electron transfer | Loss of electrons | Gain of electrons |
| Oxidation state | Increase | Decrease |
Oxidation and reduction happen together: electrons lost by one species are gained by another. In CuO(s) + H2(g) -> Cu(s) + H2O(g), copper(II) oxide loses oxygen and is reduced; hydrogen gains oxygen and is oxidised. During addition of hydrogen to an alkene, the organic compound gains hydrogen and is reduced.
Oxidation state is an accounting number. An uncombined element has state 0. A monatomic ion has its ion charge. In the common compounds here, oxygen is usually -2 and hydrogen +1; exceptions must be supplied or recognised from context. The states sum to zero in a neutral compound or to the overall charge in an ion.
Worked exampleFollow electrons and oxidation states
Explain Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s).
Try it first. Then show the worked stepsHide the worked steps
- The reacting ions give Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s).
- Zn -> Zn2+ + 2e-: oxidation state rises 0 to +2, so zinc is oxidised.
- Cu2+ + 2e- -> Cu: state falls +2 to 0, so copper ions are reduced.
Zinc transfers electrons to copper(II) ions. Sulfate is a spectator.
Check your understanding
Is H+ + OH- -> H2O a redox reaction?
Show the answer
No. Hydrogen stays +1 and oxygen stays -2, so nothing is oxidised or reduced.
Check each oxidation state before and after. Hydrogen is +1 and oxygen is -2 on both sides, so no element changes. A reaction can happen without any electron transfer; neutralisation is the classic example.
Identify the agent from what it does
The oxidising agent is itself reduced.
An oxidising agent accepts electrons and causes another species to be oxidised. It is reduced. A reducing agent supplies electrons and is oxidised. In the zinc/copper-ion reaction, Cu2+ is the oxidising agent and Zn is the reducing agent; copper metal is a product, not the agent added at the start.
| Reagent | Observation | Inference |
|---|---|---|
| Aqueous potassium iodide | Initially colourless solution becomes brown as iodine forms | Unknown oxidises I- to I2; it acts as an oxidising agent |
| Acidified potassium manganate(VII) | Purple solution becomes colourless | Unknown reduces manganate(VII); it acts as a reducing agent |
For the iodide test, 2I-(aq) -> I2(aq) + 2e- shows iodide losing electrons. For manganate(VII), use the acidified reagent specified: conditions affect its products and colours. A colour change is an observation; the assignment of oxidising or reducing behaviour is the inference.
Check your understanding
An unknown decolourises acidified potassium manganate(VII). Is the unknown oxidised or reduced?
Show the answer
Oxidised. It gives electrons to the manganate(VII), so it acts as a reducing agent.
Purple MnO4- is reduced to almost colourless Mn2+. The electrons for that reduction come from the unknown, so the unknown loses electrons: it is oxidised and is the reducing agent.
Electrolysis of a molten compound
A power supply forces chemical change using mobile ions.
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Two graphite electrodes connected to a battery are pushed into solid lead(II) bromide. Does a current flow?
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No, not until it melts
The ions are held in fixed positions in the solid lattice. They can carry charge only when melting lets them move.
Electrolysis passes electric current through a molten or dissolved ionic compound, the electrolyte, causing chemical changes at electrodes. In the solid, ions are fixed in a lattice; once molten or dissolved they can move. This supports the ionic model. Electrons move through metal wires; ions carry charge through the electrolyte.
- Negative cathode
Na+ ions move towards it and gain electrons: Na+ + e- -> Na.
- Positive anode
Cl- ions move towards it and lose electrons: 2Cl- -> Cl2 + 2e-.
- Products
Sodium metal and chlorine gas form. At the operating temperature sodium is liquid; chlorine bubbles at the anode.
Label a molten electrolysis cell
Two graphite electrodes dip into molten sodium chloride that is being heated. The left electrode joins the positive terminal of the d.c. supply and is the anode; chloride ions move to it and chlorine gas bubbles off. The right electrode joins the negative terminal and is the cathode; sodium ions move to it and sodium metal forms. Electrons flow through the wires from the anode to the supply and from the supply to the cathode.
For a molten binary ionic compound, there are only its constituent cations and anions to discharge. The metal forms at the cathode and the non-metal at the anode, with inert electrodes. For molten lead(II) bromide: Pb2+ + 2e- -> Pb and 2Br- -> Br2 + 2e-. Balance atoms and charge in each half-equation.
Check your understanding
Why does solid sodium chloride not undergo electrolysis, when molten sodium chloride does?
Show the answer
Its ions are held in the lattice and cannot move to the electrodes to carry current.
Solid NaCl already contains Na+ and Cl- ions, but they are fixed in place. Melting lets the ions move, so they can carry charge and be discharged at the electrodes.
Check your understanding
Molten calcium bromide, CaBr2, is electrolysed with inert electrodes. What forms at each electrode?
Show the answer
Calcium at the cathode and bromine at the anode
The melt contains only Ca2+ and Br- ions. Ca2+ moves to the cathode and is reduced: Ca2+ + 2e- -> Ca. Br- moves to the anode and is oxidised: 2Br- -> Br2 + 2e-.
Aqueous electrolysis: more than one possible ion
Water introduces competing products.
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Concentrated aqueous sodium chloride is electrolysed with inert electrodes. What forms at the cathode?
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Hydrogen gas, from water
Sodium is very reactive, so sodium ions are hard to reduce. Hydrogen ions from water are reduced instead, giving hydrogen gas.
In aqueous electrolysis, consider the solute ions and the water. With inert electrodes, a metal less reactive than hydrogen, such as copper, is usually deposited at the cathode. For a very reactive metal ion such as Na+, hydrogen forms instead. At the anode, sulfate is generally not discharged; oxygen forms from water/hydroxide. Halide concentration matters: concentrated chloride favours chlorine, while dilute chloride solution is treated here as giving oxygen.
- List species and electrode material
For aqueous potassium iodide with inert electrodes, consider K+, I- and water (H+ and OH-). Molten KI has no water competing.
- Choose the cathode product
Potassium is above hydrogen in the reactivity series. In the school aqueous model, hydrogen forms rather than potassium metal. For a metal below hydrogen, such as copper or silver, the metal is normally deposited.
- Choose the anode product
Halide discharge forms the corresponding halogen: bromide gives bromine; iodide gives iodine. For this iodide solution iodine forms; 2I-(aq) -> I2(aq) + 2e-. Chloride needs the dilute/concentrated comparison shown below. Sulfate is not discharged in these examples: oxygen forms from water/hydroxide.
- Check atoms, charge and observations
The iodide half-equation has two iodine atoms and total charge -2 on each side. The anode region becomes brown as iodine dissolves; hydrogen bubbles at the cathode. Do not describe every halogen product as a gas.
| Electrolyte | Cathode | Anode |
|---|---|---|
| Aqueous copper(II) sulfate | Copper | Oxygen |
| Dilute aqueous sodium chloride | Hydrogen | Oxygen |
| Concentrated aqueous sodium chloride | Hydrogen | Chlorine |
Label an aqueous electrolysis cell
Two graphite electrodes dip into blue aqueous copper(II) sulfate. Sulfate and hydroxide ions move to the anode, where oxygen bubbles form. Copper(II) and hydrogen ions move to the cathode, which becomes coated with pink-brown copper. The blue colour fades as copper(II) ions are removed.
| Electrolyte | Colour | pH | Reason |
|---|---|---|---|
| Copper(II) sulfate(aq) | Blue fades | Falls: the solution becomes acidic | Cu2+ ions are removed as copper. OH- ions are removed as oxygen, so H+ and SO42- ions are left: sulfuric acid forms. |
| Concentrated sodium chloride(aq) | Stays colourless | Rises: the solution becomes alkaline | Hydrogen forms from water at the cathode and chlorine forms at the anode, so Na+ and OH- ions are left: sodium hydroxide forms. |
| Dilute sodium chloride(aq) | Stays colourless | Stays about 7 overall | Hydrogen and oxygen both come from water, so water is used up and the solution slowly becomes more concentrated. |
| Dilute sulfuric acid | Stays colourless | Falls slowly | Hydrogen and oxygen both come from water. The same amount of acid is left in less water, so it becomes more concentrated. |
Collect the two gases to compare their volumes
Two inverted tubes full of dilute sulfuric acid sit over the two electrodes. The tube over the cathode holds twice as much gas, hydrogen, as the tube over the anode, which holds oxygen.
Why 2 : 1? Count the electrons. The same electrons that leave the anode arrive at the cathode, so both electrodes handle the same number of moles of electrons. Write both half-equations with the same number of electrons. Cathode: 4H+(aq) + 4e- → 2H2(g). Anode: 4OH-(aq) → O2(g) + 2H2O(l) + 4e-. Four moles of electrons make 2 mol of hydrogen but only 1 mol of oxygen. Equal moles of gas take up equal volumes at the same temperature and pressure, so the hydrogen volume is twice the oxygen volume. In a real experiment the oxygen volume is a little less than half, because oxygen is slightly more soluble in water than hydrogen.
Worked exampleLink the mass at one electrode to the gas at the other
Aqueous copper(II) sulfate is electrolysed with graphite electrodes. The cathode gains 0.640 g of copper. Calculate the volume of oxygen formed at the anode, at room temperature and pressure. (Ar: Cu = 64; one mole of gas occupies 24 dm3 at r.t.p.)
Try it first. Then show the worked stepsHide the worked steps
- Moles of copper deposited = 0.640 ÷ 64 = 0.0100 mol.
- Cathode: Cu2+ + 2e- → Cu, so moles of electrons = 2 × 0.0100 = 0.0200 mol.
- Anode: 4OH- → O2 + 2H2O + 4e-, so moles of oxygen = 0.0200 ÷ 4 = 0.00500 mol.
- Volume of oxygen = 0.00500 × 24 = 0.120 dm3.
0.120 dm3 (120 cm3) of oxygen. The electrons are the link: the same 0.0200 mol of electrons passes through both electrodes.
| Process | Half-equation |
|---|---|
| Copper deposition | Cu2+(aq) + 2e- -> Cu(s) |
| Hydrogen from water | 2H2O(l) + 2e- -> H2(g) + 2OH-(aq) |
| Oxygen from hydroxide | 4OH-(aq) -> O2(g) + 2H2O(l) + 4e- |
| Chlorine from chloride | 2Cl-(aq) -> Cl2(g) + 2e- |
Worked exampleConstruct a half-equation from an unfamiliar ion charge
An aqueous solution contains Ag+ ions and silver is discharged at an inert cathode. A separate molten lead(II) bromide cell produces bromine at the anode. Construct the relevant half-equations.
Try it first. Then show the worked stepsHide the worked steps
- At the cathode an Ag+ ion gains one electron to make a neutral silver atom: Ag+(aq) + e- -> Ag(s).
- At the anode bromide ions lose electrons. Bromine is Br2, so begin 2Br- -> Br2, then add 2e- on the right to balance charge.
- For the molten-cell anode use Br-(l) and Br2(g); the molten electrolyte is hot enough for bromine to be a gas. Check both atoms and net charge.
Cathode in the silver solution: Ag+(aq) + e- -> Ag(s). Anode in molten lead(II) bromide: 2Br-(l) -> Br2(g) + 2e-. Each equation must balance atoms and charge.
Worked exampleCompare dilute and concentrated brine
Explain why changing NaCl concentration can change one electrode product.
Try it first. Then show the worked stepsHide the worked steps
- Na+ is not deposited from either aqueous solution; hydrogen forms at the cathode.
- In dilute solution, oxygen forms at the inert anode.
- At high chloride concentration, chloride is preferentially discharged and chlorine forms.
Cathode product remains hydrogen; anode product changes. Always state aqueous/molten, concentration and electrode material.
Check your understanding
What happens to the blue colour during electrolysis of copper(II) sulfate solution with inert electrodes?
Show the answer
It fades, because Cu2+ ions are removed at the cathode and nothing replaces them.
Cu2+ is discharged as copper at the cathode. An inert anode makes oxygen instead of dissolving, so the Cu2+ that gives the blue colour is not replenished.
Check your understanding
Dilute sulfuric acid is electrolysed with inert electrodes. 36 cm3 of hydrogen collects at the cathode. What volume of oxygen collects at the anode in the same time?
Show the answer
18 cm3, because 4 mol of electrons make 2 mol H2 but 1 mol O2
Cathode: 2H+ + 2e- -> H2. Anode: 4OH- -> O2 + 2H2O + 4e-. The same 4 mol of electrons give 2 mol of H2 and 1 mol of O2, so H2 : O2 is 2 : 1 by volume. 36 / 2 = 18 cm3.
Check your understanding
A few drops of Universal Indicator are added to concentrated aqueous sodium chloride, which is then electrolysed with inert electrodes. What happens to the colour near the cathode?
Show the answer
It turns violet, because hydroxide ions are left there
At the cathode, hydrogen forms from water: 2H2O + 2e- -> H2 + 2OH-. Hydroxide ions build up with the Na+ ions, so the solution there becomes alkaline.
Purifying and plating copper
Copper electrodes behave differently from inert ones.
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Impure copper is the anode and pure copper is the cathode in copper(II) sulfate solution. What happens to the mass of the anode?
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It decreases, as copper dissolves
Copper atoms at the anode lose electrons and go into solution as copper(II) ions, so the anode dissolves away.
To purify copper, use impure copper as the anode, pure copper as the cathode, and aqueous copper(II) sulfate as electrolyte. At the anode, Cu(s) -> Cu2+(aq) + 2e-. At the cathode, Cu2+(aq) + 2e- -> Cu(s). Copper transfers from the impure electrode to the pure one; some impurities collect as anode sludge. Detailed industrial operation is not required.
Label the purification cell
An impure copper anode on the left and a pure copper cathode on the right dip into blue aqueous copper(II) sulfate. Copper dissolves from the anode and impurities fall below it as sludge. Copper deposits on the cathode. The solution stays blue.
The anode loses mass and the cathode gains mass. Copper ions removed at the cathode are replenished at the copper anode, so concentration remains approximately constant in the simple model. This contrasts with an inert anode, where oxygen forms and the blue solution fades.
Mass changes. The same electrons pass through both electrodes, so for every copper atom that leaves the anode, one copper atom joins the cathode. The cathode gains mass and the anode loses mass. The anode loses a little more mass than the cathode gains, because the impurities also leave it: unreactive ones such as silver fall as sludge, and more reactive ones such as zinc dissolve as ions and stay in solution.
For copper electroplating, make the object to be coated the cathode, use copper as the anode, and choose a solution containing Cu2+. Clean the object first so the coating adheres. Electroplating can improve appearance or provide a protective surface; the coating metal and intended use determine the benefit.
Label the plating cell
A copper anode on the left and a metal key on the right dip into blue aqueous copper(II) sulfate. Copper dissolves from the anode and deposits as a copper coat on the key, which is the cathode.
Check your understanding
A student makes a key the anode when trying to copper-plate it. What should change?
Show the answer
Make the key the cathode, so Cu2+ ions gain electrons and deposit on it.
Plating is reduction: Cu2+ + 2e- -> Cu happens at the cathode. A copper anode dissolves to replace the Cu2+ ions that the key removes.
Exam-style question
Impure copper can be purified by electrolysis.
(a) Draw a labelled diagram of the apparatus. [2]
(b) Write the half-equation for the reaction at each electrode. [2]
(c) During the electrolysis the pure copper electrode gains 3.20 g. Explain why the impure electrode loses more than 3.20 g. [1] [5]
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Use a reaction to generate electricity
A simple cell converts chemical energy into electrical energy.
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A simple cell uses zinc and copper in a salt solution. The zinc is replaced by magnesium. What happens to the voltage?
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It becomes larger, as Mg is more reactive than Zn
Magnesium is more reactive than zinc, so it is further from copper in the reactivity series. A bigger difference in reactivity gives a bigger voltage.
A simple cell contains two different electrodes in an electrolyte connected through an external circuit. A more reactive metal tends to lose electrons more readily. In a zinc/copper cell using dilute acid, zinc is oxidised: Zn(s) -> Zn2+(aq) + 2e-. Electrons travel through the wire towards copper, where hydrogen ions can gain them: 2H+(aq) + 2e- -> H2(g). The electrolyte carries ionic current and completes the circuit.
Electron flow in a simple cell
Zinc loses electrons that flow through the external wire to copper; hydrogen ions accept electrons at the copper surface.
Which metal is which electrode? The more reactive metal is the negative electrode. It loses electrons more readily, so it is oxidised and pushes electrons into the wire. The less reactive metal is the positive electrode, where reduction happens. In the zinc/copper cell, zinc is negative and copper is positive. Electrons in the wire always flow from the negative electrode to the positive electrode.
The voltage depends on how far apart the two metals are in the reactivity series. The bigger the difference in reactivity, the bigger the voltage. So magnesium with copper gives a bigger voltage than zinc with copper, because magnesium is more reactive than zinc. Iron with copper gives a smaller voltage than zinc with copper, because iron is less reactive than zinc. Two pieces of the same metal give no voltage at all, since neither one pushes electrons harder than the other.
| Metal pair in an electrolyte | Negative electrode | Voltage compared with zinc/copper |
|---|---|---|
| Magnesium and copper | Magnesium | Bigger: magnesium is further above copper than zinc is |
| Zinc and copper | Zinc | The reference pair |
| Iron and copper | Iron | Smaller: iron is closer to copper than zinc is |
| Magnesium and zinc | Magnesium | Smaller: magnesium and zinc are close together in the series |
| Copper and copper | Neither | Zero: identical metals give no voltage |
Hydrogen can be obtained from water by electrolysis or from hydrocarbons. In a hydrogen fuel cell it reacts with oxygen to produce water and electricity directly: 2H2(g) + O2(g) -> 2H2O(l). Water is the point-of-use product, but overall environmental impact depends on how the hydrogen and electricity were produced. Detailed fuel-cell construction is not required.
Check your understanding
How does a simple cell differ from electrolysis in the direction of energy change?
Show the answer
A cell turns chemical energy into electrical energy; electrolysis does the reverse.
In a cell, a spontaneous redox reaction pushes electrons round the circuit. In electrolysis, an external supply drives a non-spontaneous chemical change. Oxidation is electron loss in both.
Check your understanding
A simple cell made from zinc and copper in dilute sulfuric acid gives a small voltage. A student wants a bigger voltage. Which change works?
Show the answer
Replace the copper with silver and keep the zinc
Silver is less reactive than copper, so zinc and silver are further apart in the reactivity series. A bigger difference in reactivity gives a bigger voltage.
Exam-style question
A simple cell is made with a strip of magnesium and a strip of copper dipped in sodium chloride solution, joined by a wire through a voltmeter.
(a) Which strip is the negative electrode? Explain your answer. [2]
(b) The magnesium is replaced by zinc. Predict how the reading on the voltmeter changes, and explain. [1] [3]
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Quick revision
Revisit the essentials, then return to an explanation when you need it.
| Term | Definition to write |
|---|---|
| Oxidation | Gain of oxygen, loss of hydrogen, loss of electrons, or an increase in oxidation state. |
| Reduction | Loss of oxygen, gain of hydrogen, gain of electrons, or a decrease in oxidation state. |
| Oxidising agent | A substance that oxidises another substance; it is itself reduced (it gains electrons). |
| Reducing agent | A substance that reduces another substance; it is itself oxidised (it loses electrons). |
| Term | Definition to write |
|---|---|
| Electrolysis | The decomposition of an ionic compound, molten or in aqueous solution, by passing an electric current through it. |
| Electrolyte | An ionic compound, molten or in aqueous solution, that conducts electricity and is decomposed by it. |
| Anode and cathode | The anode is the positive electrode, where oxidation happens. The cathode is the negative electrode, where reduction happens. |
| Term | Meaning |
|---|---|
| Oxidation | Electron loss; oxidation state rises |
| Reduction | Electron gain; oxidation state falls |
| Oxidising agent | Accepts electrons and is reduced |
| Reducing agent | Supplies electrons and is oxidised |
Pure: list all ions, electrode material and concentration before predicting electrolysis products. At the cathode reduction occurs; at the anode oxidation occurs.
Scope
Learning outcomes
7. Redox Chemistry (6092 / K324). Use the outcome map to find the explanation for a particular syllabus requirement.
See the learning outcome map
7.1(a) Define oxygen/hydrogen redox
- Oxygen gain/loss
- Hydrogen loss/gain
7.1(b) Define electron/oxidation-state redox
- Electron loss/gain
- Oxidation-state increase/decrease
7.1(c) Identify redox reactions
- Apply all three descriptive models
- Distinguish non-redox reactions
7.1(d) Test redox agents
- KI to brown iodine
- Acidified KMnO4 purple to colourless
7.2(a) Explain electrolysis
- Molten/aqueous ionic electrolytes
- Chemical changes at electrodes
7.2(b) Use electrolysis as ionic evidence
- Fixed ions in solids
- Mobile ions in liquids/solutions
7.2(c) Explain molten NaCl electrolysis
- Ion movement
- Sodium and chlorine products
- Inert electrodes
7.2(d) Predict molten binary products
- Metal and non-metal products
- Inert electrodes
7.2(e) Apply selective discharge
- Cations and reactivity
- Halides, hydroxide, sulfate
- Copper sulfate and dilute/concentrated NaCl
- Inert electrodes
7.2(f) Predict aqueous products
- Use concentration, ions and electrode information
7.2(g) Construct electrode equations
- Atoms and charge balanced
- Relevant state symbols
Electrolysis of a molten compoundAqueous electrolysis: more than one possible ion
7.2(h) Explain copper purification
- Copper sulfate electrolyte
- Copper electrodes
- No technical detail required
7.2(i) Explain electroplating
- Copper plating setup
- A practical use
7.2(j) Explain simple cells
- Two electrodes in electrolyte
- Reactivity and electron transfer
7.2(k) Describe hydrogen fuel cells
- Hydrogen from water/hydrocarbons
- Oxygen reaction generates electricity
- No construction detail