Redox Chemistry
SEAB syllabus topic 7: how oxidation, reduction, electrolysis and cells work, with the reasons, the common mistakes and an exam check for every outcome.
Oxidation is the loss of electrons and an increase in oxidation state, and reduction is the gain of electrons and a decrease in oxidation state.
Electrolysis passes electricity through a molten or dissolved ionic compound, with oxidation at the anode and reduction at the cathode.
A simple cell produces electrical energy because the more reactive metal loses electrons, and the electrons flow through the wire to the less reactive metal.
Oxidation and reduction: oxygen and hydrogen
define oxidation and reduction (redox) in terms of oxygen/hydrogen gain/loss
Oxidation is the gain of oxygen or the loss of hydrogen, and reduction is the loss of oxygen or the gain of hydrogen, and the two always happen together in a redox reaction.
- Oxidation
- Oxidation: a reaction in which a substance gains oxygen or loses hydrogen
- Reduction
- Reduction: a reaction in which a substance loses oxygen or gains hydrogen
- Redox
- Redox: a reaction in which oxidation and reduction take place at the same time, and the word is short for reduction and oxidation
- Oxidised
- Oxidised: a substance is oxidised when it gains oxygen or loses hydrogen supports K324-7.1(a)
- Reduced
- Reduced: a substance is reduced when it loses oxygen or gains hydrogen supports K324-7.1(a)
- Combustion
- Combustion: the burning of a substance in oxygen, which is an oxidation of the substance that burns supports K324-7.1(a)
Oxygen transfer
2Mg(s) + O2(g) → 2MgO(s)Magnesium gains oxygen to form magnesium oxide, so magnesium is oxidised.
CuO(s) + H2(g) → Cu(s) + H2O(l)Copper(II) oxide loses oxygen, so it is reduced, and hydrogen gains oxygen to form water, so hydrogen is oxidised.
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)Iron(III) oxide loses oxygen and is reduced, and carbon monoxide gains oxygen and is oxidised.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)Methane gains oxygen and loses hydrogen, so methane is oxidised when it burns.
Hydrogen transfer
2H2S(g) + O2(g) → 2S(s) + 2H2O(l)Hydrogen sulfide loses hydrogen to form sulfur, so hydrogen sulfide is oxidised.
N2(g) + 3H2(g) ⇌ 2NH3(g)Nitrogen gains hydrogen to form ammonia, so nitrogen is reduced in the Haber process.
In many reactions, such as copper(II) oxide with hydrogen, one substance loses oxygen while another gains it, so oxidation and reduction occur together, and when a metal burns in oxygen only the electron-transfer definition shows that the oxygen is reduced. supports K324-7.1(a)
The SEAB glossary says that a question of the form what is meant by a term “normally implies that a definition should be given”, so write both halves of the definition, gain of oxygen or loss of hydrogen for oxidation, and loss of oxygen or gain of hydrogen for reduction.
Depth: why and how
- Why can oxidation never happen on its own? The electrons lost by the substance that is oxidised are gained by another substance, which is reduced, so oxidation and reduction always occur together.
- Why is rusting an oxidation of iron? Iron gains oxygen to form rust, which is hydrated iron(III) oxide, so iron is oxidised.
- Why is the burning of a fuel an oxidation? A fuel such as methane gains oxygen and loses hydrogen as it burns, so the fuel is oxidised.
- Why is the oxygen and hydrogen definition not the whole story? Many redox reactions, such as zinc displacing copper from copper(II) sulfate, involve no oxygen or hydrogen, so electron transfer and oxidation state are needed too.
- Cross-link to Acid-Base Chemistry: in the Haber process nitrogen gains hydrogen to form ammonia, so nitrogen is reduced.
Common mistakes
Oxidation is the loss of oxygen.
Oxidation is the gain of oxygen or the loss of hydrogen, and the loss of oxygen is reduction.
The loss of hydrogen is reduction.
The loss of hydrogen is oxidation, and the gain of hydrogen is reduction.
In the reaction of copper(II) oxide with hydrogen only the hydrogen changes.
Copper(II) oxide is reduced as it loses oxygen, and hydrogen is oxidised as it gains oxygen.
A substance can be oxidised without anything being reduced.
Oxidation and reduction always occur together, which is why the reaction is called redox.
Gain of oxygen or loss of hydrogen is oxidation, loss of oxygen or gain of hydrogen is reduction, and they always happen together.
Exam check
Q1Which change is an oxidation?
- Acopper(II) oxide, CuO, changing to copper, Cu
- Bnitrogen, N2, changing to ammonia, NH3
- Ciron(III) oxide, Fe2O3, changing to iron, Fe
- Dhydrogen sulfide, H2S, changing to sulfur, S
Show answer
MCQ answer key D (hydrogen sulfide, H2S, changing to sulfur, S): hydrogen sulfide loses hydrogen, which is oxidation, whereas the other three changes are a loss of oxygen or a gain of hydrogen, which are reductions.
Q2In the reaction CuO(s) + H2(g) → Cu(s) + H2O(l) which substance is reduced?
- Acopper(II) oxide
- Bhydrogen
- Cwater
- Dcopper
Show answer
MCQ answer key A (copper(II) oxide): copper(II) oxide loses oxygen, so it is reduced, and hydrogen gains oxygen, so it is oxidised.
Q3Which statement about redox reactions is correct?
- AOxidation is the loss of oxygen.
- BOxidation and reduction always occur together.
- CReduction is the loss of hydrogen.
- DOxidation can occur without reduction when a metal burns.
Show answer
MCQ answer key B (Oxidation and reduction always occur together.): the electrons lost by the substance that is oxidised are gained by another substance, so the two always occur together, and distractor D is wrong because a burning metal loses electrons while the oxygen gains them.
StructuredIron(III) oxide is reduced to iron by carbon monoxide: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
- State what is meant by reduction, in terms of oxygen. [1]
- Name the substance that is oxidised and give a reason. [2]
- Explain why the reaction is described as redox. [1]
Total: 4 marks
Show mark scheme
(a) 1 mark
- Mark scheme: the loss of oxygen
(b) 2 marks
- Mark scheme: carbon monoxide
- Mark scheme: it gains oxygen to form carbon dioxide
(c) 1 mark
- Mark scheme: iron(III) oxide is reduced and carbon monoxide is oxidised at the same time
Electron transfer and oxidation states
define redox in terms of electron transfer and changes in oxidation state
Oxidation is the loss of electrons and an increase in oxidation state, and reduction is the gain of electrons and a decrease in oxidation state.
- Oxidation (electrons)
- Oxidation (electrons): the loss of electrons by an atom, ion or molecule
- Reduction (electrons)
- Reduction (electrons): the gain of electrons by an atom, ion or molecule
- Oxidation state
- Oxidation state: a number, with its sign first, that shows how many electrons an atom has lost (+) or gained (–) compared with the free atom, worked out with the rules below, as in +2 or −1
- Half-equation
- Half-equation: an equation that shows only the oxidation or only the reduction, with the electrons written in the equation supports K324-7.1(b)
- Electron transfer
- Electron transfer: the movement of electrons from the substance that is oxidised to the substance that is reduced supports K324-7.1(b)
- Roman numeral
- Roman numeral: the number in a name such as iron(III) oxide, which gives the oxidation state of the metal supports K324-7.1(b)
The four definitions together
| Definition in terms of | Oxidation | Reduction |
|---|---|---|
| oxygen | Oxidation is the gain of oxygen. | Reduction is the loss of oxygen. |
| hydrogen | Oxidation is the loss of hydrogen. | Reduction is the gain of hydrogen. |
| electrons | Oxidation is the loss of electrons. | Reduction is the gain of electrons. |
| oxidation state | Oxidation is an increase in oxidation state. | Reduction is a decrease in oxidation state. |
Half-equations and electron transfer
Zn(s) → Zn2+(aq) + 2e–Zinc loses two electrons to form zinc ions, so zinc is oxidised.
Cu2+(aq) + 2e– → Cu(s)Copper(II) ions gain two electrons to form copper, so copper(II) ions are reduced.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)Adding the two half-equations, with the electrons cancelling, gives the overall ionic equation.
Cl2(g) + 2e– → 2Cl–(aq)Chlorine gains electrons to form chloride ions, so chlorine is reduced.
2Br–(aq) → Br2(aq) + 2e–Bromide ions lose electrons to form bromine, so bromide ions are oxidised.
Rules for oxidation states
- 1The oxidation state of an atom in an element is 0.
- 2The oxidation state of a simple ion is the charge on the ion, so sodium ions, Na+, are +1 and chloride ions, Cl–, are −1.
- 3In a neutral compound the oxidation states of all the atoms add up to 0, and in an ion they add up to the charge on the ion.
- 4Oxygen is usually −2 and hydrogen is usually +1 in compounds.
- 5Work out the unknown oxidation state by making the total equal to zero for a compound or to the charge for an ion.
| Atom | Rule applied and working |
|---|---|
| magnesium, Mg | The oxidation state of an atom in an element is 0, so the oxidation state of magnesium is 0. |
| carbon in carbon dioxide, CO2 | C + 2 × (−2) = 0, so the oxidation state of C is +4. |
| sulfur in sulfur dioxide, SO2 | S + 2 × (−2) = 0, so the oxidation state of S is +4. |
| sulfur in the sulfite ion, SO32– | S + 3 × (−2) = −2, so the oxidation state of S is +4. |
| sulfur in the sulfate ion, SO42– | S + 4 × (−2) = −2, so the oxidation state of S is +6. |
| nitrogen in ammonia, NH3 | N + 3 × (+1) = 0, so the oxidation state of N is −3. |
| nitrogen in the nitrate ion, NO3– | N + 3 × (−2) = −1, so the oxidation state of N is +5. |
| manganese in the manganate(VII) ion, MnO4– | Mn + 4 × (−2) = −1, so the oxidation state of Mn is +7. |
| iron in iron(III) oxide, Fe2O3 | 2Fe + 3 × (−2) = 0, so the oxidation state of Fe is +3. |
SEAB writes oxidation state, so write oxidation state in your answers and not oxidation number. supports K324-7.1(b)
The Roman numeral in a name gives the oxidation state, so iron(III) oxide contains iron with an oxidation state of +3 and the manganate(VII) ion contains manganese with an oxidation state of +7. supports K324-7.1(b)
Worked example
Magnesium reacts with hydrochloric acid, Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) Show that this is a redox reaction by using oxidation states.
- Magnesium is an element, so its oxidation state is 0 on the left, and in magnesium chloride it is +2 because the magnesium ion is Mg2+.
- Hydrogen in hydrochloric acid is +1 and hydrogen gas is an element, so its oxidation state falls to 0.
- The oxidation state of chlorine stays at −1 on both sides, so chlorine is neither oxidised nor reduced.
- Magnesium rises from 0 to +2 so magnesium is oxidised, and hydrogen falls from +1 to 0 so hydrogen is reduced.
The reaction is redox because magnesium is oxidised and hydrogen is reduced at the same time.
The SEAB glossary says “Calculate is used when a numerical answer is required”, so when you are asked to calculate an oxidation state, give the sign with the number, such as +4, and show the working.
Depth: why and how
- Why is the sign written before the number? The sign shows whether the atom has lost or gained electrons, so +2 is not the same as 2−, which is the charge of an ion.
- Why does the oxidation state of an element equal zero? The atoms in an element are all the same, so no atom has a share of electrons that differs from its neighbour.
- How can you tell oxidation from the half-equation? Electrons appear on the right in an oxidation, as in zinc forming zinc ions, and on the left in a reduction.
- Why must the electrons cancel when two half-equations are added? Electrons lost by the substance that is oxidised are gained by the substance that is reduced, so none remain in the overall equation.
- Cross-link to Chemical Bonding and Structure: when sodium reacts with chlorine, each sodium atom transfers an electron to a chlorine atom, so sodium is oxidised and chlorine is reduced.
Common mistakes
Oxidation is the gain of electrons.
Oxidation is the loss of electrons and reduction is the gain of electrons.
The oxidation state of oxygen in every compound is −2.
Oxygen is usually −2 in compounds, so use −2 unless the information given says otherwise, but it is −1 in hydrogen peroxide, H2O2, and 0 in the element oxygen, O2.
The oxidation state of sulfur in the sulfate ion is −2.
The oxidation state of sulfur in the sulfate ion is +6, because the total for the ion is −2 and the four oxygen atoms contribute −8.
The electrons are left in the overall equation.
The electrons cancel when the half-equations are added, so the overall equation shows no electrons.
Loss of electrons and an increase in oxidation state are oxidation, gain of electrons and a decrease in oxidation state are reduction, and the oxidation state of an element is 0.
Exam check
Q1What is the oxidation state of sulfur in the sulfate ion, SO42–?
- A+4
- B+6
- C−2
- D+8
Show answer
MCQ answer key B (+6): the ion has a total of −2 and four oxygen atoms at −2 add up to −8, so sulfur is +6.
Q2What is the oxidation state of manganese in the manganate(VII) ion, MnO4–?
- A+7
- B+4
- C−1
- D−7
Show answer
MCQ answer key A (+7): the ion has a total of −1 and four oxygen atoms at −2 add up to −8, so manganese is +7, which the Roman numeral VII also shows.
Q3In the half-equation Zn(s) → Zn2+(aq) + 2e– which statement is correct?
- AZinc loses electrons and is oxidised.
- BZinc gains electrons and is oxidised.
- CZinc loses electrons and is reduced.
- DZinc gains electrons and is reduced.
Show answer
MCQ answer key A (Zinc loses electrons and is oxidised.): the electrons are on the right-hand side, so zinc loses them, and loss of electrons is oxidation.
StructuredZinc reacts with aqueous copper(II) sulfate, and the ionic equation is Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
- Write the half-equation for the oxidation, with state symbols. [2]
- State the change in the oxidation state of copper. [1]
- Explain, in terms of electrons, why copper(II) ions are reduced. [1]
Total: 4 marks
Show mark scheme
(a) 2 marks
- Mark scheme: Zn(s) → Zn2+(aq) + 2e–
- Mark scheme: the electrons are shown and the state symbols are correct
(b) 1 mark
- Mark scheme: copper falls from +2 to 0
(c) 1 mark
- Mark scheme: copper(II) ions gain electrons
Identifying redox reactions and agents
identify redox reactions in terms of oxygen/hydrogen gain/loss, electron gain/loss and changes in oxidation state
A reaction is redox when something gains or loses oxygen or hydrogen, gains or loses electrons, or changes its oxidation state, and the oxidising agent is the substance that is reduced while the reducing agent is the substance that is oxidised.
- Oxidising agent
- Oxidising agent: a substance that takes electrons from another substance and so oxidises it, and it is itself reduced in the reaction
- Reducing agent
- Reducing agent: a substance that gives electrons to another substance and so reduces it, and it is itself oxidised in the reaction
- Redox reaction
- Redox reaction: a reaction in which at least one oxidation state changes supports K324-7.1(c)
- Displacement reaction
- Displacement reaction: a reaction in which a more reactive element takes the place of a less reactive element in a compound, and it is a redox reaction supports K324-7.1(c)
- Spectator ion
- Spectator ion: an ion that is present in the mixture but does not change in the reaction, so it does not appear in the ionic equation supports K324-7.1(c)
- Not redox
- Not redox: a reaction in which no oxidation state changes, such as neutralisation, precipitation and the thermal decomposition of calcium carbonate supports K324-7.1(c)
Is it redox?
| Reaction | What changes | Redox? |
|---|---|---|
| CuO(s) + H2(g) → Cu(s) + H2O(l) | Copper(II) oxide loses oxygen, hydrogen gains oxygen, copper falls from +2 to 0 and hydrogen rises from 0 to +1. | Yes, it is redox. |
| Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) | Zinc loses electrons and copper(II) ions gain electrons, so zinc rises from 0 to +2 and copper falls from +2 to 0. | Yes, it is redox. |
| Cl2(g) + 2KI(aq) → 2KCl(aq) + I2(aq) | Chlorine falls from 0 to −1 and iodine rises from −1 to 0, and potassium stays at +1. | Yes, it is redox. |
| HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) | No atom changes its oxidation state, because hydrogen stays at +1 and chlorine stays at −1. | No, it is neutralisation. |
| Ag+(aq) + Cl–(aq) → AgCl(s) | Silver stays at +1 and chlorine stays at −1, so there is no transfer of electrons. | No, it is precipitation. |
| CaCO3(s) → CaO(s) + CO2(g) | Calcium stays at +2, carbon stays at +4 and oxygen stays at −2. | No, it is thermal decomposition without a change in oxidation state. |
Method
- 1Write the oxidation state of every element above each formula in the equation.
- 2Look for an element whose oxidation state changes, because if none changes the reaction is not redox.
- 3The element whose oxidation state rises is oxidised, and the element whose oxidation state falls is reduced.
- 4The substance that contains the element that is oxidised is the reducing agent, and the substance that contains the element that is reduced is the oxidising agent.
Oxidising and reducing agents
| Reaction | Oxidising agent | Reducing agent |
|---|---|---|
| copper(II) oxide and hydrogen | Copper(II) oxide is the oxidising agent because it is reduced. | Hydrogen is the reducing agent because it is oxidised. |
| zinc and copper(II) ions | Copper(II) ions are the oxidising agent because they are reduced. | Zinc is the reducing agent because it is oxidised. |
| chlorine and aqueous potassium iodide | Chlorine is the oxidising agent because it is reduced. | Iodide ions are the reducing agent because they are oxidised. |
Worked example
Zinc reacts with dilute hydrochloric acid, Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) Identify the oxidising agent and the reducing agent.
- Zinc rises from 0 to +2, so zinc is oxidised.
- Hydrogen falls from +1 in hydrochloric acid to 0 in hydrogen gas, so the hydrogen ions are reduced.
- Chlorine stays at −1, so it is not oxidised or reduced and the chloride ions are spectator ions.
- The substance that is reduced is the oxidising agent, which is hydrochloric acid, and the substance that is oxidised is the reducing agent, which is zinc.
Zinc is the reducing agent and hydrochloric acid is the oxidising agent.
The SEAB glossary says “Predict or deduce implies that the candidate is not expected to produce the required answer by recall but by making a logical connection between other pieces of information”, so when you identify the oxidising agent, deduce it from the change in oxidation state and name the substance, not only the element.
Depth: why and how
- Why is the oxidising agent the substance that is reduced? It takes electrons from the other substance and so causes its oxidation, and by taking them it is reduced itself.
- Why is a displacement reaction a redox reaction? A more reactive metal loses electrons to the ions of a less reactive metal, so oxidation states change.
- Why is neutralisation not redox? Hydrogen ions and hydroxide ions combine and no atom changes its oxidation state.
- Why is decomposition not always redox? The thermal decomposition of calcium carbonate changes no oxidation states, although some decompositions, such as water by electricity, do.
- Cross-link to Patterns in the Periodic Table: in a halogen displacement reaction the more reactive halogen is the oxidising agent.
Common mistakes
The oxidising agent is the substance that is oxidised.
The oxidising agent is the substance that is reduced, and the reducing agent is the substance that is oxidised.
Every reaction in which a compound forms is redox.
A reaction is redox only if an oxidation state changes, so neutralisation and precipitation are not redox.
The oxidising agent is the element, such as zinc, that loses electrons.
The element that loses electrons is oxidised, so the substance that contains it is the reducing agent.
The spectator ion chloride is reduced.
A spectator ion does not change its oxidation state, so chloride ions are neither oxidised nor reduced.
Write the oxidation states, look for a change, and remember that the substance that is reduced is the oxidising agent while the substance that is oxidised is the reducing agent.
Exam check
Q1Which reaction is not a redox reaction?
- AZn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
- BCuO(s) + H2(g) → Cu(s) + H2O(l)
- CHCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
- DCl2(g) + 2KI(aq) → 2KCl(aq) + I2(aq)
Show answer
MCQ answer key C (HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)): no oxidation state changes in the neutralisation of hydrochloric acid by sodium hydroxide, whereas the other three reactions each involve a change in oxidation state.
Q2In Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) which substance is the reducing agent?
- Azinc
- Bhydrochloric acid
- Czinc chloride
- Dhydrogen
Show answer
MCQ answer key A (zinc): zinc rises from 0 to +2, so zinc is oxidised and is the reducing agent.
Q3Which statement about an oxidising agent is correct?
- AIt is oxidised in the reaction.
- BIt loses electrons in the reaction.
- CIts oxidation state rises in the reaction.
- DIt is reduced in the reaction.
Show answer
MCQ answer key D (It is reduced in the reaction.): an oxidising agent takes electrons from another substance, so it gains electrons, its oxidation state falls and it is reduced.
StructuredChlorine gas is bubbled into aqueous potassium iodide, and the equation is Cl2(g) + 2KI(aq) → 2KCl(aq) + I2(aq)
- State the change in the oxidation state of chlorine. [1]
- Name the reducing agent and give a reason. [2]
- Identify the oxidising agent. [1]
Total: 4 marks
Show mark scheme
(a) 1 mark
- Mark scheme: chlorine falls from 0 to −1
(b) 2 marks
- Mark scheme: potassium iodide, or iodide ions
- Mark scheme: the iodide ions are oxidised because their oxidation state rises from −1 to 0
(c) 1 mark
- Mark scheme: chlorine
Tests for oxidising and reducing agents
describe the use of aqueous potassium iodide and acidified potassium manganate(VII) in testing for oxidising and reducing agents from the resulting colour changes.
Aqueous potassium iodide goes from colourless to brown with an oxidising agent, and acidified potassium manganate(VII) goes from purple to colourless with a reducing agent.
- Aqueous potassium iodide
- Aqueous potassium iodide: a colourless solution that contains iodide ions, which are oxidised to brown iodine by an oxidising agent
- Acidified potassium manganate(VII)
- Acidified potassium manganate(VII): a purple solution that is made colourless by a reducing agent, and the acid is dilute sulfuric acid
- Test for an oxidising agent
- Test for an oxidising agent: add the unknown to aqueous potassium iodide, and a change from colourless to brown shows that iodine has formed supports K324-7.1(d)
- Test for a reducing agent
- Test for a reducing agent: add the unknown to acidified potassium manganate(VII), and a change from purple to colourless shows that the manganate(VII) ions have been reduced supports K324-7.1(d)
- Iodine
- Iodine: I2, a non-metal that makes the solution brown when it forms in aqueous potassium iodide supports K324-7.1(d)
- Manganate(VII) ion
- Manganate(VII) ion: MnO4–, the purple ion in which manganese has an oxidation state of +7 supports K324-7.1(d)
The two tests
| Reagent | Colour before | Colour change with the agent | What the result shows |
|---|---|---|---|
| aqueous potassium iodide | Aqueous potassium iodide is colourless. | With an oxidising agent the colour changes from colourless to brown. | An oxidising agent is present, because iodide ions have been oxidised to iodine. |
| acidified potassium manganate(VII) | Acidified potassium manganate(VII) is purple. | With a reducing agent the colour changes from purple to colourless. | A reducing agent is present, because manganate(VII) ions have been reduced. |
SEAB's reagent list gives aqueous potassium manganate(VII) as approximately 0.02 mol/dm³ and aqueous potassium iodide as approximately 0.1 mol/dm³. supports K324-7.1(d)
- 1To test for an oxidising agent, add a few drops of aqueous potassium iodide to a small volume of the unknown solution.
- 2Watch for a colour change from colourless to brown, which shows that an oxidising agent is present.
- 3To test for a reducing agent, add the unknown solution slowly to acidified potassium manganate(VII).
- 4Watch for a colour change from purple to colourless, which shows that a reducing agent is present.
Equation
Cl2(aq) + 2I–(aq) → 2Cl–(aq) + I2(aq)Chlorine water oxidises colourless iodide ions to brown iodine, so chlorine is the oxidising agent and iodide ions are the reducing agent.
Sulfur dioxide as a reducing agent
- turns aqueous acidified potassium manganate(VII) from purple to colourless
SEAB lists sulfur dioxide as a gas that gives this result with acidified potassium manganate(VII), which shows that sulfur dioxide is a reducing agent. supports K324-7.1(d)
Write potassium manganate(VII) and not permanganate, because this is the name SEAB uses. supports K324-7.1(d)
Memory hook
Bees Dance
Brown in aqueous potassium iodide shows an oxidising agent, and a decolourised acidified potassium manganate(VII) shows a reducing agent.Key: B is the brown colour with aqueous potassium iodide, which means an oxidising agent, and D is the decolourised purple with acidified potassium manganate(VII), which means a reducing agent.Brown with aqueous potassium iodide means an oxidising agent, and purple to colourless with acidified potassium manganate(VII) means a reducing agent.
The SEAB glossary says “Describe requires candidates to state in words (using diagrams where appropriate) the main points of the topic”, so when you describe a test, name the reagent, state the colour before and after, and say what the result shows.
Depth: why and how
- Why is the test for an oxidising agent done with potassium iodide? Iodide ions are easily oxidised, so an oxidising agent turns them into brown iodine.
- Why is the potassium manganate(VII) acidified? The reaction that makes it colourless needs hydrogen ions, so the acid is added with the reagent.
- Why does the purple colour disappear? The purple manganate(VII) ions are reduced, and the product is almost colourless in dilute solution.
- Why can one reagent test only one kind of agent? Aqueous potassium iodide can only be oxidised, so it detects an oxidising agent, and manganate(VII) is easily reduced, so it detects a reducing agent.
- Cross-link to Qualitative Analysis: sulfur dioxide gas is tested with acidified potassium manganate(VII), and the same colour change shows a reducing agent.
Common mistakes
Aqueous potassium iodide turns brown with a reducing agent.
Aqueous potassium iodide turns brown with an oxidising agent, because iodide ions are oxidised to iodine.
Acidified potassium manganate(VII) turns colourless with an oxidising agent.
Acidified potassium manganate(VII) turns from purple to colourless with a reducing agent.
The colour change in the test for a reducing agent is from colourless to purple.
The colour change is from purple to colourless.
The test uses permanganate.
SEAB names the reagent as acidified potassium manganate(VII).
Exam check
Q1A solution is added to aqueous potassium iodide and the mixture turns brown. What does this show?
- AThe solution contains a reducing agent.
- BThe solution is acidic.
- CThe solution contains a precipitate.
- DThe solution contains an oxidising agent.
Show answer
MCQ answer key D (The solution contains an oxidising agent.): iodide ions are oxidised to brown iodine, so an oxidising agent is present.
Q2Which colour change shows a reducing agent with acidified potassium manganate(VII)?
- Apurple to colourless
- Bcolourless to purple
- Cpurple to brown
- Dcolourless to brown
Show answer
MCQ answer key A (purple to colourless): the purple manganate(VII) ions are reduced and the mixture becomes colourless.
Q3Which gas turns acidified potassium manganate(VII) from purple to colourless?
- Achlorine
- Bhydrogen
- Csulfur dioxide
- Doxygen
Show answer
MCQ answer key C (sulfur dioxide): SEAB gives this as the test for sulfur dioxide, which shows that sulfur dioxide is a reducing agent.
StructuredA student tests two solutions, X and Y. Aqueous potassium iodide turns brown when X is added, and acidified potassium manganate(VII) turns from purple to colourless when Y is added.
- State what the result shows about X. [1]
- State what the result shows about Y. [1]
- Name the substance that makes the mixture brown in the test with X. [1]
- State, in terms of oxidation state, what happens to the iodide ions in the test with X. [1]
Total: 4 marks
Show mark scheme
(a) 1 mark
- Mark scheme: X is an oxidising agent
(b) 1 mark
- Mark scheme: Y is a reducing agent
(c) 1 mark
- Mark scheme: iodine
(d) 1 mark
- Mark scheme: the oxidation state of iodine rises from −1 to 0
Electrolysis and the electrolytic cell
describe electrolysis as the conduction of electricity through an ionic compound (an electrolyte), when molten or dissolved in water, leading to chemical changes (including decomposition) at the electrodes
Electrolysis is the conduction of electricity through an ionic compound, called the electrolyte, when it is molten or dissolved in water, and it causes chemical changes, including decomposition, at the electrodes.
- Electrolysis
- Electrolysis: the conduction of electricity through a molten or dissolved ionic compound, which causes chemical changes, including decomposition, at the electrodes
- Electrolyte
- Electrolyte: a molten or dissolved ionic substance that conducts electricity because its ions are free to move, and the current causes chemical changes in it
- Electrode
- Electrode: a conductor, such as graphite or a metal, through which the current enters or leaves the electrolyte
- Anode
- Anode: the positive electrode, where oxidation takes place, because negative ions lose electrons or the atoms of a metal anode lose electrons
- Cathode
- Cathode: the negative electrode, where positive ions gain electrons and reduction takes place
- Inert electrode
- Inert electrode: an electrode, such as graphite or platinum, that conducts but does not take part in the reaction supports K324-7.2(a)
- Cation
- Cation: a positive ion, which moves towards the cathode supports K324-7.2(a)
- Anion
- Anion: a negative ion, which moves towards the anode supports K324-7.2(a)
The electrolytic cell
What happens in the cell
- 1Electrons leave the negative terminal of the power supply and flow through the wire to the cathode.
- 2At the cathode, cations gain electrons and are reduced.
- 3At the anode, anions lose electrons and are oxidised, or the atoms of an active metal anode lose electrons.
- 4The electrons from the anode flow through the wire to the positive terminal of the power supply.
- 5Ions carry the charge through the electrolyte, with cations moving to the cathode and anions moving to the anode, and electrons do not travel through the electrolyte.
In the wires, the electrons leave the negative terminal of the power supply and reach the cathode, and the electrons released at the anode return to the positive terminal, while ions, not electrons, carry the charge through the electrolyte. supports K324-7.2(a)
Cu2+(aq) + 2e– → Cu(s)Copper(II) ions gain electrons at the cathode, so they are reduced.
2Cl–(aq) → Cl2(g) + 2e–Chloride ions lose electrons at the anode, so they are oxidised.
Nine Red Crabs
In electrolysis the cathode is the negative electrode, reduction happens at the cathode and cations move to the cathode.Key: N is the negative electrode, R is reduction and C is the cations, and all three belong to the cathode of an electrolytic cell.Electrolysis passes electricity through a molten or dissolved ionic compound, with reduction at the cathode, oxidation at the anode, and ions carrying the charge through the electrolyte.
The SEAB glossary says “Describe requires candidates to state in words (using diagrams where appropriate) the main points of the topic”, so when you describe electrolysis, name the electrolyte, say that it is molten or dissolved in water, and state what happens at each electrode.
Depth: why and how
- Why must the compound be molten or dissolved? The ions of a solid are held in a lattice and cannot move, so the solid cannot conduct.
- Why is graphite a good inert electrode? Graphite conducts electricity and does not react with most electrolytes, so it only carries the electrons.
- Why are the changes at the electrodes redox reactions? Electrons are gained at the cathode and lost at the anode, so reduction and oxidation take place at the same time.
- How do you know which electrode is the cathode? The cathode is connected to the negative terminal of the power supply, and the positive ions are attracted to it.
- Cross-link to Chemical Bonding and Structure: the ions of an ionic compound are held in a giant lattice by strong electrostatic attraction.
Common mistakes
Electrons travel through the electrolyte.
Ions carry the charge through the electrolyte, and electrons travel only through the wires and the electrodes.
Reduction takes place at the anode.
Oxidation takes place at the anode and reduction takes place at the cathode.
The cathode is the positive electrode.
In an electrolytic cell the cathode is connected to the negative terminal of the power supply.
Electrolysis only works with a molten compound.
Electrolysis works when an ionic compound is molten or dissolved in water.
Exam check
Q1Which description of the anode in an electrolytic cell is correct?
- Athe positive electrode, where oxidation takes place
- Bthe negative electrode, where oxidation takes place
- Cthe positive electrode, where reduction takes place
- Dthe negative electrode, where reduction takes place
Show answer
MCQ answer key A (the positive electrode, where oxidation takes place): the anode is connected to the positive terminal and anions lose electrons there, which is oxidation.
Q2What carries the electric charge through the electrolyte during electrolysis?
- Aelectrons
- Batoms
- Cions
- Dmolecules
Show answer
MCQ answer key C (ions): ions are free to move in a molten or dissolved electrolyte, and electrons only flow through the wires.
Q3Which statement about the flow of electrons in an electrolytic cell is correct?
- AElectrons flow through the electrolyte from the anode to the cathode.
- BElectrons flow through the wire from the positive terminal of the power supply to the cathode.
- CElectrons flow through the wire from the cathode to the anode.
- DElectrons flow through the wire from the anode to the positive terminal of the power supply.
Show answer
MCQ answer key D (Electrons flow through the wire from the anode to the positive terminal of the power supply.): electrons leave the negative terminal and go to the cathode, and the electrons lost at the anode return to the positive terminal.
StructuredA student electrolyses a molten ionic compound using graphite electrodes.
- State what is meant by the term electrolyte. [1]
- Name the positive electrode and state the type of reaction that takes place there. [2]
- Describe the direction of the flow of electrons in the wires. [2]
Total: 5 marks
Show mark scheme
(a) 1 mark
- Mark scheme: a substance, such as a molten or dissolved ionic compound, that conducts electricity
(b) 2 marks
- Mark scheme: the anode
- Mark scheme: oxidation
(c) 2 marks
- Mark scheme: electrons flow from the negative terminal of the power supply to the cathode
- Mark scheme: electrons flow from the anode to the positive terminal of the power supply
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Still to read in this chapter:
- Electrolysis as evidence for ions
- Electrolysis of molten sodium chloride
- Molten binary ionic compounds
- Selective discharge of ions
- Products of aqueous electrolytes
- Ionic equations at the electrodes
- Purifying copper by electrolysis
- Electroplating
- Simple cells
- The hydrogen fuel cell
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Notes version 1.0 · Last updated 2 Oct 2026 · Chapter 7 of 13