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Cambridge IGCSE0620

Physical and chemical changes

Chemistry 0620 Chapter Notes

What this chapter covers

Physical and chemical changesRate of reactionReversible reactions and equilibriumRedox
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In this chapter

  1. 1Physical changes
  2. 2Chemical changes
  3. 3Telling them apart
  4. 4What changes the rate of a reaction
  5. 5Catalysts
  6. 6Measuring the rate in the laboratory
  7. 7Reading rate graphs
  8. 8Collision theory and activation energy
  9. 9Explaining the factors with collision theory
  10. 10How a catalyst works
  11. 11Evaluating practical methods
  12. 12Reversible reactions
  13. 13Hydrated and anhydrous compounds
  14. 14The colour tests for water
  15. 15Equilibrium in a closed system
  16. 16Changing the position of equilibrium
  17. 17The Haber process
  18. 18The Contact process
  19. 19Why these conditions are used
  20. 20Roman numerals and oxidation numbers
  21. 21Oxidation and reduction as oxygen transfer
  22. 22Redox: oxidation and reduction together
  23. 23Oxidation and reduction as electron transfer
  24. 24Oxidation numbers
  25. 25Colour changes that show redox
  26. 26Oxidising agents and reducing agents

1. Physical changes

In a physical change, no new substance is formed. The same substance is there before and after.

Melting, boiling and sublimation are changes of state, so they are physical changes. Dissolving is a physical change too.

Examiner insight

“Candidates knew that melting and boiling were physical changes but not dissolving.” “The commonest error was to suggest that salt dissolving in water is a chemical change.”

Worked example 12 marks

A student melts some ice. State whether this is a physical change or a chemical change. Give a reason for your answer.

  1. 1

    It is a physical change. [1]

  2. 2

    No new substance is formed. Ice and liquid water are the same substance. [1]

2. Chemical changes

In a chemical change, a new substance is formed. This is a chemical reaction.

Look for the observations of a reaction: effervescence, a colour change, or a gas that passes a test.

Examiner insight

“All the options were popular. This suggests that the expected observations for chemical reactions is not well known.”

Worked example 11 mark

Which process involves a chemical change? A adding sodium to water · B boiling water · C dissolving sodium chloride in water · D producing water from aqueous sodium chloride

  1. 1

    A: adding sodium to water. A new substance is formed. B, C and D make no new substance. [1]

3. Telling them apart

Ask one question: is a new substance made? If yes, the change is chemical. If no, it is physical.

A change of state does not decide it. A chemical change sometimes has a change of state. A physical change never makes a new substance.

Examiner insight

“It is expected that all candidates know that temperature remains constant during a physical change such as boiling for example.” “Similarly, all candidates should recognise that as matter does not appear or disappear, the mass of all the substance before and after a chemical or physical change is constant.”

Worked example 11 mark

Which row describes a chemical change?

new substances are madethere is a change of state
Aalwaysalways
Balwayssometimes
Cneveralways
Dneversometimes
  1. 1

    B: new substances are always made, and there is sometimes a change of state. [1]

Recap

  • Physical change: no new substance is formed.
  • Melting, boiling, sublimation and dissolving are physical changes.
  • Chemical change: a new substance is formed.
  • Observations of a reaction: effervescence, colour change, gas tests.
  • A chemical change always makes new substances. It only sometimes changes state.
  • Mass is constant in a chemical or a physical change. Temperature is constant while a substance boils.

4. What changes the rate of a reaction

A higher concentration of a solution, a higher pressure of a gas, a larger surface area of a solid and a higher temperature each make a reaction faster. A faster reaction takes a shorter time to finish.

Smaller pieces of a solid have a larger total surface area than one large piece, so a powder reacts faster than a lump or ribbon. A lower concentration or a lower temperature makes a reaction slower.

The volume of a solution is not its concentration. Adding more acid of the same concentration gives no change in the initial rate.

Examiner insight

“Candidates clearly thought that increasing the volume of the sulfuric acid would increase the rate of reaction.” “Many candidates confused time taken with rate in this question.”

Worked example 11 mark

Small pieces of calcium carbonate react with excess dilute hydrochloric acid of different concentrations. The time taken for each reaction to finish is recorded. The concentrations used are 0.75 mol / dm³, 1.50 mol / dm³ and 3.00 mol / dm³. All other conditions stay the same. Complete the table by writing the concentrations in the first column.

concentration of dilute hydrochloric acid in mol / dm³time taken for reaction to finish in s
16
32
8
  1. 1

    higher concentration, faster reaction, shorter time: 8 s is 3.00, 16 s is 1.50, 32 s is 0.75 (mol / dm³) [1]

5. Catalysts

A catalyst increases the rate of a reaction and is unchanged at the end of the reaction. Adding a catalyst makes a reaction faster. Without the catalyst the same reaction is slower. Enzymes are catalysts.

To show that a solid, insoluble catalyst is unchanged, separate it by filtration, dry it and weigh it. Its mass is the same as at the start.

Examiner insight

“However, some answers stated that catalysts alter the rate of a reaction without saying that they speed up a reaction.” “The most common error was to state ‘catalysts do not take part in the reaction’.”

Worked example 12 marks

Aqueous hydrogen peroxide, H₂O₂, decomposes to produce water and oxygen gas. Manganese(IV) oxide is a catalyst in this reaction. State what is meant by the term catalyst.

  1. 1

    a catalyst increases the rate of a reaction [1]

  2. 2

    and is unchanged at the end of the reaction [1]

6. Measuring the rate in the laboratory

A rate is found by measuring a change against time. When a gas is made, collect it in a gas syringe, or in a measuring cylinder over water, and time it. With the volume in cm³ and the time in s, the unit of the mean rate is cm³ / s.

When the gas escapes from the flask, the mass of the flask and contents falls. Record the mass at regular time intervals. A stop-watch is needed either way.

mean rate of reaction = (volume of gas collected)/(time taken to collect the gas)

Examiner insight

“Some candidates thought that a measuring cylinder could be used but did not know that it must be over water.” “A minority of candidates could describe how the results could be used to find the rate of the reaction and vague references to the rate being the same as the time of the reaction were common.”

Worked example 11 mark

A student measures the rate of two reactions. In one reaction, there is a change in mass of the reactants during the reaction. In the second reaction, there is a change in temperature during the reaction. Which piece of apparatus would be essential in both experiments?

A balance · B stop-watch · C pipette · D thermometer

  1. 1

    B: a stop-watch. A rate needs a change measured against time. [1]

7. Reading rate graphs

On a graph of gas volume against time, the line is steepest at the start, where the reaction is fastest. It gets less steep as the reaction slows.

When the line becomes horizontal the reaction has finished, because all of the reactant that was not in excess has been used up. The time taken is where the horizontal line starts.

A steeper line means a faster rate. A change that only speeds the reaction up gives a steeper gradient from (0,0) and levels off at the same final volume.

Examiner insight

“The commonest errors were 26 s or 140 s; the latter being the end of the horizontal line, rather than the point at which the horizontal line starts.” “The reaction is fastest at the start of the reaction because the concentration of hydrogen peroxide is at its highest at the start.”

Worked example 12 marks

A student investigates the reaction of small pieces of magnesium with excess dilute hydrochloric acid. The graph shows the total volume of hydrogen gas released as the reaction proceeds. Use the graph to determine the total volume of hydrogen gas released in 10 s, and to deduce the time taken for the reaction to finish.

  1. 1

    read up from 10 s to the curve: 16 cm³ [1]

  2. 2

    the line becomes horizontal at 40 s: the reaction finishes at 40 s [1]

8. Collision theory and activation energy

Collision theory describes rate using four ideas: the number of particles per unit volume, the frequency of collisions between particles, the kinetic energy of the particles and the activation energy, Ea.

The activation energy is the minimum energy colliding particles must have to react. A collision leads to a reaction only if its energy is greater than or equal to Ea. It is not the energy given out or taken in during the reaction.

Examiner insight

“Some candidates confused the activation energy with the overall energy change of reaction, choosing option A.” “This suggests that many candidates were unclear about the role of activation energy in reaction rate.”

Worked example 11 mark

Which statement describes activation energy?

A It is the maximum energy given out during a reaction. · B It is the maximum energy taken in during a reaction. · C It is the minimum energy colliding particles must have to react. · D It is the minimum energy particles must have to make bonds.

  1. 1

    C: the minimum energy colliding particles must have to react. [1]

9. Explaining the factors with collision theory

A higher concentration means more particles per unit volume. The particles are closer together, so the frequency of collisions rises: more collisions per second. A higher pressure pushes the particles of a gas closer together, so they also collide more frequently. A larger surface area of a solid gives more frequent collisions too.

These changes do not give the particles more energy, so the proportion of collisions with energy greater than or equal to Ea does not change. Pressure changes the rate only when a reactant is a gas.

A higher temperature does three things. The particles have more kinetic energy, they collide more frequently, and a greater proportion of collisions have energy greater than or equal to the activation energy.

Examiner insight

“A common error was to state that there were ‘less collisions’ or there was ‘less chance of a collision’ rather than stating that the frequency of collisions decreases.” “Just stating there are more successful collisions is insufficient as if there are more frequent collisions there must be more successful collisions.”

Worked example 13 marks

Aqueous hydrogen peroxide, H₂O₂, decomposes to produce water and oxygen gas. The reaction is catalysed by manganese(IV) oxide. The experiment is repeated at a higher temperature. All other conditions remain the same. Explain why the rate of reaction is greater at the higher temperature.

  1. 1

    the kinetic energy of the particles increases [1]

  2. 2

    the frequency of collisions between particles increases [1]

  3. 3

    a higher proportion of collisions have energy greater than or equal to the activation energy [1]

10. How a catalyst works

A catalyst decreases the activation energy, Ea, of a reaction. A greater proportion of collisions then have energy greater than or equal to Ea, so the rate increases.

A catalyst does not change the enthalpy change, Δ H, or the amount of product made. On a reaction pathway diagram, only the arrow for Ea gets smaller.

Examiner insight

“Lowering activation energy was also occasionally seen, which gained credit.” “A catalyst increases the rate of a reaction without having any effect on the amount of substance produced.”

Worked example 11 mark

Which row describes the effect of a catalyst on a reaction?

effect on activation energy, Eaeffect on enthalpy change, Δ H
Adecreasesdecreases
Bdecreasesno change
Cincreasesdecreases
Dincreasesno change
  1. 1

    B: Ea decreases and Δ H shows no change. [1]

11. Evaluating practical methods

To evaluate a method, decide how each fault changes the reading. Gas that escapes before the stopper is replaced makes the volume collected in a fixed time smaller, and the time taken to collect a fixed volume longer. The most gas escapes in the experiment that makes gas most quickly.

To stop the leak, keep the reagents apart inside the closed flask, then tip the flask to start the reaction.

Air already in the flask has no effect on accuracy. The carbon dioxide pushes out an equal volume of air, so the volume of gas collected is the same as the volume of carbon dioxide made.

Examiner insight

“Many candidates incorrectly thought that the air in the apparatus would increase the volume of gas collected and so make the results inaccurate.” “Many candidates suggested collecting the gas in a gas syringe, this will not remove the need to take the stopper out of the flask and so will not prevent some of the gas leaking.”

Worked example 12 marks

The gas collected in the inverted measuring cylinder contains air from the conical flask. State what effect, if any, this air will have on the accuracy of the results of the investigation. Explain your answer.

  1. 1

    effect on accuracy: none [1]

  2. 2

    the volume of air collected is the same as the volume of carbon dioxide left in the flask [1]

Recap

  • Higher concentration, pressure, surface area or temperature, or a catalyst: the reaction is faster and takes less time.
  • Volume of a solution is not concentration. More acid of the same concentration gives no change in the initial rate.
  • A catalyst increases the rate of a reaction and is unchanged at the end of the reaction.
  • Follow a rate by gas volume or by loss in mass, against time. Mean rate = volume of gas ÷ time.
  • On a volume–time graph the steepest part is the fastest. The reaction finishes where the line starts to be horizontal.
  • Extended: activation energy, Ea, is the minimum energy colliding particles must have to react.
  • Extended: concentration, pressure and surface area change collision frequency. Temperature also raises kinetic energy and the proportion of collisions with energy greater than or equal to Ea.
  • Extended: a catalyst decreases Ea. It does not change Δ H or the amount of product.

12. Reversible reactions

Some chemical reactions are reversible. The products can react to make the reactants again.

A reversible reaction is shown by the symbol ⇌. It has a forward reaction and a reverse reaction.

A + B ⇌ C + D

Examiner insight

“The commonest error here was to draw a single arrow, sometimes pointing backwards.” “There were a few candidates that called it a ‘reverse’ reaction and these candidates did not gain credit.”

Worked example 11 mark

What does the symbol ⇌ show in the equation for a chemical reaction? A Heat is needed for the reaction to occur · B Water is needed for the reaction to occur · C A catalyst is needed to increase the rate of reaction · D The chemical reaction is reversible

  1. 1

    D: the chemical reaction is reversible. [1]

13. Hydrated and anhydrous compounds

Hydrated means ‘with water’. Anhydrous means ‘with no water’. Hydrated crystals are chemically combined with water.

Adding water to an anhydrous compound makes the hydrated compound. Heating the hydrated compound drives the water off as steam and makes the anhydrous compound again. The change is reversible.

CuSO₄(s) + 5H₂O(l) ⇌ CuSO₄.5H₂O(s)

Examiner insight

“The word hydrated should be understood as ‘with water’ and anhydrous as ‘with no water’.”

Worked example 11 mark

Substance X is a blue solid. When X is heated, steam is produced and the solid turns white. What is X? A anhydrous cobalt(II) chloride · B anhydrous copper(II) sulfate · C hydrated cobalt(II) chloride · D hydrated copper(II) sulfate

  1. 1

    D: hydrated copper(II) sulfate is blue and turns white when it loses water. [1]

14. The colour tests for water

Anhydrous copper(II) sulfate is white. It turns blue when water is added.

Anhydrous cobalt(II) chloride is blue. It turns pink when water is added.

Both tests must use the anhydrous form. Heating the hydrated form reverses each change.

Examiner insight

“Most candidates knew that the halide used to detect water is cobalt(II) chloride, but many omitted the key point that it should be the anhydrous form of the compound.” “Many confused it with the test for water using anhydrous copper sulfate, often giving the colours the wrong way round.”

Worked example 11 mark

Which statement about anhydrous salts is correct? A Anhydrous cobalt(II) chloride turns from white to blue when added to water. · B Anhydrous cobalt(II) chloride loses water on gentle heating. · C Anhydrous copper(II) sulfate reacts with water to make hydrated copper(II) sulfate. · D Anhydrous copper(II) sulfate turns from blue to white when water is added.

  1. 1

    C: anhydrous copper(II) sulfate reacts with water to make the hydrated salt. [1]

15. Equilibrium in a closed system

In a closed system no reactants or products can enter or leave.

A reversible reaction in a closed system is at equilibrium when two things are true. The rate of the forward reaction is equal to the rate of the reverse reaction. The concentrations of reactants and products are no longer changing.

The reaction has not stopped. Both reactions carry on at the same rate.

Examiner insight

“The main error seen in the first statement was ‘forward reaction is equal to reverse reaction’ without mention of rates.” “The main error seen in the second statement was the concentrations of reactants and products are the same or are equal.”

Worked example 12 marks

H⁺(aq) ions convert CrO₄²⁻(aq) ions to Cr₂O₇²⁻(aq) ions. An equilibrium is reached.

2H⁺(aq) + 2CrO₄²⁻(aq) ⇌ Cr₂O₇²⁻(aq) + H₂O(l)

Describe this equilibrium in terms of the rate of the forward reaction and the rate of the reverse reaction, and the concentration of the reactants and of the products.

  1. 1

    The rates of the forward and reverse reactions are equal. [1]

  2. 2

    The concentrations of reactants and products are no longer changing. [1]

16. Changing the position of equilibrium

A change of condition moves the position of equilibrium to the left or to the right. State which way it moves.

Temperature: raising it moves the equilibrium in the endothermic direction. Lowering it moves it in the exothermic direction.

Pressure: raising it moves the equilibrium towards the side with fewer molecules of gas. If no gases are present, or both sides have the same number of molecules of gas, the position does not change.

Concentration: adding a substance moves the equilibrium away from it. Removing one moves it towards that substance, to replace it.

A catalyst speeds up the forward and the reverse reaction equally. The position of equilibrium shows no change.

Examiner insight

“A significant number of candidates focused on the rate of the reaction, despite the question asking about the position of equilibrium.” “It should be noted that there will be an endothermic direction and an exothermic direction; there is not an endothermic side or an exothermic side.”

Worked example 13 marks

A mixture of sulfur dioxide, SO₂, oxygen, O₂, and sulfur trioxide, SO₃, is allowed to reach equilibrium in a closed system in the presence of a catalyst.

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

The forward reaction is exothermic. Complete Table 6.1 using the words increases, decreases or no change.

change in conditioneffect on the equilibrium yield of SO₃(g)
decrease the pressure
increase the temperature
remove the catalyst
  1. 1

    decrease the pressure: decreases. The equilibrium moves left, to the side with more molecules of gas. [1]

  2. 2

    increase the temperature: decreases. The equilibrium moves left, in the endothermic direction. [1]

  3. 3

    remove the catalyst: no change. The position of equilibrium is the same. [1]

17. The Haber process

The Haber process makes ammonia. The hydrogen comes from methane. The nitrogen comes from air.

The typical conditions are 450 °C, 20 000 kPa (200 atm) and an iron catalyst.

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Examiner insight

“Air was well known; methane, far less so.” “200 000 kPa or 0.2 kPa were the most frequent errors.”

Worked example 15 marks

Ammonia is manufactured by reacting hydrogen with nitrogen in the Haber process.(i) State one source of hydrogen and one source of nitrogen.(ii) Write the symbol equation for the reaction in the Haber process.(iii) State the typical pressure, in kPa, used in the Haber process.

  1. 1

    hydrogen: methane [1]

  2. 2

    nitrogen: (liquid) air [1]

  3. 3

    species correct: N₂, H₂ and NH₃ [1]

  4. 4

    equation balanced: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) [1]

  5. 5

    pressure: 20 000 kPa [1]

18. The Contact process

The Contact process converts sulfur dioxide to sulfur trioxide. The sulfur dioxide comes from burning sulfur or roasting sulfide ores. The oxygen comes from air.

The typical conditions are 450 °C, 200 kPa (2 atm) and a vanadium(V) oxide catalyst.

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

Examiner insight

“As vanadium has many oxides, it was the specific vanadium(V) oxide which was required as the catalyst.” “Candidates picked the correct catalyst but chose the conditions for the Haber process.”

Worked example 11 mark

Four statements about the Haber process and the Contact process are listed. 1 Both processes are carried out at 200 kPa. 2 Both processes are carried out at 450 °C. 3 Both processes use a reactant that is extracted from air. 4 Both processes use iron as a catalyst. Which two statements are correct? A 1 and 2 · B 1 and 4 · C 2 and 3 · D 3 and 4

  1. 1

    C: both use 450 °C and both take a reactant from air. The pressures and catalysts differ. [1]

19. Why these conditions are used

Each condition is a compromise between the rate of reaction and the position of equilibrium.

For an exothermic forward reaction, a higher temperature moves the equilibrium to the left, so the yield is too low. A lower temperature makes the rate too low. About 450 °C balances the two.

A higher pressure moves the equilibrium towards the side with fewer molecules of gas, so the yield rises. Very high pressure costs more and is a safety risk, so the pressure is a compromise too.

Examiner insight

“It was acceptable to say that equilibrium shifted to the left-hand side at higher temperatures.” “There was confusion between equilibrium and rate in some cases.”

Worked example 14 marks

In the Contact process, 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), the forward reaction is exothermic. Explain why a temperature of 450 °C is used, and not a much higher or a much lower one.

  1. 1

    At a much higher temperature the equilibrium moves to the left, the endothermic direction. [1]

  2. 2

    so the yield of sulfur trioxide is too low. [1]

  3. 3

    At a much lower temperature the rate of reaction is too low. [1]

  4. 4

    450 °C is a compromise: a fast enough rate and an acceptable yield. [1]

Recap

  • Reversible reactions are shown by ⇌.
  • Hydrated: with water. Anhydrous: with no water. Heat drives off water; adding water reverses it.
  • Anhydrous to hydrated: copper(II) sulfate turns white to blue; cobalt(II) chloride turns blue to pink.
  • Extended: at equilibrium the rates are equal and the concentrations are no longer changing.
  • Extended: temperature, pressure and concentration move the position of equilibrium. A catalyst does not.
  • Extended: Haber, N₂(g) + 3H₂(g) ⇌ 2NH₃(g). 450 °C, 20 000 kPa, iron.
  • Extended: Contact, 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). 450 °C, 200 kPa, vanadium(V) oxide.

20. Roman numerals and oxidation numbers

A Roman numeral in the name of a compound indicates the oxidation number of the element it follows.

In nitrogen(II) oxide, the (II) shows that the oxidation number of nitrogen is +2. In copper(II) oxide, the oxidation number of copper is +2. In potassium manganate(VII), the oxidation number of manganese is +7.

The numeral is not the number of atoms in the formula. It says nothing about the other element.

Examiner insight

“Vague responses such as ‘its oxidation number is 2’ received no credit.” “Many candidates referred to charges on chlorine or to the valency of chlorine and did not appreciate that the use of Roman numerals indicates a positive oxidation number only.”

Worked example 11 mark

Nitrogen and oxygen can form the compound nitrogen(II) oxide.

Which statements about the Roman numeral in nitrogen(II) oxide are correct?

1 It indicates that the oxidation number of nitrogen is +2.

2 It indicates that the oxidation number of oxygen is +2.

3 It indicates that the formula for nitrogen(II) oxide contains two nitrogen atoms.

A 1 and 3 · B 1 only · C 2 and 3 · D 2 only

  1. 1

    Statement 1 is correct: the numeral gives the oxidation number of nitrogen.

  2. 2

    Statement 2 is wrong: the numeral belongs to nitrogen, not oxygen.

  3. 3

    Statement 3 is wrong: the numeral is not a number of atoms.

  4. 4

    B: 1 only. [1]

21. Oxidation and reduction as oxygen transfer

Oxidation is gain of oxygen. Reduction is loss of oxygen.

A reaction in which one substance gains oxygen and another loses oxygen is a redox reaction. The combustion of a hydrocarbon to form carbon dioxide and water is one.

2Mg + O₂ → 2MgO

CuO + H₂ → Cu + H₂O

Examiner insight

“Candidates should recall that if species are gaining and losing one or more oxygen atoms, a redox reaction is taking place. Many candidates did not recognise that combustion of a hydrocarbon to form carbon dioxide and water is an example of a redox reaction.” “Another common error was to suggest that oxygen is reduced.”

Worked example 11 mark

When heated strongly, silicon(IV) oxide reacts with carbon.

SiO₂ + 2C → Si + 2CO

Which term describes what happens to silicon(IV) oxide?

A thermal decomposition · B neutralisation · C oxidation · D reduction

  1. 1

    Silicon(IV) oxide, SiO₂, becomes silicon, Si. It loses oxygen.

  2. 2

    Loss of oxygen is reduction. The answer is D: reduction. [1]

22. Redox: oxidation and reduction together

A redox reaction involves simultaneous oxidation and reduction. One substance cannot be oxidised unless another is reduced.

Redox is a type of reaction. Oxidation and reduction each describe only what one reactant undergoes.

To identify them, look at the reactants on the left of the arrow. The reactant that gains oxygen is oxidised. The reactant that loses oxygen is reduced.

ZnO + C → Zn + CO

Examiner insight

“Questions about the oxidising agent and the reducing agent always require the reactants to be chosen rather than the products.” “Neither oxidation nor reduction should be described as a type of reaction.”

Worked example 11 mark

The equations show two chemical reactions involved in the production of iron from its ore.

1 C + CO₂ → 2CO

2 Fe₂O₃ + 3CO → 2Fe + 3CO₂

Which substance is reduced in reaction 1 and in reaction 2?

A CO₂ and CO · B CO₂ and Fe₂O₃ · C C and CO · D C and Fe₂O₃

  1. 1

    Reaction 1: CO₂ loses oxygen to become CO, so CO₂ is reduced.

  2. 2

    Reaction 2: Fe₂O₃ loses oxygen to become Fe, so Fe₂O₃ is reduced.

  3. 3

    B: CO₂ and Fe₂O₃. [1]

23. Oxidation and reduction as electron transfer

Oxidation is loss of electrons. Reduction is gain of electrons.

A reaction in which one species loses electrons and another gains them is a redox reaction.

Name the particle that changes. In the half-equation for copper, the copper(II) ions gain the electrons, not copper atoms.

Mg → Mg²⁺ + 2e⁻

Cu²⁺ + 2e⁻ → Cu

Examiner insight

“Most related oxidation to a loss of electrons; very few candidates could state what had been oxidised.” “Most candidates knew the change was reduction but erroneously attributed the electron gain to the Cu (atoms) rather than Cu2+ (ions).”

Worked example 11 mark

The ionic equation for the reaction of aqueous iron(II) sulfate with acidified aqueous potassium manganate(VII) is:

5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

Which statement about the reaction is correct?

A The iron(II) ions gain electrons. · B The iron(II) ions act as an oxidising agent. · C The manganate(VII) ions are reduced. · D The oxidation state of the iron(II) ions decreases.

  1. 1

    Fe²⁺ becomes Fe³⁺. Each ion loses an electron and its oxidation number rises from +2 to +3, so A and D are wrong.

  2. 2

    Iron(II) ions are oxidised, so they are the reducing agent. B is wrong.

  3. 3

    C: the manganate(VII) ions are reduced. [1]

24. Oxidation numbers

Oxidation is an increase in oxidation number. Reduction is a decrease in oxidation number.

The oxidation number of an element in its uncombined state is zero. The oxidation number of a monatomic ion is the same as the charge on the ion.

Find other values by sum. The sum in a compound is zero. The sum in an ion is equal to the charge on the ion. Oxygen is –2. Always write the sign before the number.

compound: sum(oxidation numbers) = 0

ion: sum(oxidation numbers) = charge on the ion

Examiner insight

“Candidates need to be aware that a key part of an oxidation number is the positive or negative sign preceding the integer. Integers alone are not oxidation numbers.” “Many of these answers were incorrect because they referred to sulfur dioxide as a whole rather than the sulfur.”

Worked example 14 marks

Use oxidation numbers to show that the reaction below is a redox reaction. The oxidation number of oxygen is –2.

Fe₂O₃ + 3CO → 2Fe + 3CO₂

  1. 1

    Iron in Fe₂O₃: 2x + 3(-2) = 0, so x = +3. Iron in Fe is 0. [1]

  2. 2

    Oxidation number of iron decreases from +3 to 0, so iron(III) oxide is reduced. [1]

  3. 3

    Carbon in CO: x + (-2) = 0, so x = +2. Carbon in CO₂: x + 2(-2) = 0, so x = +4. [1]

  4. 4

    Oxidation number of carbon increases from +2 to +4, so carbon monoxide is oxidised. Oxidation and reduction occur together: redox. [1]

25. Colour changes that show redox

Acidified aqueous potassium manganate(VII) is an oxidising agent. It is purple. When it is reduced it goes purple to colourless, so the substance added to it has been oxidised.

Aqueous potassium iodide is colourless and is a reducing agent. When its iodide ions are oxidised to iodine, the solution goes colourless to brown, so the substance added to it has been reduced.

Name a reagent with its numeral: potassium manganate(VII). State that it is acidified.

Examiner insight

“The acidic nature of the solution and the observation with red litmus was well known but the colour change when aqueous iodide ions are oxidised to form aqueous iodine was not.” “Most candidates knew the acidified reagent but often did not give the correct oxidation state or placed it in the wrong position such as potassium(VII) manganate.”

Worked example 13 marks

Aqueous chlorine is added to colourless aqueous potassium iodide. The solution turns brown.

Cl₂ + 2I⁻ → 2Cl⁻ + I₂

Explain why this is a redox reaction.

  1. 1

    Iodide ions lose electrons to form iodine, I₂. Iodide ions are oxidised. The brown colour is iodine. [1]

  2. 2

    Chlorine gains electrons to form chloride ions. Chlorine is reduced. [1]

  3. 3

    Oxidation and reduction happen at the same time, so it is a redox reaction. [1]

26. Oxidising agents and reducing agents

An oxidising agent is a substance that oxidises another substance and is itself reduced.

A reducing agent is a substance that reduces another substance and is itself oxidised.

Both agents are reactants, on the left of the arrow. The oxidising agent gains electrons and its oxidation number falls. The reducing agent loses electrons and its oxidation number rises.

Examiner insight

“‘A reducing agent is oxidised’ is the best way to express one version of the correct answer.” “Some candidates used ambiguous phrases such as: ‘a substance that helps reduce another substance’ or ‘ … and oxidises itself’ (instead of ‘… is itself oxidised’).”

Worked example 11 mark

The equation represents the reaction between sulfur dioxide and acidified potassium dichromate(VI).

3SO₂ + Cr₂O₇²⁻ + 2H⁺ → 3SO₄²⁻ + 2Cr³⁺ + H₂O

What is oxidised and what is the oxidising agent in this reaction?

A Cr₂O₇²⁻ and H⁺ · B Cr₂O₇²⁻ and Cr₂O₇²⁻ · C SO₂ and Cr₂O₇²⁻ · D SO₂ and H⁺

  1. 1

    Sulfur: +4 in SO₂ to +6 in SO₄²⁻. The oxidation number rises, so SO₂ is oxidised.

  2. 2

    Chromium: +6 in Cr₂O₇²⁻ to +3 in Cr³⁺. The oxidation number falls, so Cr₂O₇²⁻ is reduced and is the oxidising agent.

  3. 3

    C: SO₂ and Cr₂O₇²⁻. [1]

Recap

  • A Roman numeral in a name gives the oxidation number of that element: copper(II) oxide, copper is +2.
  • Oxidation is gain of oxygen. Reduction is loss of oxygen.
  • Redox is simultaneous oxidation and reduction. Name the reactants on the left of the arrow.
  • Extended: oxidation is loss of electrons. Reduction is gain of electrons.
  • Extended: oxidation is an increase in oxidation number. Reduction is a decrease.
  • Extended: uncombined element is 0. Monatomic ion equals its charge. Sum in a compound is 0. Sum in an ion equals its charge. Write the sign.
  • Extended: acidified potassium manganate(VII) goes purple to colourless. Potassium iodide goes colourless to brown as iodine forms.
  • Extended: a reducing agent is itself oxidised. An oxidising agent is itself reduced.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. State what is formed in a chemical change that is not formed in a physical change.1 mark
  2. Wax melts when it is warmed. State whether this is a physical change or a chemical change. Give a reason.2 marks
  3. A student stirs sugar into water until it disappears. State whether this is a physical change or a chemical change. Give a reason.2 marks
  4. A student says: “A change of state always means a chemical change.” Explain why she is wrong.2 marks
  5. A student adds a solid to a liquid. Give two observations that would suggest a chemical reaction is taking place.2 marks
  6. Classify each change as physical or chemical. | Change | Physical or chemical? | | --- | --- | | Adding sodium to water | | | A solid subliming | | | Dissolving sodium chloride in water | | | Producing water from aqueous sodium chloride | |4 marks
  7. A sealed flask and its contents have a mass of 125.0 g. The contents react and a gas forms. The flask stays sealed. State the mass of the flask and its contents after the reaction.1 mark
  8. Water is heated steadily and boils at 100 °C. A thermometer stays in the water. State the reading while the water is boiling and name the type of change.2 marks
  9. A student stirs 5.0 g of sodium chloride into 50.0 g of water until no solid can be seen. She then heats the solution until all the water has boiled away. A white solid is left in the dish. (a) State the mass of the solution before it is heated. (b) State the mass of the white solid left in the dish. (c) Another student says both steps are chemical changes, because a solid disappeared and then a solid appeared. Explain why he is wrong.4 marks
  10. An open flask of water stands on a balance. The reading is 152.40 g. A student adds a tablet of mass 2.50 g. The tablet disappears and bubbles of gas form. When the bubbling stops, the reading is 154.35 g. (a) Another student says the tablet has only dissolved, so the change is physical. Give the observation that shows a chemical change took place, and state what it shows. (b) Calculate the mass of gas that left the flask. (c) The student says that this mass of matter was destroyed. Explain why she is wrong.5 marks
  11. 1 g of manganese(IV) oxide speeds up the decomposition of aqueous hydrogen peroxide. After the reaction it is filtered off, dried and weighed. State the mass found and explain.2 marks
  12. Equal masses of zinc powder and zinc pieces react with excess dilute sulfuric acid. State which reacts faster and explain why.2 marks
  13. 1.0 g of magnesium powder reacts with 25 cm³ of dilute hydrochloric acid. The acid is in excess. The experiment is repeated with 50 cm³ of the same acid. State the effect on the initial rate and give a reason.2 marks
  14. 40 cm³ of dilute acid and 30 cm³ of water are mixed and a solid is added. 50 cm³ of gas is collected in 125 s. Calculate the mean rate of reaction. Give the units.3 marks
  15. Calcium carbonate reacts with dilute hydrochloric acid in an open flask on a balance. The mass of the flask and contents is recorded every 20 s. (a) State why the mass decreases. (b) State the other piece of apparatus that is essential to find the rate.2 marks
  16. A student measures the volume of hydrogen gas made when magnesium reacts with excess dilute hydrochloric acid. The graph shows the results. (a) Use the graph to find the volume of gas made in 20 s. (b) Deduce the time taken for the reaction to finish.2 marks
  17. Explain, in terms of particles and activation energy, why a reaction is faster at 40 °C than at 20 °C.3 marks
  18. Zinc carbonate is added to dilute nitric acid in a flask closed with a stopper. A delivery tube leads to an inverted measuring cylinder of water. The student times the collection of 50 cm³ of gas. (a) Some gas escapes while the stopper is replaced. State the effect on the time taken. (b) Describe a change to the apparatus so that no gas escapes. (c) The cylinder also collects the air from the flask. State the effect on accuracy and explain.5 marks
  19. Magnesium carbonate powder reacts with dilute nitric acid. MgCO₃(s) + 2HNO₃(aq) → Mg(NO₃)₂(aq) + CO₂(g) + H₂O(l) A student adds 2.0 g of the powder, an excess, to four mixtures of the acid and water. All four are at the same temperature. The student times the collection of 40 cm³ of carbon dioxide. The results are in the table. (a) Calculate the mean rate of reaction in experiment C. Give the units. (b) Calculate how many times greater the mean rate is in experiment A than in experiment D. (c) Explain, using collision theory, why the rate is lower in experiment D than in experiment A. (d) Another student says the acid particles in experiment A have more energy than those in experiment D. State whether this is correct. Give a reason. | experiment | volume of acid / cm³ | volume of water / cm³ | time to collect 40 cm³ of gas / s | | --- | --- | --- | --- | | A | 50 | 0 | 16 | | B | 40 | 10 | 20 | | C | 25 | 25 | 32 | | D | 10 | 40 | 80 |7 marks
  20. Aqueous hydrogen peroxide decomposes. 2H₂O₂(aq) → 2H₂O(l) + O₂(g) A student compares two solid catalysts. In each experiment, 0.5 g of the powdered oxide is added to 30 cm³ of the same aqueous hydrogen peroxide, at the same temperature. The flask has a loose cotton wool plug and stands on a balance. The student records the total loss in mass. The results are in the table. (a) State which oxide is the better catalyst. Use the results to justify your answer. (b) Explain why the loss in mass with manganese(IV) oxide stays at 0.48 g after 60 s. (c) Predict the final loss in mass with copper(II) oxide. Explain your answer. (d) Explain, using collision theory, why a catalyst increases the rate of this reaction. (e) Another student replaces the cotton wool plug with a tightly fitting stopper. Explain why the balance reading then does not change. | time / s | loss in mass with manganese(IV) oxide / g | loss in mass with copper(II) oxide / g | | --- | --- | --- | | 0 | 0.00 | 0.00 | | 20 | 0.30 | 0.10 | | 40 | 0.42 | 0.19 | | 60 | 0.48 | 0.27 | | 80 | 0.48 | 0.34 | | 100 | 0.48 | 0.40 |7 marks
  21. Write the symbol used in an equation to show that a reaction is reversible.1 mark
  22. State the colour of anhydrous copper(II) sulfate and the colour it turns when water is added.2 marks
  23. Describe a test for water using cobalt(II) chloride.3 marks
  24. Blue crystals of hydrated copper(II) sulfate are heated strongly in a test-tube. (a) State two observations. (b) State how the white solid can be turned blue again.3 marks
  25. Water is added to anhydrous copper(II) sulfate to form CuSO₄.5H₂O. Write the balanced equation for this reversible reaction. Include state symbols.2 marks
  26. A student writes about equilibrium in a closed system: ‘The forward and reverse reactions are equal. The reaction has stopped. The concentrations are equal.’ Correct the three mistakes.3 marks
  27. A student writes notes on the Haber process: ‘Hydrogen comes from air. The pressure is 200 000 kPa. The catalyst is iron. The temperature is 450 °C.’ (a) Write the symbol equation for the reaction. (b) Correct the two mistakes in the notes.4 marks
  28. H₂(g) + I₂(g) ⇌ 2HI(g) The forward reaction is exothermic. State the effect on the equilibrium concentration of HI(g) of: (a) increasing the pressure, (b) increasing the temperature, (c) adding a catalyst. Use the words increases, decreases or no change.3 marks
  29. X(g) + 3Y(g) ⇌ 2Z(g) X and Y react in a closed system. The table gives the percentage of Z at equilibrium. (a) Is the forward reaction exothermic or endothermic? Use the table to explain. (b) Use the equation to explain why more Z forms at 30 000 kPa than at 10 000 kPa. (c) A catalyst is added at 500 °C and 10 000 kPa. State and explain the percentage of Z. (d) Some Z is removed. State which way the position of equilibrium moves. | temperature / °C | pressure / kPa | percentage of Z at equilibrium | | --- | --- | --- | | 300 | 10 000 | 52 | | 300 | 30 000 | 71 | | 500 | 10 000 | 11 | | 500 | 30 000 | 26 |7 marks
  30. Sulfur dioxide is converted to sulfur trioxide in the Contact process. The forward reaction is exothermic. A manager suggests raising the temperature from 450 °C to 700 °C and the pressure from 200 kPa to 20 000 kPa. (a) Write the symbol equation for the reaction. State one source of the sulfur dioxide. (b) State the effect of 700 °C on the rate and on the yield. Explain the effect on the yield. (c) Use the equation to explain the effect of 20 000 kPa on the yield. Give one reason why 200 kPa is still used.9 marks
  31. State the oxidation number of iron in iron(III) chloride.1 mark
  32. Sulfur forms sulfur(IV) oxide and sulfur(VI) oxide. State what the Roman numerals tell you.2 marks
  33. 2Cu + O₂ → 2CuO. State what happens to the copper in terms of oxygen.1 mark
  34. Copper(II) oxide is heated in a stream of hydrogen. CuO + H₂ → Cu + H₂O Name the substance that is reduced and explain your answer.2 marks
  35. Zinc oxide is heated with carbon. ZnO + C → Zn + CO Explain, using the equation, why this is a redox reaction.3 marks
  36. Hot magnesium burns in carbon dioxide. 2Mg + CO₂ → 2MgO + C Name the substance oxidised and the substance reduced.2 marks
  37. Chlorine reacts with aqueous potassium bromide. Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂ Describe, in terms of electrons, what happens to each reactant. State which reactant is oxidised.3 marks
  38. Acidified aqueous potassium manganate(VII) is added to aqueous iron(II) sulfate. MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺ (a) State the colour change of the potassium manganate(VII). (b) Calculate the oxidation number of manganese in MnO₄⁻. Oxygen is –2. (c) Identify the reducing agent. Explain your answer in terms of electrons.5 marks
  39. Vanadium(V) oxide, V₂O₅, takes part in two reactions. 1 V₂O₅ + 2NaOH → 2NaVO₃ + H₂O 2 V₂O₅ + SO₂ → 2VO₂ + SO₃ In these compounds oxygen is –2, sodium is +1 and hydrogen is +1. (a) Calculate the oxidation number of vanadium in NaVO₃ and in VO₂. (b) Use oxidation numbers to decide which reaction is redox. Explain for both reactions.5 marks
  40. A solution of iodate ions, IO₃⁻, and dilute acid is added to aqueous potassium iodide. IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O (a) State the colour change seen. (b) Calculate the oxidation number of iodine in IO₃⁻. Oxygen is –2. (c) Use oxidation numbers to identify the oxidising agent and the reducing agent.6 marks

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