Topic 4 Chemical changes
The 'Chemical changes' topic covers the reactivity of metals, how they are extracted from their ores, and their reactions with acids to form salts. It also explores the process of electrolysis, which uses electricity to break down ionic compounds.
When a metal reacts with oxygen, it forms a metal oxide. This chemical reaction is a type of oxidation, as the metal gains oxygen. A common example is the rusting of iron, where iron reacts with oxygen in the presence of water to form iron oxide.
A general word equation for this process is:
Metal + Oxygen ⟶ Metal Oxide
For instance, when magnesium burns in the air, it reacts vigorously with oxygen to produce a white powder, magnesium oxide. The balanced symbol equation for this reaction is:
2Mg(s) + O₂(g) ⟶ 2MgO(s)
The opposite of oxidation is reduction, which is the loss of oxygen. To extract a pure metal from its oxide, the metal oxide must be reduced. This involves removing the oxygen from it.
The ease with which a metal oxide can be reduced is linked to the metal’s position in the reactivity series. Metals that are high up in the series, such as potassium or sodium, are very reactive and form very stable oxides. A large amount of energy is needed to break these strong bonds and reduce the oxides.
Metals that are less reactive, like copper or zinc, form less stable oxides which are much easier to reduce to extract the metal.
Metals that are less reactive than carbon can be extracted from their oxides by heating them with carbon
The general equation is:
Metal Oxide + Carbon ⟶ Metal + Carbon Dioxide
For example, copper can be extracted from copper(II) oxide by heating it with carbon:
2CuO(s) + C(s) ⟶ 2Cu(s) + CO₂(g)
Exam tip – For questions on this topic, remember to define oxidation as the gain of oxygen and reduction as the loss of oxygen
. When asked if a metal can be extracted using carbon, always check the metal’s position in the reactivity series relative to carbon. If the metal is below carbon, it can be extracted this way!
The reactivity series is a list that ranks metals in order of their reactivity.
The series you need to know is:
Potassium > Sodium > Lithium > Calcium > Magnesium > (Carbon) > Zinc > Iron > (Hydrogen) > Copper
Carbon and hydrogen are included as non-metal reference points. Carbon’s position helps predict how a metal can be extracted from its ore, while hydrogen’s position indicates whether a metal will react with dilute acid
A metal’s position in the series predicts how it will react with water and acids.
Metals at the top, like potassium and sodium, react vigorously with cold water, producing a metal hydroxide and hydrogen gas. Metal + Water ⟶ Metal Hydroxide + Hydrogen
Metals lower down, such as magnesium, react very slowly with cold water but will react with steam to form a metal oxide and hydrogen.
Metals above hydrogen in the series will react with dilute acids (like hydrochloric or sulfuric acid) to produce a salt and hydrogen gas. The higher the metal is in the series, the faster the reaction and the more vigorous the fizzing will be. Metal + Acid ⟶ Salt + Hydrogen
A displacement reaction occurs when a more reactive metal displaces (pushes out) a less reactive metal from a solution of its compound. For example, if you place a piece of zinc (more reactive) into a solution of blue copper(II) sulfate, the zinc will displace the copper. You would observe the blue colour of the solution fading as the zinc forms zinc sulfate, and a layer of solid brown copper would form on the surface of the zinc.
The balanced symbol equation for this reaction is: Zn(s) + CuSO₄(aq) ⟶ ZnSO₄(aq) + Cu(s)
Exam tip – In displacement reaction questions, a more reactive metal will always displace a less reactive one from its salt solution. If you are asked to predict the outcome of a reaction, simply check the positions of the two metals in the reactivity series. The metal higher up will end up in the compound, and the one lower down will become the solid element.
How a metal is extracted depends on its reactivity. Unreactive metals, such as gold, are so stable that they do not easily react with other elements to form compounds. As a result, they can be found in the Earth’s crust as the pure metal itself, in their native state.
Most metals, however, are more reactive and exist as compounds within ores. To obtain the pure metal, these compounds must be broken down through chemical reactions. This process often involves reduction, which is the removal of oxygen.
For metals that are less reactive than carbon, a common extraction method is heating the metal oxide with carbon. In this reaction, carbon acts as a reducing agent, pulling the oxygen away from the metal to form carbon dioxide. This leaves the pure metal behind.
A classic example is the extraction of copper from copper(II) oxide:
2CuO(s) + C(s) ⟶ 2Cu(s) + CO₂(g)
Here, the copper(II) oxide is reduced to copper, and the carbon is oxidized to carbon dioxide.
Exam tip – A key takeaway is the definition of reduction in this context: the loss of oxygen. When you’re given a reaction, identify which substance loses oxygen—that’s the one being reduced. Also, remember that only metals below carbon in the reactivity series can be extracted this way.
While oxidation and reduction can be described in terms of oxygen gain or loss, at a Higher Tier level, you need to understand them in terms of electron transfer. This provides a more detailed picture of what happens in reactions like displacement.
The definitions are simple:
Oxidation Is Loss of electrons.
Reduction Is Gain of electrons.
A helpful way to remember this is the acronym OILRIG.
In a redox reaction, one substance must be oxidized (lose electrons) and another must be reduced (gain electrons).
The substance that gets oxidized is the reducing agent (it donates electrons).
The substance that gets reduced is the oxidizing agent (it accepts electrons).
We can split a redox reaction into two half-equations to clearly show what happens to each substance.
Let’s look at the displacement reaction between magnesium and copper(II) sulfate solution. The overall ionic equation is: Mg(s) + Cu²⁺(aq) ⟶ Mg²⁺(aq) + Cu(s)
We can break this down:
Magnesium’s Half-Equation (Oxidation): The magnesium atom loses two electrons to become a magnesium ion (Mg²⁺). Since it loses electrons, it is oxidized. Mg ⟶ Mg²⁺ + 2e⁻
Copper’s Half-Equation (Reduction): The copper(II) ion (Cu²⁺) gains the two electrons lost by magnesium to become a neutral copper atom. Since it gains electrons, it is reduced. Cu²⁺ + 2e⁻ ⟶ Cu
Notice that the number of electrons lost in oxidation must equal the number of electrons gained in reduction. The electrons are transferred from the magnesium to the copper ion.
Exam tip – Always remember OILRIG (Oxidation Is Loss, Reduction Is Gain) when dealing with electron transfer. In an exam, if you are asked for half-equations, make sure the charges and the number of electrons (e⁻) are balanced on both sides of the arrow.
When a metal reacts with an acid, a chemical reaction takes place that produces a salt and hydrogen gas. However, this reaction only happens with metals that are more reactive than hydrogen. You can check a metal’s position in the reactivity series to predict if it will react. Metals like copper, which are below hydrogen, will not react with dilute acids.
The general word equation for this reaction is:
Metal + Acid ⟶ Salt + Hydrogen
You can observe this reaction by looking for several key signs:
Fizzing or bubbling, which is the production of hydrogen gas.
The piece of metal getting smaller as it reacts and dissolves.
An increase in temperature, as the reaction is exothermic.
The hydrogen gas produced can be confirmed using the ‘squeaky pop’ test.
The rate of the reaction is a good indicator of the metal’s reactivity. A more reactive metal, like magnesium, will react much more vigorously and quickly with an acid compared to a less reactive metal like zinc.
The name of the salt produced depends on both the metal and the acid used.
Hydrochloric acid produces chloride salts.
Sulfuric acid produces sulfate salts.
Nitric acid produces nitrate salts.
For example, if magnesium reacts with hydrochloric acid, the salt formed is magnesium chloride. The balanced symbol equation for this is:
Mg(s) + 2HCl(aq) ⟶ MgCl₂(aq) + H₂(g)
Exam tip – When describing the reaction between a metal and an acid, always state the observation clearly (e.g., “fizzing was observed”) and then give the scientific reason for it (“which shows a gas was produced”). Also, be careful to use the correct ending for the salt name (-chloride, -sulfate, or -nitrate) based on the acid used in the reaction.
Neutralisation is a chemical reaction between an acid and a base (or an alkali). In these reactions, the acid’s properties are ‘cancelled out’ by the base. Neutralisation always produces a salt and usually water.
There are three main types of neutralisation reactions you need to know:
Acid + Base (like a metal oxide or insoluble hydroxide) ⟶ Salt + Water For example, reacting sulfuric acid with copper(II) oxide produces copper(II) sulfate and water. H₂SO₄(aq) + CuO(s) ⟶ CuSO₄(aq) + H₂O(l)
Acid + Alkali (a soluble base) ⟶ Salt + Water The ionic equation for any reaction between an acid and an alkali is always the same: it shows a hydrogen ion reacting with a hydroxide ion to form water. H⁺(aq) + OH⁻(aq) ⟶ H₂O(l)
Acid + Metal Carbonate ⟶ Salt + Water + Carbon Dioxide This reaction is easy to spot because it’s the only one that produces three products, including carbon dioxide gas, which causes fizzing.
The name of the salt produced depends on the acid and the metal-containing base used:
Hydrochloric acid (HCl) always produces chloride salts.
Sulfuric acid (H₂SO₄) always produces sulfate salts.
Nitric acid (HNO₃) always produces nitrate salts.
For instance, if calcium carbonate reacts with nitric acid, the salt formed is calcium nitrate.
You need to know how to prepare a pure, dry sample of a soluble salt, like copper(II) sulfate, from an acid and an insoluble base, like copper(II) oxide.
The key steps are:
Gently warm the dilute acid (e.g., sulfuric acid).
Add the insoluble base (e.g., copper(II) oxide) to the acid a little at a time, stirring until no more dissolves. This means the base is in excess and all the acid has been neutralised.
Filter the mixture to remove the unreacted excess base. The filtrate is the pure salt solution (e.g., copper(II) sulfate solution).
Gently heat the solution to evaporate some of the water until crystals start to form (the crystallisation point).
Leave the solution to cool slowly, allowing larger crystals to form.
Filter the crystals from the solution and leave them in a warm, dry place.
Exam tip – A common exam question asks why the insoluble base is added in excess. The reason is to make sure that all of the acid has reacted. This is crucial so that the final salt solution is not contaminated with any leftover acid.
A common method for making a soluble salt is by reacting a suitable acid with an insoluble substance, such as a metal, a metal oxide, a metal hydroxide, or a metal carbonate. For example, to make copper(II) sulfate, you would react sulfuric acid with an insoluble copper compound like copper(II) oxide.
To prepare a pure, dry sample of a soluble salt, you need to follow a precise method to ensure the final product is not contaminated.
The key steps are:
Gently warm the dilute acid. This speeds up the reaction.
Add the insoluble base to the acid in small amounts, stirring continuously, until no more of the base reacts or dissolves. At this point, the base is in excess.
Use filtration to separate the unreacted excess base from the salt solution. The solution that passes through the filter paper is called the filtrate, and it contains only the dissolved salt and water.
Gently heat the filtrate in an evaporating basin to boil off some of the water. Stop heating when small crystals start to form around the edge. This is the point of crystallisation.
Leave the concentrated solution to cool down slowly. As it cools, the salt becomes less soluble, and pure crystals will form.
Once crystallisation is complete, filter the crystals from the remaining solution, wash them with a small amount of cold, distilled water, and then leave them to dry.
Exam tip – You are often asked why the insoluble base is added in excess. This is a crucial step to ensure that all of the acid is neutralised. If the acid were in excess, the final salt crystals would be contaminated with unreacted acid. Filtering removes the excess solid base, leaving a pure salt solution.
The pH scale measures how acidic or alkaline a solution is, running from 0 to 14.
Solutions with a pH less than 7 are acidic.
A solution with a pH of 7 is neutral.
Solutions with a pH greater than 7 are alkaline.
Universal indicator is used to estimate pH. It turns red for strong acids (pH 0-2), yellow for weak acids (pH 3-6), green for neutral solutions (pH 7), and blue/purple for alkalis (pH 8-14).
Acids produce hydrogen ions (H⁺) in a solution. The higher the concentration of H⁺ ions, the lower the pH. Alkalis produce hydroxide ions (OH⁻) in a solution.
During neutralisation, the H⁺ ions from an acid react with the OH⁻ ions from an alkali to form neutral water (H₂O). This removes the acidic and alkaline properties from the solution. The ionic equation for this reaction is always:
H⁺(aq) + OH⁻(aq) ⟶ H₂O(l)
Exam tip – Do not confuse the terms strong and concentrated. A strong acid fully releases its hydrogen ions in water. A concentrated acid has a large amount of acid particles in a given volume. You can have a dilute strong acid or a concentrated weak acid.
A titration is a precise experimental technique used to find out the exact volume of an acid needed to neutralise a measured volume of an alkali. This point of neutralisation is called the end point. By knowing the volumes that react and the concentration of one of the solutions, you can calculate the unknown concentration of the other.
Performing a titration requires careful use of specific apparatus to ensure the results are accurate and reliable.
Use a pipette to accurately measure a set volume of the alkali (e.g., 25.0 cm³) and transfer it to a conical flask.
Add a few drops of a suitable indicator, such as phenolphthalein (pink in alkali, colourless in acid) or methyl orange (yellow in alkali, red in acid).
Fill a burette with the acid and record the initial reading.
Slowly add the acid from the burette to the flask, swirling constantly, until the indicator shows a permanent colour change.
Record the final burette reading and calculate the volume of acid added (the titre).
Repeat the titration until you get at least two concordant results (titres that are within 0.10 cm³ of each other).
To find the concentration of an unknown solution:
Balanced Equation: Write the balanced symbol equation for the reaction (e.g., HCl + NaOH ⟶ NaCl + H₂O).
Calculate Moles: Use the known solution’s concentration and volume to find the moles: moles = concentration () × volume (). (Remember to convert cm³ to dm³ by dividing by 1000).
Use the Molar Ratio: Use the ratio from the balanced equation to find the moles of the other substance.
Calculate Concentration: Rearrange the formula to find the unknown concentration: concentration () = moles / volume ().
Exam tip – To get full marks for describing a titration, you must name the key equipment: pipette for measuring the alkali and burette for adding the acid. A common mistake is to average all your results, including the rough one. You should only calculate the mean from your concordant titres.
The difference between a strong acid and a weak acid is all about how they behave in water. It is determined by the degree of ionisation – how many of the acid molecules split up (ionise) to release hydrogen ions (H⁺).
A strong acid is one that completely ionises when dissolved in water. This means that every single acid molecule releases its H⁺ ion. Examples of strong acids include hydrochloric acid (HCl), sulfuric acid (H₂SO₄), and nitric acid (HNO₃).
A weak acid is one that only partially ionises in water. This means only a small fraction of the acid molecules release their H⁺ ions. This is a reversible reaction, and an equilibrium is established where most of the molecules remain un-ionised. Examples of weak acids include ethanoic acid, citric acid, and carbonic acid.
For two acid solutions of the same concentration:
pH: A strong acid will have a much lower pH than a weak acid. This is because the complete ionisation of the strong acid results in a much higher concentration of H⁺ ions. Remember, a change of one unit on the pH scale represents a tenfold change in H⁺ ion concentration.
Reactivity: A strong acid will react more vigorously than a weak acid. For instance, when reacting with magnesium ribbon, a strong acid will produce hydrogen gas much faster because the higher concentration of H⁺ ions means there are more frequent collisions with the metal.
Conductivity: A strong acid solution will conduct electricity better than a weak acid solution. Electrical conductivity in solutions depends on the concentration of mobile ions. Since strong acids produce a higher concentration of ions (H⁺ and the negative ion), they are better conductors.
It is crucial not to confuse acid strength with concentration.
Strength describes the degree of ionisation (what proportion of acid molecules release H⁺ ions).
Concentration describes the amount of acid substance dissolved in a given volume of water (e.g., in mol/dm³).
You can have a dilute solution of a strong acid or a concentrated solution of a weak acid.
Exam tip – When comparing strong and weak acids in an exam, always link your explanation to the degree of ionisation. State that strong acids fully ionise, leading to a higher concentration of H⁺ ions at the same concentration. This higher H⁺ concentration is the reason for the lower pH, faster reaction rate, and better electrical conductivity.
Electrolysis is a process that uses a direct electric current to break down an ionic compound. For electrolysis to happen, the ionic substance must be in a state where its ions are free to move. This means it must either be molten (melted) or dissolved in water (an aqueous solution). Solid ionic compounds cannot be electrolysed because their ions are held in a fixed crystal lattice and cannot move to carry an electrical charge. The substance that is broken down and conducts electricity is called the electrolyte.
During electrolysis, two conducting rods called electrodes are placed into the electrolyte. The positive electrode is the anode, and the negative electrode is the cathode.
Opposite charges attract, so the positively charged ions (cations) move to the negative cathode. At the cathode, they gain electrons. This process is called reduction.
The negatively charged ions (anions) move to the positive anode. At the anode, they lose electrons. This process is called oxidation. A good way to remember this is OILRIG: Oxidation Is Loss, Reduction Is Gain of electrons.
We can show the reactions happening at each electrode using half-equations. For example, in the electrolysis of molten lead(II) bromide (PbBr₂), the electrolyte contains lead ions (Pb²⁺) and bromide ions (Br⁻).
At the cathode (negative), lead ions gain electrons to form lead metal: Pb²⁺ + 2e⁻ ⟶ Pb (Reduction)
At the anode (positive), bromide ions lose electrons to form bromine gas: 2Br⁻ ⟶ Br₂ + 2e⁻ (Oxidation)
The overall result is that lead(II) bromide is broken down into its elements, lead and bromine.
Exam tip – When answering questions about electrolysis, always identify the ions present in the electrolyte first. Then, state which ion moves to which electrode (positive ions to the negative cathode, negative ions to the positive anode). Finally, describe what happens at each electrode in terms of gaining or losing electrons (reduction or oxidation).
When a simple ionic compound is heated until it melts, its ions break free from the fixed lattice and are able to move. This allows the molten compound to conduct electricity and be broken down by electrolysis. As the substance is purely molten, the only ions present are those from the compound itself, making the products easy to predict.
The positive metal ions (cations) are attracted to the negative electrode (cathode), where they gain electrons (are reduced) to form the pure metal. The negative non-metal ions (anions) are attracted to the positive electrode (anode), where they lose electrons (are oxidised) to form the non-metal element.
In molten sodium chloride, the electrolyte contains sodium ions (Na⁺) and chloride ions (Cl⁻). At the cathode (negative), positive sodium ions gain electrons to form molten sodium metal (Na⁺ + e⁻ ⟶ Na). At the anode (positive), negative chloride ions lose electrons to form pale green chlorine gas (2Cl⁻ ⟶ Cl₂ + 2e⁻).
Exam tip – For the electrolysis of any simple molten ionic compound, the products are always the two elements that make it up. The metal is always produced at the negative cathode, and the non-metal is always produced at the positive anode. Remember that you must not include H⁺ or OH⁻ ions in your answers, as no water is present.
Some metals are too reactive to be extracted from their ores using carbon. If a metal is above carbon in the reactivity series (like aluminium), it means its oxide is too stable for carbon to reduce it. For these reactive metals, we must use a more powerful method: electrolysis. This process uses a lot of energy to melt the compounds and provide the electricity, which makes it very expensive.
Aluminium is a very useful metal, but it’s found in an ore called bauxite, which is purified into aluminium oxide (Al₂O₃).
The Problem: Aluminium oxide has an extremely high melting point (over 2000°C). Melting it would require a huge amount of energy, making the process unaffordable.
The Solution: To solve this, the aluminium oxide is dissolved in molten cryolite. This lowers the operating temperature to around 900°C, which significantly reduces energy costs.
The Electrolysis Process:
The electrolysis cell has carbon electrodes. The lining of the cell acts as the negative cathode, and large blocks of graphite are the positive anodes.
At the cathode (negative), positive aluminium ions (Al³⁺) are attracted. They gain three electrons each and are reduced to form molten aluminium, which sinks to the bottom of the cell. Al³⁺ + 3e⁻ ⟶ Al (l)
At the anode (positive), negative oxide ions (O²⁻) are attracted. They lose two electrons each and are oxidised to form oxygen gas. 2O²⁻ ⟶ O₂ (g) + 4e⁻
A problem arises because the oxygen produced is very hot and reacts with the carbon anodes, burning them away to form carbon dioxide (C + O₂ ⟶ CO₂). Because of this, the anodes have to be replaced regularly, which adds to the overall cost of the process.
Exam tip – Two key facts about aluminium extraction frequently appear in exams:
Why is cryolite used? To lower the melting point of the aluminium oxide, saving energy and money.
Why do the anodes need replacing? Because the oxygen produced reacts with the hot carbon anodes, burning them away.
When an ionic compound is dissolved in water, the solution contains ions from the compound and ions from the water. Water molecules slightly break down to form hydrogen ions (H⁺) and hydroxide ions (OH⁻). This means there are now four different ions in the electrolyte, leading to a competition at each electrode to see which ion is discharged.
To work out what will be produced, you just need to follow two simple rules.
At the Cathode (Negative Electrode): The positive ions (the metal ion and H⁺) are attracted here.
Rule: The least reactive positive ion will be discharged. If the metal is more reactive than hydrogen, then hydrogen gas will be produced. If the metal is less reactive than hydrogen (like copper), then the metal will be deposited.
At the Anode (Positive Electrode): The negative ions (the non-metal ion and OH⁻) are attracted here.
Rule: If the solution contains halide ions (chloride, bromide, or iodide), then the halogen (chlorine, bromine, or iodine) will be produced. If there are no halide ions, then oxygen gas will be produced from the hydroxide ions.
Ions present: Na⁺, Cl⁻, H⁺, and OH⁻.
At the cathode (-): Sodium is more reactive than hydrogen, so hydrogen gas is produced. (2H⁺ + 2e⁻ ⟶ H₂)
At the anode (+): Chloride is a halide ion, so chlorine gas is produced. (2Cl⁻ ⟶ Cl₂ + 2e⁻)
The solution left behind contains Na⁺ and OH⁻ ions, which is sodium hydroxide.
Ions present: Cu²⁺, SO₄²⁻, H⁺, and OH⁻.
At the cathode (-): Copper is less reactive than hydrogen, so copper metal is deposited. (Cu²⁺ + 2e⁻ ⟶ Cu)
At the anode (+): Sulfate is not a halide, so oxygen gas is produced. (4OH⁻ ⟶ O₂ + 2H₂O + 4e⁻)
Exam tip – The most common mistake is forgetting that water provides H⁺ and OH⁻ ions. Always start by listing all four ions present in the solution before applying the rules. Remember the simple rules: at the cathode, the least reactive metal (or hydrogen) is produced. At the anode, if you see a halide, you get a halogen; if not, you get oxygen.
Half-equations are a great way to show exactly what’s happening at each electrode during electrolysis. They focus on the transfer of electrons, which is the key to redox reactions.
Remember that in electrolysis:
At the cathode (negative), positive ions gain electrons. This is Reduction.
At the anode (positive), negative ions lose electrons. This is Oxidation.
A half-equation shows one of these processes, including the electrons (e⁻) that are either gained or lost.
Let’s break down how to write them for a simple molten compound like lead(II) bromide (PbBr₂).
1. At the Cathode (Reduction):
Identify the ion: The positive lead ion (Pb²⁺) moves to the cathode.
What happens? It gains electrons to become a neutral lead atom (Pb).
Write the equation: Start with the ion and the product: Pb²⁺ ⟶ Pb.
Balance the charge: To balance the 2+ charge on the left, add two negative electrons to the left side. Pb²⁺ + 2e⁻ ⟶ Pb
2. At the Anode (Oxidation):
Identify the ion: The negative bromide ion (Br⁻) moves to the anode.
What happens? It loses electrons to become a neutral bromine molecule (Br₂).
Balance the atoms first: We need two bromide ions to make one bromine molecule: 2Br⁻ ⟶ Br₂.
Balance the charge: The left side has a total 2- charge. To balance this, add two negative electrons to the right side.2Br⁻ ⟶ Br₂ + 2e⁻
To get the overall ionic equation, you combine the two half-equations. The number of electrons lost and gained must be the same so they can be cancelled out. In the PbBr₂ example, two electrons were lost and two were gained, so we can simply add them together.
Overall equation: Pb²⁺ + 2Br⁻ ⟶ Pb + Br₂
Exam tip – When writing half-equations for exam questions, always clearly state which reaction is happening at which electrode. Forgetting to balance the atoms before balancing the charge is a common slip-up, especially for halogens that form diatomic molecules (like Cl₂ or Br₂). Remember, if the electrons don’t cancel out when you combine the half-equations, you’ll need to multiply one or both equations to make the electron numbers match.