This mini-lesson walks you through the whole of Edexcel Topic 5 — Separate chemistry 1: transition metals, corrosion & alloys, titrations & concentration, the molar gas volume, percentage yield and atom economy, empirical formulae, dynamic equilibrium (the Haber process) and chemical & fuel cells.
Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. Screens marked HT are Higher-tier only. Press Start when you're ready.
5.1 — Transition metals
The block in the middle
Most metals are transition metals — they sit in the central block of the periodic table, between Group 2 and Group 3. Their typical properties are:
High melting points and high densities (iron melts at ~1538 °C).
They form coloured compounds (e.g. blue Cu²⁺, green Fe²⁺, orange Fe³⁺).
They (and their compounds) are useful as catalysts (e.g. iron in the Haber process).
They have variable oxidation states, forming more than one ion (e.g. Fe²⁺ and Fe³⁺).
Transition metals fill the central block. Iron is the spec's named example of a catalyst.5.1 — vs Group 1
Compared with Group 1
A favourite exam comparison: transition metals against the Group 1 alkali metals.
Transition metals are harder, denser, higher-melting and far less reactive than Group 1.
Watch out: a common error is to call transition metals "more reactive" because they're famous. They are actually less reactive than Group 1 — sodium and potassium fizz and burn on water; iron and copper do not.
Quick check
Spot the transition-metal property
?Which property is typical of a transition metal but not of a Group 1 metal?
5.2–5.3 — Corrosion
Rusting needs air AND water
Corrosion is the oxidation of a metal at its surface. For iron this is rusting, which forms hydrated iron(III) oxide. Rusting needs BOTH oxygen and water:
iron + oxygen + water → hydrated iron(III) oxidethe classic experiment: nail in water + air rusts; nails sealed from air OR from water stay shiny
Remove the air (oil layer on boiled water) OR the water (drying agent) and the iron will not rust.5.3 — Preventing rust
Barriers and sacrifice
Edexcel lists two ideas: keep oxygen/water out, or use a more reactive metal to corrode instead.
Barrier methods — paint, oil/grease, or plastic coating exclude oxygen and water.
Galvanising — coating iron with zinc. Zinc is a barrier and, because it is more reactive, it gives sacrificial protection.
Sacrificial protection — attach blocks of a more reactive metal (zinc or magnesium). It oxidises in preference to the iron, so the iron is protected even if scratched.
The sacrificial metal must be more reactive than iron — that's the whole point.
Watch out: the sacrificial metal is the more reactive one, not the less reactive one. A less reactive coating (like tin on a tin can) protects only as a barrier — scratch it and the iron rusts faster.
Quick check
Protecting a steel pipe
?A buried steel pipe is protected by bolting on blocks of metal that corrode instead of it. For this sacrificial protection to work, the blocks must be made of a metal that is…
5.5–5.7 — Alloys
Why alloys are harder
An alloy is a mixture of a metal with other elements. Steels are alloys of iron with carbon (and sometimes other metals). Alloys are usually harder than the pure metal.
Different-sized atoms disrupt the regular layers, so they can no longer slide over each other — the alloy is harder.
Iron is alloyed to make alloy steels (e.g. adding chromium and nickel makes stainless steel, which resists corrosion). Other metals are matched to uses by their properties: aluminium (low density), copper (conducts), gold (unreactive); alloys include brass (copper + zinc) and magnalium (aluminium + magnesium).
5.8 — Concentration · HT only
Concentration of a solution
Concentration says how much solute is dissolved per dm³ (1 dm³ = 1000 cm³ = 1 litre). You meet two units:
To convert g/dm³ → mol/dm³, divide by the molar mass (Mr in g/mol). To go back the other way, multiply.
Worked example
0.50 mol of NaOH is dissolved to make 0.25 dm³ of solution.
conc = 0.50 ÷ 0.25 = 2.0 mol/dm³
In g/dm³: 2.0 × 40 = 80 g/dm³ (Mr of NaOH = 23+16+1 = 40)
Calculate · HT only
Your turn — concentration
10.40 mol of sodium chloride is dissolved to make 2.0 dm³ of solution. Calculate the concentration in mol/dm³.
mol/dm³
Hint: conc = moles ÷ volume = 0.40 ÷ 2.0.
5.9 — Core practical
The titration
A titration finds an unknown concentration accurately, e.g. a strong acid against a strong alkali. You measure a fixed volume of one solution with a pipette into a conical flask, add an indicator, then run in the other from a burette until the colour just changes (the end-point).
Repeat until two or more concordant titres (within 0.10 cm³) are obtained, then average only those.
Watch out: the mean titre uses only the concordant results — discard the rough trial and any anomalies. Never just average every reading.
5.10 — Titration calc · HT only
Titration calculations
Use the balanced equation. For a strong acid–strong alkali like HCl + NaOH → NaCl + H₂O the ratio is 1 : 1.
moles = conc (mol/dm³) × volume (dm³)find moles of the known solution → use the ratio → divide by the other volume
Worked example
25.0 cm³ of 0.100 mol/dm³ NaOH is exactly neutralised by 20.0 cm³ of HCl. Find the concentration of the HCl.
moles NaOH = 0.100 × (25.0/1000) = 0.00250 mol
ratio 1:1, so moles HCl = 0.00250 mol
conc HCl = 0.00250 ÷ (20.0/1000) = 0.125 mol/dm³
Calculate · HT only
Your turn — titration
2In HCl + NaOH → NaCl + H₂O, 25.0 cm³ of 0.200 mol/dm³ NaOH is neutralised by 20.0 cm³ of HCl. Calculate the concentration of the HCl in mol/dm³.
% yield vs atom economy: they are different. Percentage yield is about how much you actually got in the lab; atom economy is about how much of the atoms in your equation are useful — even a 100%-yield reaction can have a poor atom economy.
Calculate
Your turn — atom economy
4Iron is produced in the blast furnace by the reaction Fe₂O₃ + 3CO → 2Fe + 3CO₂. The desired product is Fe. Total Mr of products: 2Fe = 112, 3CO₂ = 132. Calculate the atom economy for making iron.
At room temperature and pressure (rtp), one mole of any gas occupies the same volume — the molar volume:
1 mol of gas = 24 dm³ (24 000 cm³) at rtpvolume of gas (dm³) = moles × 24 · moles = volume ÷ 24
By Avogadro's law, equal volumes of gases (at the same temperature and pressure) contain equal numbers of molecules, so the mole ratio in the equation is also the volume ratio.
Worked example
What volume does 0.50 mol of carbon dioxide occupy at rtp?
volume = 0.50 × 24 = 12 dm³
Calculate · HT only
Your turn — gas volume
5Calculate the volume occupied by 0.25 mol of hydrogen gas at rtp. (Molar volume = 24 dm³/mol.)
dm³
Hint: volume = moles × 24 = 0.25 × 24.
Empirical & molecular formula
Empirical formula from masses
The empirical formula is the simplest whole-number ratio of atoms. From masses (or % composition):
Divide each element's mass ÷ its Ar to get moles.
Divide each by the smallest to get the simplest ratio.
The molecular formula is a whole-number multiple of the empirical formula (find it by comparing the empirical Mr with the real Mr).
Worked example
4.6 g sodium combines with 1.6 g oxygen. (Ar: Na = 23, O = 16)
Na: 4.6 ÷ 23 = 0.20 mol · O: 1.6 ÷ 16 = 0.10 mol
divide each by the smallest (0.10): Na = 2, O = 1 → ratio 2 : 1
empirical formula = Na₂O
Calculate
Your turn — empirical formula
?2.4 g of magnesium combines with 1.6 g of oxygen. Using Ar(Mg) = 24 and Ar(O) = 16, what is the empirical formula?
5.19 — Dynamic equilibrium
The Haber process
In a sealed container a reversible reaction reaches dynamic equilibrium: forward and backward reactions still occur, but at equal rates, so concentrations stay constant. The Haber process makes ammonia for fertilisers:
Typical industrial conditions: about 450 °C, 200 atmospheres, with an iron catalyst.
5.20–5.21 — Le Chatelier · HT only
Choosing the conditions
Changing the conditions shifts the equilibrium position (Le Chatelier's idea). For the Haber process the chosen conditions are a compromise between yield, rate and cost:
Pressure ~200 atm — high pressure favours the side with fewer gas molecules (the NH₃ side, 4 → 2), giving more ammonia; but very high pressure is expensive and unsafe.
Temperature ~450 °C — the forward reaction is exothermic, so a lower temperature gives a higher yield, but too low is far too slow. 450 °C is a compromise giving an acceptable yield at an acceptable rate.
Iron catalyst — speeds up equilibrium being reached; it does not change the position/yield.
Watch out: a catalyst speeds up how fast equilibrium is reached but does not change the equilibrium yield. The 450 °C is a compromise, not the temperature that gives the most ammonia.
Quick check
Reading the equilibrium
?In the Haber process, why is an iron catalyst used?
5.25–5.27 — Cells & fuel cells
Chemical cells and fuel cells
A simple chemical cell is two different metal electrodes in an electrolyte. The more reactive metal loses electrons more readily, so a voltage is produced — until one reactant is used up (then it goes flat).
A hydrogen–oxygen fuel cell is fed hydrogen and oxygen continuously and produces a voltage with water as the only product:
2H₂ + O₂ → 2H₂Ohydrogen + oxygen → water · the cell keeps working while fuel is supplied
Hydrogen at the (−) electrode, oxygen at the (+); electrons flow round the external circuit doing the work.5.27 — Evaluate fuel cells
Fuel cells vs rechargeable cells
Unlike a rechargeable cell, a fuel cell does not store charge — it runs as long as hydrogen and oxygen are supplied.
Compared with rechargeable cells: fuel cells don't run down or need recharging and have no toxic metals, but hydrogen is hard to store/transport and is often produced from fossil fuels.
Quick check
The fuel-cell product
?What is the only product of a hydrogen–oxygen fuel cell?
Recap
Topic 5 at a glance
Transition metals: high mp/density, coloured compounds, variable oxidation states, catalysts (vs soft, +1, reactive Group 1)
Corrosion: rusting needs oxygen AND water; prevent by barriers or sacrificial protection (more reactive metal) / galvanising
Alloys: different-sized atoms disrupt layers → harder; e.g. steels