The Sydney Harbour Bridge is painted every year to stop it rusting — but the zinc bolts holding the steel panels together corrode first and are replaced regularly. Understanding why zinc sacrifices itself to protect iron is exactly what this lesson is about.
Understand the core concepts covered in this lesson.
Apply your knowledge to solve problems and explain phenomena.
Evaluate and analyse scientific information and data.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
The Sydney Harbour Bridge contains both steel (iron alloy) and zinc. When both metals are exposed to oxygen and moisture, zinc corrodes preferentially — the iron stays intact while the zinc degrades.
(1) Both metals are reacting with the same oxygen and water — why does zinc react faster than iron? (2) If the zinc bolts were replaced with copper bolts, what do you predict would happen to the iron? Write your prediction before reading on — what does “more reactive” actually mean chemically?
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end.
📚 Content
Metal reactivity is not a fixed property — it is a consequence of how easily a metal atom loses its valence electrons, which is determined by atomic structure. A metal reacts chemically by losing electrons to form positive ions (cations). The easier it is for a metal atom to lose electrons, the more reactive it is.
Three atomic properties determine how easily electrons are lost:
Down Group 1 of the periodic table, atomic radius increases, ionisation energy decreases, and electronegativity decreases — reactivity increases. This is why caesium is more reactive than lithium, and why potassium reacts more violently with water than sodium.
The activity series is constructed experimentally by comparing how vigorously different metals react with the same reagents under the same conditions. Three standard investigations are used:
| Metal | Reaction with O₂ | Reaction with Cold Water | Reaction with Dilute HCl |
|---|---|---|---|
| Potassium (K) | Burns vigorously | Explosive | Explosive (not used) |
| Sodium (Na) | Burns vigorously | Very vigorous | Explosive (not used) |
| Calcium (Ca) | Burns | Vigorous, bubbles | Vigorous |
| Magnesium (Mg) | Burns brightly | Very slow (reacts with steam) | Vigorous |
| Aluminium (Al) | Burns | No reaction (passivation) | Moderate (slowed by oxide layer) |
| Zinc (Zn) | Burns | No reaction | Moderate |
| Iron (Fe) | Burns slowly (as powder) | No reaction | Slow |
| Lead (Pb) | Tarnishes | No reaction | Very slow |
| Copper (Cu) | Surface oxide only | No reaction | No reaction |
| Silver (Ag) | No reaction | No reaction | No reaction |
| Gold (Au) | No reaction | No reaction | No reaction |
The NESA standard activity series, from most reactive to least reactive:
Mnemonic: Please Stop Calling Me A Zombie, I Like Having Copper Silverware Guaranteed. (K, Na, Ca, Mg, Al, Zn, Fe, Pb, H, Cu, Ag, Au)
A more reactive metal will always displace a less reactive metal from its salt solution — the more reactive metal has a greater tendency to form ions, so it “takes” the electron-release role away from the less reactive metal’s ions.
Prediction rule: Is the metal being added higher (more reactive) than the metal ion in solution? If yes → reaction occurs. If no → no reaction.
| Metal Added | Solution | Reaction? | Reason |
|---|---|---|---|
| Zn(s) | CuSO₄(aq) | Yes ✓ | Zn more reactive than Cu |
| Fe(s) | CuSO₄(aq) | Yes ✓ | Fe more reactive than Cu |
| Cu(s) | ZnSO₄(aq) | No ✗ | Cu less reactive than Zn |
| Mg(s) | FeSO₄(aq) | Yes ✓ | Mg more reactive than Fe |
| Ag(s) | HCl(aq) | No ✗ | Ag below H in series |
| Zn(s) | HCl(aq) | Yes ✓ | Zn above H in series |
For Zn(s) + CuSO₄(aq): The blue colour of CuSO₄ solution fades (Cu²⁺ ions removed); reddish-brown solid copper deposits on the zinc surface; the zinc gradually dissolves.
Equation: Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s). Atom check: 1Zn, 1Cu, 1S, 4O each side. ✓
Steel (iron alloy) and zinc are both susceptible to oxidation in the presence of oxygen and moisture. When they are in electrical contact — as they are when zinc bolts fasten steel panels — zinc preferentially loses electrons because it is higher in the activity series than iron. The zinc corrodes while the iron is protected. This is called sacrificial protection or cathodic protection.
The same principle is used in galvanised steel (zinc-coated steel), where the zinc coating sacrificially corrodes even if scratched, continuing to protect the underlying iron.
If copper bolts were used instead, the situation reverses — copper is below iron in the activity series, so iron would corrode preferentially. This is a galvanic corrosion problem.
📐 Worked Examples
Activity series check: Fe is above Cu → Fe will displace Cu²⁺. Reaction occurs.
Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)
Atom check: 1Fe, 1Cu, 1S, 4O each side. ✓
Observable: blue colour of solution fades (Cu²⁺ removed); reddish-brown copper deposits on iron nail surface.
Activity series check: Cu is above Ag → Cu will displace Ag⁺. Reaction occurs.
Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)
Atom check: 1Cu, 2Ag, 2N, 6O each side. ✓
Observable: solution gradually becomes blue (Cu²⁺ ions form); silver crystals deposit on copper wire surface.
Activity series check: Ag is below Cu → Ag cannot displace Cu²⁺. No reaction occurs.
Observable: no change in solution colour; no solid deposits on the silver wire.
(a) Reaction: Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s) — blue fades, copper deposits on nail.
(b) Reaction: Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s) — solution turns blue, silver crystals deposit.
(c) No reaction.
K is in Period 4, Group 1. Mg is in Period 3, Group 2. Potassium has a larger atomic radius than magnesium because it has more electron shells and its valence electron (4s) is further from the nucleus than Mg’s 3s electrons.
K has a lower first ionisation energy (419 kJ/mol) compared to Mg (738 kJ/mol). The larger atomic radius means the outermost electron is held less tightly by the nucleus, requiring less energy to remove. K loses its valence electron more readily under the same conditions.
K has a lower electronegativity (0.82) than Mg (1.31). This means K has less tendency to attract electrons back toward the nucleus, reinforcing its greater tendency to exist as K⁺ rather than as neutral K.
Potassium reacts more vigorously because it has a larger atomic radius, lower ionisation energy, and lower electronegativity than magnesium. These properties mean K loses its outermost electron more readily, making it more reactive toward water.
✏️ Activities
For each of the following, predict whether a displacement reaction will occur (use the activity series). Where a reaction occurs, write the balanced equation with state symbols and describe one observable change.
Type your prediction, equation, and observable for each.
Write your predictions and equations in your workbook.
A steel jetty is protected from corrosion by attaching blocks of zinc metal at regular intervals. After 18 months, the zinc blocks are significantly corroded but the steel is intact.
Question A: Explain why the zinc corrodes preferentially using the activity series and the concept of electron loss.
Question B: A maintenance worker proposes replacing the zinc blocks with copper blocks to save money (copper is cheaper in this scenario). Evaluate this proposal using your knowledge of the activity series.
Type your responses to both questions.
Write your responses in your workbook.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Wrong: A metal will displace any other metal from a compound if it is higher in the activity series.
Right: Displacement only occurs if the metal is higher in the activity series AND the reaction occurs in solution (aqueous). Solid metals cannot displace ions from solid compounds — the ions must be free to move in solution for electron transfer to occur.
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
8. (4 marks) Explain, using the concepts of atomic radius, ionisation energy, and electronegativity, why sodium reacts more vigorously with water than lithium, even though both are in Group 1. (4 marks)
Type your response — aim for all three properties.
Write your response in your book.
9. (4 marks) A student places a piece of iron metal into a solution of copper(II) sulfate. (a) Predict whether a reaction will occur, with reference to the activity series. (b) Write the balanced equation with state symbols. (c) Describe two observable changes the student would see. (1 + 2 + 1 marks)
Type your prediction, equation, and observations.
Write your response in your book.
10. (5 marks) The Sydney Harbour Bridge uses zinc bolts to fasten steel panels. (a) Explain, using the activity series, why the zinc corrodes preferentially while the iron is protected. (b) Calculate the mass loss if 0.050 mol of Zn corrodes (Zn = 65.4 g/mol). (c) A maintenance engineer proposes replacing the zinc bolts with stainless steel bolts (iron-based) to reduce replacement frequency. Evaluate this proposal with reference to the activity series and the consequences for the bridge structure. (2 + 1 + 2 marks)
Type your full extended response.
Write your full response in your book.
Go back to your Think First response. Can you now explain precisely why zinc corrodes first using atomic radius, ionisation energy, and electronegativity? And what would happen to the iron if copper bolts replaced the zinc?
1. Reaction occurs (Mg > Fe). Mg(s) + FeSO₄(aq) → MgSO₄(aq) + Fe(s). Observable: grey iron metal deposits on magnesium surface; magnesium gradually dissolves.
2. Reaction occurs (Cu > Ag). Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s). Observable: silver crystals form on copper wire; solution gradually turns blue.
3. No reaction (Pb < Zn). Lead is below zinc in the activity series; Pb cannot displace Zn²⁺ from solution.
4. Reaction occurs (Zn > H). Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g). Observable: zinc dissolves; bubbles of hydrogen gas form.
A: Zinc is above iron in the activity series, meaning zinc has a lower reduction potential and a greater tendency to lose electrons (be oxidised) than iron. When in electrical contact with steel, zinc preferentially loses electrons to form Zn²⁺ ions. The iron acts as a cathode (electrons are delivered to it), so iron is not oxidised. The zinc “sacrifices” itself.
B: The proposal should be rejected. Copper is below iron in the activity series (Cu has a higher reduction potential than Fe). When copper and iron are in electrical contact in seawater, iron would be preferentially oxidised (acts as the anode), and copper would be protected. This reverses the intended protection effect — the steel structure would corrode, not the bolts. Using copper would accelerate the corrosion of the jetty rather than protecting it.
Q1 B: Zn is above Pb in the activity series → Zn displaces Pb²⁺. Grey lead deposits on zinc; zinc dissolves. Option A describes Fe into CuSO₄. Option C describes metal reacting with acid.
Q2 B: Al forms Al₂O₃ passivation layer. This is passivation, not low reactivity. Al is genuinely reactive (above Zn) but appears less reactive due to the protective oxide.
Q3 B: The correct NESA order is Na > Mg > Zn > Fe > Cu.
Q4 C: Mg is above Fe in the activity series, so Mg displaces Fe²⁺. Mg(s) + FeSO₄(aq) → MgSO₄(aq) + Fe(s) is the correct equation. Option B is the reverse reaction (non-spontaneous).
Q5 A: K has lower first ionisation energy than Ca (419 vs 590 kJ/mol), meaning K loses its valence electron more readily, driving a faster, more exothermic reaction with water.
Q6 C: Iron is higher in the activity series than copper. In contact with seawater (an electrolyte), iron and copper form a galvanic pair where iron is the anode (corrodes). This is the galvanic corrosion problem.
Q7 D: A sacrificial anode must be higher in the activity series (lower reduction potential) than the metal being protected. With Ti at −1.63 V, magnesium (E° = −2.37 V) is lower and would be preferentially oxidised, protecting the titanium.
Q8: Na has a larger atomic radius than Li (Na: Period 3; Li: Period 2 — Na has more electron shells). Larger radius means the outermost electron (3s) is further from the nucleus and more shielded → lower ionisation energy for Na (496 kJ/mol vs 520 kJ/mol for Li) → Na loses its valence electron more readily. Na also has lower electronegativity than Li (0.93 vs 0.98) → less tendency to hold onto electrons. Combined effect: Na loses electrons more readily under the same conditions, producing a more vigorous reaction with water.
Q9: (a) Fe is above Cu in the activity series, so Fe will displace Cu²⁺ — reaction occurs. (b) Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s). Atom check: 1Fe, 1Cu, 1S, 4O each side ✓. (c) Blue colour of the CuSO₄ solution fades (Cu²⁺ ions are removed); reddish-brown copper metal deposits on the iron nail surface.
Q10: (a) Zn is above Fe in the activity series (lower reduction potential), meaning Zn has a greater tendency to be oxidised (lose electrons). When Zn and Fe are in electrical contact in the presence of an electrolyte (moisture), Zn acts as the anode and is preferentially oxidised (Zn → Zn²⁺ + 2e⁻). Fe acts as the cathode and is protected from oxidation. (b) Mass = 0.050 mol × 65.4 g/mol = 3.27 g. (c) The proposal has a significant flaw: replacing zinc bolts with stainless steel (iron-based) removes the sacrificial anode mechanism entirely. With no more reactive metal present, both bolt and panel corrode at the same rate (no electrochemical protection). More critically, if there is any compositional difference between the bolt alloy and the panel alloy, the slightly less noble alloy would corrode preferentially — potentially the structural panel rather than the replaceable bolt. The zinc bolt system is designed so the easily-replaced bolt corrodes rather than the structural panel. The engineer’s proposal should be rejected.
Answer questions on Metal Activity Series & Reactions of Metals before your opponents cross the line. Fast answers = faster car. Pool: lessons 1–7.