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Corrosion and its prevention: TCE Chemistry (Tasmania)

Syllabus dot point

“Explain corrosion as an electrochemical process and evaluate methods of prevention.”

9 min read

Quick answer

Corrosion of iron is an electrochemical process: iron is oxidised at anodic regions while oxygen is reduced at cathodic regions, requiring both water and oxygen. It is prevented by barrier coatings that exclude water and oxygen, or by cathodic protection that forces the iron to be the cathode, often using a more reactive metal as a sacrificial anode.

What this dot point is asking

TASC expects you to explain rusting in electrochemical terms, including half-equations, and to evaluate the methods used to prevent it.

Corrosion as electrochemistry

Rusting is the oxidation of iron in the presence of water and oxygen. Different regions of the metal act as the electrodes of a tiny galvanic cell.

  • At the anodic region, iron is oxidised: Fe(s)→Fe2+(aq)+2e−\text{Fe}(s) \rightarrow \text{Fe}^{2+}(aq) + 2e^-.
  • At the cathodic region, oxygen is reduced in the presence of water: O2(g)+2H2O(l)+4e−→4OH−(aq)\text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4e^- \rightarrow 4\text{OH}^-(aq).

The iron(II) ions are then further oxidised by oxygen and combine with hydroxide and water to form hydrated iron(III) oxide, the flaky orange-brown solid we call rust.

Rusting needs both water and oxygen

Iron does not rust in dry air or in oxygen-free water; both must be present. Salt and acids speed corrosion because dissolved ions improve electrical conductivity, completing the electrochemical circuit more easily. This is why cars rust faster in coastal and salted-road environments.

Prevention methods

  • Barrier methods keep water and oxygen away from the metal: painting, greasing, plastic coating and galvanising (a zinc layer) all form a physical barrier.
  • Sacrificial protection uses a more reactive metal (zinc or magnesium) connected to the iron. The more reactive metal is preferentially oxidised, corroding instead of the iron. The block is a sacrificial anode and is replaced periodically.
  • Cathodic protection forces the iron to be the cathode so it cannot be oxidised, either with a sacrificial anode or with an impressed current from an external supply.
Sacrificial anode

A block of a more reactive metal attached to a structure so it is oxidised in preference to the protected metal, forcing the protected metal to act as a cathode.

Galvanising combines two effects: the zinc layer is a barrier, and because zinc is more reactive than iron, even a scratched layer still protects the exposed iron sacrificially.

Choosing a sacrificial metal

State the requirement

A steel ship's hull is protected with attached metal blocks. For sacrificial protection the attached metal must be more reactive than iron, so it is oxidised in preference.

Test zinc

Zinc is more reactive than iron (more easily oxidised), so it corrodes instead of the hull and continues to protect even when scratched.

Reject copper

Copper is less reactive than iron, so coupling it to iron would make the iron the anode and accelerate corrosion. Therefore zinc is the correct choice.

Factors that speed corrosion

Several conditions accelerate the electrochemical process. Electrolytes such as dissolved salt or acid increase the conductivity of the surface water film, completing the circuit faster, which is why coastal and road-salted environments are so corrosive. Contact with a less reactive metal sets up a bimetallic (galvanic) couple in which the iron becomes the anode and corrodes faster, as in the copper-plated hull example. Mechanical stress and scratches expose fresh metal and concentrate anodic activity, and acidic rain or industrial pollutants lower the pH of the surface film, again speeding oxidation.

Stainless steel and passivation

Some metals resist corrosion by forming a thin, adherent oxide layer that seals the surface, a process called passivation. Aluminium forms a tough Al2O3\text{Al}_2\text{O}_3 film, and stainless steel contains chromium that forms a protective chromium(III) oxide layer. Unlike rust, which is flaky and lets corrosion continue underneath, these oxide films are dense and self-repairing, so they protect the metal beneath. This contrasts with iron, whose oxide flakes away and exposes fresh metal to further attack.

Saying a less reactive coating always protects

A less reactive metal coating such as tin protects only while intact. If it is scratched, the exposed iron becomes the anode and corrodes faster than unprotected iron (because the tin acts as a cathode). A more reactive coating like zinc continues to protect even when scratched. A second slip is forgetting that both water and oxygen are required, so airtight or fully dry conditions prevent rust.

In the exam, write the anode and cathode half-equations for rusting, state that both water and oxygen are required, note the accelerating effect of salt, and justify each prevention method by whether it forms a barrier or makes the iron a cathode.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice6 marks
An iron nail attaches a sign to a wooden post. (a) Explain how the nail corrodes over time, including relevant half-equations. (b) Explain how a zinc coating protects the iron even when the coating is scratched.
Show worked answer →

(a) Where the iron contacts water and dissolved oxygen, iron is oxidised at anodic regions: Fe(s)→Fe2+(aq)+2e−\text{Fe}(s) \rightarrow \text{Fe}^{2+}(aq) + 2e^-. The electrons travel through the metal to cathodic regions where oxygen is reduced: O2(g)+2H2O(l)+4e−→4OH−(aq)\text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4e^- \rightarrow 4\text{OH}^-(aq). The Fe2+\text{Fe}^{2+} and OH−\text{OH}^- ions combine and are further oxidised by oxygen to hydrated iron(III) oxide (rust). Water acts as the electrolyte, so corrosion is fastest where the nail stays damp. (4 marks)

(b) Zinc is more easily oxidised than iron, so even if the coating is scratched the zinc acts as a sacrificial anode and is oxidised in preference; the iron becomes the cathode and is protected. This is sacrificial (galvanic) protection, the principle of galvanising. (2 marks)

Practice6 marks
Three identical steel boats: Boat A on a freshwater beach at the water line, Boat B on a saltwater beach at the water line, Boat C 100 m100\ \text{m} underwater in a freshwater lake. (a) List them from fastest to slowest predicted corrosion and explain. (b) Explain how zinc attached to a hull would affect corrosion.
Show worked answer →

(a) Fastest to slowest: B, then A, then C. Corrosion is electrochemical and needs water, oxygen and an electrolyte. Boat B (saltwater, water line) has the most dissolved ions (good electrolyte) plus ready oxygen, so it corrodes fastest. Boat A (freshwater, water line) has oxygen but a poorer electrolyte, so it is slower. Boat C is deep underwater where dissolved oxygen is very low, so the cathode reaction (oxygen reduction) is starved and corrosion is slowest. (4 marks)

(b) Zinc is more easily oxidised than iron, so it acts as a sacrificial anode: it is oxidised in preference to the steel, forcing the hull to behave as the cathode, protecting the iron even where the zinc is worn. (2 marks)

Practice3 marks
An 18th-century wooden hull was covered with copper plates attached by iron nails. Within a few years the copper detached because the iron nails were heavily corroded. Explain this, including appropriate half-equations.
Show worked answer →

Iron and copper in contact in seawater form a galvanic (bimetallic) couple. Iron is the stronger reductant, so it becomes the anode and is oxidised, while copper acts as an inert cathode. Because the small iron nails are coupled to a large area of copper, the iron corrodes rapidly and the nails are eaten away, releasing the plates. (1 mark)

Anode (iron oxidised): Fe(s)→Fe2+(aq)+2e−\text{Fe}(s) \rightarrow \text{Fe}^{2+}(aq) + 2e^-. Cathode (oxygen reduced on the copper): O2(g)+2H2O(l)+4e−→4OH−(aq)\text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4e^- \rightarrow 4\text{OH}^-(aq). (2 marks)

Practise this

Sources & how we know this

ExamExplained