The rusting of iron can be summarised by the equation
4 Fe(s) + 3 O2(g) + H2O(l) ⇾ 2 Fe2O3•xH2O(s)
In reality, the process takes place via a series of half reactions that take place in the iron itself and the water droplet.
- Iron metal is oxidised to iron(II) ions in the centre of the water droplet, where the dissolved oxygen concentration is lower. The Fe2+ ions pass into solution.
Fe(s) ⇾ Fe2+(aq) + 2e– E⦵ = – 0.44V
- In regions of the water droplet with a higher concentration of dissolved oxygen (the edges), oxygen is reduced to hydroxide ions by electrons that flow from the oxidation of iron through the metal.
O2(g) + 2H2O(l) + 4e– ⇾ 4OH–(aq) E⦵ = +0.40V
- Fe2+ and OH– ions react to form Fe(OH)2, a green precipitate, away from the surface of the iron so there is no protective oxide layer formed.
- Fe2+(aq) ions are oxidised to Fe3+(aq), which then react with OH– ions and water to form the insoluble hydrated oxide known as rust.
Sea water has a greater conductivity because of the high concentration of dissolved ions and so iron rusts faster in the sea than in rain water.
Preventing corrosion
- oil / paint / grease will all help prevent water and oxygen from reaching the surface of the iron
- galvanising is a process in which the iron surface is coated in zinc which forms an impermeable oxide layer
Zn2+(aq) + 2e– ⇌ Zn(s) E⦵ = – 0.76V
Fe2+(aq) + 2e– ⇌ Fe(s) E⦵ = – 0.44V
O2(g) + 2H2O(l) + 2e– ⇌ 4OH–(aq) E⦵= + 0.40V
If scratched, the iron (Fe) is oxidised to Fe2+ by O2(g) / H2O(l) / OH–(aq) half cell which has the more positive E⦵. The Fe2+ is then reduced back to Fe by the Zn2+(aq) / Zn(s) half cell which has the less positive E⦵ of the two and so forms an oxidation half cell.
- cathodic protection is where the iron is attached by a wire to a block of a more easily oxidised metal such as magnesium
Mg2+(aq) + 2e– ⇌ Mg(s) E⦵ = – 2.37V
Fe2+(aq) + 2e– ⇌ Fe(s) E⦵ = – 0.44V
The magnesium is a sacrificial anode, reacting with in preference to iron with O2(g) / H2O(l), and it needs to be replaced periodically.
Practice question
- A student submerged an iron nail in boiled, distilled water in a test tube and a layer of oil was added to the top of the water. The experiment was repeated using tap water and without the oil layer. The test tubes were left for 1 week.
(a) Describe the appearance of the nails after 1 week.
(b) Use the standard electrode data below to fully explain the expected observations.
| Half equation | E⦵ / V |
|---|---|
| O2(g) + 2H2O(l) + 2e– ⇌ 4OH–(aq) | +0.40 |
| Cu2+(aq) + 2e– ⇌ Cu(s) | +0.34 |
| Fe2+(aq) + 2e– ⇌ Fe(s) | -0.44 |
| Zn2+(aq) + 2e– ⇌ Zn(s) | -0.76 |
| 2H2O(l) + 2e– ⇌ H2(g) + 2OH–(aq) | -0.83 |
(c) In a third experiment, the student wrapped a strip of zinc around one iron nail and a strip of copper around a second iron nail. Each nail was placed in a test tube of tap water and left for 1 week. Predict and explain, using standard electrode data, the results of this experiment.
- A rusted iron bar found on a beach was covered in pits containing sea water and a greenish precipitate that had turned orange at the surface.
(a) Give the formulae of the green and orange precipitates.
(b) Give the half equations for the two redox reactions that take place when iron rusts.
(c) Explain why the iron bar rusts faster in sea water than if exposed only to rain water.
Answer
- (a) There will be no visible change to the nail submerged in the boiled water (boiling removes all dissolved oxygen and the oil layer prevents oxygen from the air dissolving). The nail submerged in tap water will show signs of rusting (orange brown precipitate).
(b) The Fe2+/Fe half cell has a less positive standard electrode potential than the O2/H2O half cell so the Fe2+/Fe half cell will be the oxidation half cell. Fe will be oxidised to Fe2+ and O2/H2O will be reduced to OH– ions. The Fe2+ ions are oxidised to Fe3+ and a precipitate of Fe2O3•xH2O is seen on the nail.
The nail submerged in boiled water does not rust because the Fe2+/Fe half cell has a more positive standard electrode potential than the H2O/H2 half cell. The H2O/H2 half cell is now the oxidation half cell and the reaction would be
H2(g) + 2OH–(aq) + Fe2+(aq) ⇾ 2H2O(l) + Fe(s) E⦵cell = -0.44 – (-0.83) = +0.39V
(c) The Zn2+/Zn half cell has a less positive standard electrode potential than the Fe2+/Fe half cell so zinc will be oxidised in preference to iron. Zinc is a sacrificial metal and the nail will show no signs of rusting.
The Fe2+/Fe half cell has a less positive standard electrode potential than the Cu2+/Cu half cell so iron will be oxidised in preference to copper. The nail will show accelerated signs of rusting compared with the nail in tap water from the second experiment.
2. (a) Green precipitate is Fe(OH)2 and orange precipitate is Fe2O3•xH2O
(b) Fe(s) ⇾ Fe2+(aq) + 2e- / O2(g) + 2H2O(l) + 4e- ⇾ 4OH-(aq)
(c) Seawater contains a greater concentration (not more!) of dissolved ions than rainwater so it is a better conductor.