Sacrificial Anodes on a Steel Pier: Zinc Versus Aluminium
A steel pier is protected from seawater corrosion by blocks of zinc bolted to the submerged structure. The zinc blocks are replaced when about 60% of their mass has been lost. After one maintenance cycle, an engineer replaces one zinc block with an aluminium block of similar size because aluminium has a more negative standard reduction potential (Al3+/Al, E° = -1.66 V; Zn2+/Zn, E° = -0.76 V; Fe2+/Fe, E° = -0.44 V). Six months later, rust is found on the steel near the aluminium block, but not near the remaining zinc blocks. The seawater contains dissolved oxygen and chloride ions. The aluminium block is covered with a thin, adherent layer that is not visible on the zinc blocks. A current of 0.50 A is measured flowing from one zinc block to the steel. (1 F = 96 500 C; M(Zn) = 65.4 g mol-1.)
1. Explain, in terms of electrode potentials and electron flow, why the steel near an intact zinc block does not rust even when the steel is exposed to seawater.
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Zinc has a more negative standard reduction potential than iron, so zinc is the stronger reducing agent and is oxidised preferentially: Zn(s) → Zn2+(aq) + 2e-. The electrons flow through the steel to cathodic regions on the steel surface. There, dissolved oxygen is reduced: O2(g) + 2H2O(l) + 4e- → 4OH-(aq). The steel is forced to act as the cathode, so iron is not oxidised to Fe2+ and rust cannot form. The seawater acts as the electrolyte and completes the circuit. The protection continues even where the steel is exposed, because the zinc and steel are in electrical contact and the zinc continues to dissolve sacrificially. The zinc block is gradually consumed and must be replaced before it is used up.
2. Explain why the aluminium block failed to protect the steel despite aluminium having a more negative E° than zinc. Refer to the surface layer and to the marine environment.
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Aluminium reacts readily with oxygen to form a thin, adherent layer of aluminium oxide on its surface. This passivating layer is electrically insulating and prevents the aluminium underneath from continuing to oxidise freely. As a result, the aluminium does not behave as a simple sacrificial anode with its theoretical E° value; the oxide layer reduces the galvanic current between aluminium and steel. In seawater, chloride ions can also cause local pitting and breakdown of the oxide film, so any protection is uneven. The steel near the aluminium may therefore remain exposed to oxygen and water and rust, even though aluminium is more reducing than zinc on paper. A zinc coating or zinc block avoids this problem because zinc corrosion products are less protective and zinc continues to dissolve sacrificially.
3. Calculate the mass of zinc lost if a steady current of 0.50 A flows for 90 days. State one assumption in your calculation.
Read the worked answer
Time = 90 × 24 × 3600 = 7.776 × 106 s. Charge passed = I × t = 0.50 × 7.776 × 106 = 3.888 × 106 C. Moles of electrons = 3.888 × 106 / 96 500 = 40.29 mol. The zinc half-equation is Zn → Zn2+ + 2e-, so moles of Zn = 40.29/2 = 20.15 mol. Mass of zinc = 20.15 × 65.4 = 1.32 × 103 g, i.e. about 1.3 kg. This assumes that essentially all the measured current comes from zinc oxidation, with no significant side reactions or current from other parts of the structure, and that the current remains steady over the 90 days.