Choosing a sacrificial anode for a buried steel pipeline
A steel pipeline buried in moist soil at pH 6.5 is protected by connecting blocks of a more reactive metal to it. The engineer must choose between magnesium (Mr 24.3, density 1.74 g cm⁻³) and zinc (Mr 65.4, density 7.14 g cm⁻³). Standard electrode potentials: Mg²⁺/Mg = −2.37 V, Zn²⁺/Zn = −0.76 V, Fe²⁺/Fe = −0.44 V; O2 + 4H⁺ + 4e⁻ → 2H2O has E° = +1.23 V. The pipeline coating is epoxy and is damaged at three points. The budget allows 20 kg of either metal per year. A salesman recommends magnesium because it gives 'the biggest voltage, so the best protection'. The steel is high-strength, and the soil in some sections has high electrical resistivity.
1. Calculate the cell emf for the Mg–Fe and Zn–Fe couples, and explain, with half-equations, what actually happens at each metal in the corroding cell.
Read the worked answer
Using the metal-ion couples E(Mg–Fe) = E(cathode) − E(anode) = −0.44 − (−2.37) = +1.93 V E(Zn–Fe) = −0.44 − (−0.76) = +0.32 V Both are positive, so both metals can, in principle, oxidise preferentially and drive electrons to the iron.
What actually happens in the soil The dominant cathodic reaction on the wet steel is oxygen reduction, not Fe²⁺ reduction. At pH 6.5: E(O2/H2O) = 1.23 − 0.059 × 6.5 = 1.23 − 0.38 = +0.85 V Anode (sacrificial block): Mg(s) → Mg²⁺(aq) + 2e⁻ (or Zn → Zn²⁺ + 2e⁻) Cathode (steel surface): O2(g) + 2H2O(l) + 4e⁻ → 4OH⁻(aq) So the real driving voltages are roughly 0.85 − (−2.37) = 3.22 V for magnesium and 0.85 − (−0.76) = 1.61 V for zinc.
Key point The steel is never oxidised while the connection holds: it is the cathode, and electrons are supplied to it from the block. The iron half-equations above merely show the relative tendency to oxidise.
2. Evaluate the salesman's claim quantitatively and qualitatively, and recommend a metal, allowing for the high-resistivity soil sections.
Read the worked answer
Quantitative comparison Charge supplied per kg: Mg: (2 × 96 500 C mol⁻¹) ÷ 0.0243 kg mol⁻¹ = 7.9 × 10⁶ C kg⁻¹ Zn: (2 × 96 500) ÷ 0.0654 = 3.0 × 10⁶ C kg⁻¹ So 20 kg of magnesium supplies about 2.7 times as much charge as 20 kg of zinc - a genuine advantage.
But a bigger voltage is not automatically better At 3.2 V driving force, the magnesium blocks deliver a much larger current density than needed, so they are consumed faster than the charge-capacity figure suggests. Severe overprotection has two consequences: (i) the cathodic reaction at the steel can shift to hydrogen evolution (2H2O + 2e⁻ → H2 + 2OH⁻), and hydrogen can embrittle high-strength steel; (ii) the alkaline environment produced at the coating holidays can disbond epoxy coatings, extending the exposed area. Zinc's ~1.6 V is usually enough to polarise steel to a protective potential, and zinc corrodes in a more self-limiting way.
Recommendation Use zinc for the well-coated, lower-resistivity sections. For the high-resistivity sections, the higher driving voltage is genuinely useful because soil resistance limits the current, so magnesium (or a magnesium-rich alloy) is justified there. This is a case where 'biggest voltage' is right for one part of the route and wasteful for the rest.
3. Explain why the steel does not rust while the block is attached, and what changes when the block is exhausted.
Read the worked answer
Protection mechanism Magnesium or zinc is a stronger reducing agent than iron. At the sacrificial block, M → M²⁺ + 2e⁻ occurs; the electrons travel through the soil/backfill and the pipeline wall to the damaged coating points. There the steel surface is the cathode, where O2 + 2H2O + 4e⁻ → 4OH⁻ consumes them. Because electrons arrive at the steel instead of leaving it, Fe → Fe²⁺ + 2e⁻ cannot occur, so no Fe²⁺ is available to form the rust that needs both Fe²⁺ and O2. The pipeline is protected not because it is coated everywhere, but because it is cathodically polarised.
When the block is used up Once the block has fully corroded, the electron supply stops. The steel reverts to being the anode at the coating holidays, Fe²⁺ forms, and with dissolved oxygen Fe(OH)2/Fe(OH)3 and hydrated Fe2O3 develop - rust appears preferentially at the three damaged points, because that is where bare metal meets the electrolyte. Protection is also lost if the electrical connection breaks or if the backfill dries out, both of which are routine inspection points. The engineer should size the blocks so they outlast the inspection interval, not just the year's budget.