Sacrificial anode choice and anode mass for a buried steel pipeline
A 2.0 km buried mild-steel pipeline carrying cooling water runs through wet clay soil of pH 6.5 and moderate chloride content. A coating holiday of 0.02 m2 exposes bare steel. Corrosion is to be controlled by a galvanic anode bed. The available anodes are zinc (E° Zn2+/Zn = −0.76 V; A = 65.4 g mol−1) and magnesium (E° Mg2+/Mg = −2.37 V; A = 24.3 g mol−1); the steel has E° Fe2+/Fe = −0.44 V. The protection current required is 20 mA. The design life is 10 years. A contractor proposes magnesium because it gives a larger driving voltage and therefore “will last longer”, and also suggests a copper bond to “earth” the pipe because copper is more noble. The soil resistivity is low enough that either anode can operate. Faraday constant = 96500 C mol−1; assume zinc current efficiency 90% and magnesium 50%; 1 year = 3.15×107 s.
1. Show whether zinc and magnesium are both thermodynamically able to protect steel. Why is copper different?
Read the worked answer
Both zinc and magnesium are anodic to steel. For Zn–Fe: E°cell = E°cathode − E°anode = −0.44 − (−0.76) = +0.32 V. For Mg–Fe: E°cell = −0.44 − (−2.37) = +1.93 V.
Both couples are spontaneous in the protection direction, so thermodynamics alone does not choose between them. Copper would be cathodic: E°cell = +0.34 − (−0.44) = +0.78 V with copper as cathode, so connecting copper makes steel the anode and accelerates corrosion.
The practical judgement is that magnesium is too driving: it can overprotect the steel, generate hydrogen at the pipe surface, disbond the coating and waste current in self-corrosion. Zinc gives adequate protection with less overprotection in low-resistivity wet clay.
2. Calculate the mass of zinc consumed at 20 mA for 10 years at 90% current efficiency. Compare with magnesium at 50% efficiency and comment on the contractor’s claim.
Read the worked answer
Charge = I t = 0.020 A × (10 × 3.15×107 s) = 6.30×106 C. Moles of electrons = 6.30×106 / 96500 = 65.3 mol. Zn2+ + 2e− → Zn, so moles Zn = 65.3/2 = 32.6 mol. Mass Zn = 32.6 × 65.4 = 2.13 kg. At 90% current efficiency, actual zinc consumed = 2.13/0.90 = 2.37 kg.
For magnesium: mass = (65.3/2) × 24.3 = 0.79 kg; at 50% efficiency it is 1.59 kg.
Magnesium is lighter, but that does not make it the better choice: its higher driving voltage causes overprotection, and its self-corrosion is already included as the 50% efficiency. The design should specify about 2.4 kg of zinc per 20 mA for 10 years, with a corrosion allowance.
3. The contractor proposes a copper bond to earth the pipe. Explain why this would accelerate corrosion, including the area-ratio effect.
Read the worked answer
Copper is more noble than steel (E° Cu2+/Cu = +0.34 V vs Fe2+/Fe = −0.44 V), so in a copper–steel couple copper becomes the cathode and steel becomes the anode. The pipe is then forced to corrode faster.
Because the copper bond is small compared with the pipe, the cathode-to-anode area ratio is unfavourable: the anodic current density at the coating holiday becomes very high and pitting is severe. Earthing must use a galvanic anode or an impressed-current system, not a bare copper bond. The contractor’s reasoning reverses the meaning of “noble”: a noble metal is cathodic, not protective, when coupled to steel.