21-Mat-A7 Environmental Degradation of Materials · May 2018
Question 7 of 8: Pilling-Bedworth Ratio — Al and Mg
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2018 — 10-Met-A7, Corrosion and Oxidation. Three hours, open book, approved Casio/Sharp calculator only. Eight questions of 20 marks each; the rubric states that the first five questions as they appear in the answer book constitute a complete paper (100 marks). All eight are answered here, because this set is a study resource rather than an exam script. The rubric also flags that answers take one of three forms — essay, calculation, or a comparison table — and marks clarity and organisation accordingly.
Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:
M. G. Fontana, Corrosion Engineering, 3rd ed. — electrochemical corrosion theory, polarization and passivity (Ch. 6), forms of corrosion: galvanic, crevice, pitting, intergranular, stress corrosion cracking (Ch. 3), cathodic and anodic protection (Ch. 4–9).
D. A. Jones, Principles and Prevention of Corrosion, 2nd ed. — thermodynamics and the Nernst equation (Ch. 2–3), mixed-potential and kinetics/polarization diagrams (Ch. 4), passivity (Ch. 6), environmental cracking (Ch. 10–11), coatings and cathodic protection (Ch. 13–14).
W. D. Callister and D. G. Rethwisch, Materials Science and Engineering: An Introduction, 10th ed. — corrosion and degradation of materials, incl. degradation of polymers (Ch. 17).
ASM Handbook, Vol. 13A (Corrosion: Fundamentals, Testing, and Protection) — Pilling–Bedworth ratio and high-temperature oxidation, forms of corrosion, cathodic protection practice.
Question 7: Pilling-Bedworth Ratio — Al and Mg (20 marks)
Find. $PBR_{Mg}$, $PBR_{Al}$, and which metal's oxide is the more protective.
Approach. $PBR = \dfrac{M_{oxide}\,\rho_{metal}}{n\,M_{metal}\,\rho_{oxide}}$, where $n$ is the number of metal atoms per formula unit of the oxide ($n=1$ for MgO; $n=2$ for Al$_2$O$_3$) — the ratio of the oxide's molar volume to the molar volume of the metal it replaced.
Aluminum is the more protective of the two. $PBR_{Al}\approx1.28$ falls inside the classically "protective" window of roughly $1less volume than the metal consumed to form it; the film is therefore necessarily porous, cracked, or discontinuous (it physically cannot cover the metal surface without gaps), letting the oxidizing atmosphere reach fresh metal continuously and giving magnesium's oxide a much poorer, often linear (non-protective) oxidation rate law compared with aluminum's parabolic, diffusion-controlled behaviour.
7(c) — Three limitations of the Pilling-Bedworth ratio
It says nothing about scale adhesion or stress relief. A PBR well above 1 (e.g. Fe$_2$O$_3$ at $\approx2.1$) can develop such large compressive growth stresses that the scale buckles, cracks or spalls off despite having "enough" volume, which is just as non-protective as the porous scale a low-PBR oxide gives.
It ignores the oxide's own defect chemistry and diffusion mechanism. Two oxides with similar PBR can have vastly different oxidation rates because ionic/electronic transport through the scale (via vacancies, interstitials, or grain boundaries) — not just whether the scale is geometrically continuous — controls the actual parabolic rate constant.
It is a single-value, room-temperature volume ratio applied to a high-temperature, evolving system — it ignores thermal-expansion mismatch between oxide and metal on heating/cooling (which drives spallation on thermal cycling even for a "protective" PBR), multi-layered or multi-phase scales (common in alloys, where several oxides form simultaneously), and plastic/creep relaxation of stress in the scale at service temperature, none of which the simple volume ratio captures.
