21-Mat-A7 Environmental Degradation of Materials · May 2018
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
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:
Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.
Approach. All four parts share the same reference curve — a stainless steel's anodic branch rising from $E_{corr}$ (where it carries $i_{corr}$) through the active peak at ($i_{cc}$, $E_{pp}$), holding an approximately flat passive plateau at the low passive current density $i_p$, then rising again past the breakdown/transpassive potential $E_b$. The exam's own symbols are used throughout: $i_{cc}$ is the critical current density for passivation (elsewhere written $i_{crit}$) and $i_p$ the passive current density (elsewhere $i_{pass}$). Each sub-part shifts a different feature of that curve (solid, blue = reference; dashed, red = the changed system) and the physical reasoning behind each shift is what earns the marks, not the sketch itself.
Chloride ion is the specific species that locally breaks down the passive film (via adsorption and complexation with the film's own cations) to nucleate pits and drive transpassive/pitting breakdown. Lowering its concentration raises the potential $E_b$ at which the film fails and can also lower the steady passive current $i_p$ slightly (a marginally more stable film), while the active peak ($i_{cc}$, $E_{pp}$) and $E_{corr}$ shift only slightly, since chloride's main role here is on film stability, not on the active-region dissolution kinetics.
Elements such as Cr, Mo, and N (the same PREN-raising elements from Question 2) make the passive film form more easily and leak less current once formed. This lowers both the critical current density $i_{cc}$ needed to reach passivity (an easier active-to-passive transition, so a system that might not spontaneously passivate now does) and the passive current density $i_p$ itself (a more protective film), which together also make $E_{corr}$ more noble (shift up along the now-lower passive branch) — the classic goal of alloy design for corrosion resistance.
Higher temperature accelerates every rate process on the curve: the active-region exchange current and $i_{cc}$ both rise (faster active dissolution and a harder passivation transition), $i_p$ rises (the film itself becomes a poorer barrier, and any diffusion-limited cathodic partner reaction also speeds up), and the breakdown potential $E_b$ typically falls because pit nucleation and film breakdown are themselves thermally activated. The whole curve effectively shifts to higher current density at every potential.
Lower pH accelerates the active-region dissolution reaction directly (many active-metal dissolution reactions are proton-assisted) and, by Nernst-shifting the hydrogen-evolution cathodic partner reaction to more noble potentials, raises both $E_{corr}$ and $i_{corr}$ — the corrosion potential now sits squarely in the active region rather than on the passive plateau. The passive current density rises and the passive range narrows because the oxide/hydroxide film is itself less thermodynamically stable at low pH (closer to its own dissolution boundary on a Pourbaix diagram) — at sufficiently low pH the passive plateau can disappear altogether and the metal corrodes actively across the whole potential range.
