Question 2 of 6: Electrochemical Corrosion Theory

Question 2: Electrochemical Corrosion Theory (20 marks)

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.

2.1 — (a) Reading the polarization curve: regions, $E_{corr}$, $i_{corr}$, and passivation

Given. The printed figure plots applied potential $E$ (mV) against $\log(i)$ ($i$ in A/cm²) for ANSI 321 stainless steel in 0.5 M H₂SO₄. Reading the curve from the printed figure: a near-flat branch enters from the left at $E\approx-170$ mV and meets a sharp cusp at $\log(i)\approx-6.5$; the curve then rises steeply (the active branch) to a first dashed reference line at $E\approx70$ mV, $\log(i)\approx-5.1$; between the two dashed reference lines the curve is nearly vertical, i.e. $E$ rises by some 830 mV for less than one decade of current, up to the second dashed line at $E\approx900$ mV, $\log(i)\approx-4.3$; beyond that the curve bends to the right again, rising to $E\approx1100$–$1200$ mV at $\log(i)\approx-2.7$ at the top of the plotted range.

Find. The three corrosion-behavior regions, the corrosion potential $E_{corr}$ and corrosion current density $i_{corr}$, and a description of the active-to-passive transition (passivation).

The corrosion potential and corrosion current density are read at the sharp cusp where the descending (cathodic, hydrogen-evolution) branch and the rising (anodic, active-dissolution) branch meet — the point of sharpest curvature, which is the conventional graphical estimate of the mixed potential on a single combined polarization trace of this kind: $\boxed{E_{corr}\approx-170\ \text{mV}}$, $\boxed{i_{corr}\approx10^{-6.5}\approx3\times10^{-7}\ \text{A/cm}^2}$ (about 0.3 $\mu$A/cm²).

From $E_{corr}$, increasing the applied potential first traces the active region (labelled "Active" on the figure): current rises steeply and almost linearly in $\log(i)$ with $E$, the classic anodic Tafel behaviour of a freely dissolving metal, from the cusp up to the primary passivation potential $E_{pp}\approx70$ mV, where the current reaches its local maximum (the critical anodic current density, $i_{crit}\approx8\times10^{-6}$ A/cm² on this curve). Beyond $E_{pp}$ the curve enters the passive region (labelled "Passive"): $E$ climbs from about 70 mV to about 900 mV while $i$ stays confined to roughly $1\times10^{-5}$ A/cm², essentially flat on this log scale — the near-vertical segment between the two dashed reference lines. Past the breakdown/transpassive potential $E_{tp}\approx900$ mV the curve bends to the right into the transpassive region (labelled "Transpassive"): current rises again with increasing $E$, up to roughly $2\times10^{-3}$ A/cm² at the top of the scanned range, as the protective film itself becomes oxidised (e.g. $Cr^{3+}\rightarrow CrO_4^{2-}$) and/or oxygen evolution begins.

Passivation, described. ANSI 321 is an austenitic stainless steel whose corrosion resistance comes from a thin ($\sim$1–3 nm), adherent, chromium-rich oxide film. In the active region that film has not yet formed (or has been dissolved by the strong acid), so bare metal is exposed and dissolves by ordinary anodic Tafel kinetics: rate rises exponentially with overpotential, exactly as plotted. As the potential is raised past $E_{pp}$, the surface chemistry crosses the point at which $Cr_2O_3\cdot nH_2O$ (with some iron oxide/hydroxide) becomes thermodynamically stable and kinetically able to nucleate and spread faster than the acid can dissolve it; once it covers the surface, further metal dissolution has to occur by solid-state ion transport through the film rather than by direct dissolution of bare metal, and that transport step is far slower and almost independent of the applied potential — which is exactly why $i$ collapses by roughly an order of magnitude at $E_{pp}$ and then stays nearly constant for the next 800 mV. This self-limiting, potential-independent film-growth-controlled current is what makes the passive state so protective: the corrosion rate in the passive region ($\sim10^{-5}$ A/cm²) is more than an order of magnitude below the active-region peak, and the metal stays in that state over a wide potential window until either the film is chemically attacked (chloride-induced pitting/crevice attack, Question 4) or is driven, by a sufficiently oxidising environment, into the transpassive region where the film itself breaks down electrochemically.

2.2 — (b) Corrosion rate from $i_{corr}$

Given. The corrosion current density read from part (a), the active dissolution reaction $Fe\rightarrow Fe^{2+}+2e^-$, and iron's atomic weight and density.

Find. The corrosion (metal-loss) rate in $\mu$m/yr.

Approach. Faraday's law converts the measured current density directly into a molar dissolution rate per unit area; dividing by density converts mass loss into a thickness loss, and the result is then rescaled from seconds to years.

1. Molar (and mass) dissolution rate per unit area. Each Fe atom that dissolves releases 2 electrons, so the molar flux is $i_{corr}/(nF)$ and the mass flux is $$\dot{m} \;=\; \frac{i_{corr}\,M_{Fe}}{nF} \;=\; \frac{(3\times10^{-7}\ \text{A/cm}^2)(55.85\ \text{g/mol})}{(2)(96\,490\ \text{C/mol})} \;=\; 8.68\times10^{-14}\ \text{g/(cm}^2\text{s)}$$ using $A=C/s$ so the units resolve to g$\cdot$cm$^{-2}\cdot$s$^{-1}$.
2. Convert mass loss to thickness loss. Dividing by the metal density removes the mass unit and leaves a linear penetration rate, $$\dot{x} \;=\; \frac{\dot{m}}{\delta_{Fe}} \;=\; \frac{8.68\times10^{-14}\ \text{g/(cm}^2\text{s)}}{7.86\ \text{g/cm}^3} \;=\; 1.104\times10^{-14}\ \text{cm/s}$$
3. Rescale to $\mu$m per year. With $1\ \text{cm}=10^4\ \mu\text{m}$ and $1\ \text{yr}=3.1536\times10^{7}\ \text{s}$, $$CR \;=\; \dot{x}\times10^4\ \frac{\mu\text{m}}{\text{cm}}\times3.1536\times10^{7}\ \frac{\text{s}}{\text{yr}} \;=\; (1.104\times10^{-14})(10^4)(3.1536\times10^{7})$$ $$\boxed{CR \;\approx\; 3.7\ \mu\text{m/yr}}$$ (equivalently $3.7\times10^{-3}$ mm/yr, or about 0.14 mils per year).