Cathodic Protection via Tafel Extrapolation: 21-Mat-A7 Environmental Degradation of Materials, May 2018, Q8

Question 8: Cathodic Protection via Tafel Extrapolation (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.

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The paper prints the target current density in part (b) as the ambiguous string "1 0.5 $\mu$A/cm$^2$", and part (c) asks for a "corrosion potential" while giving current-density units. This solution reads the target as $i_2=0.5\ \mu\text{A/cm}^2$, since $500\rightarrow0.5\ \mu\text{A/cm}^2$ is an exact, clean three-decade reduction (the natural "nice number" a Tafel-extrapolation exam question is built around), and reads part (c) as asking for the corrosion current density (the quantity actually given and used throughout the rest of the question). The alternative reading $i_2=1.5\ \mu\text{A/cm}^2$ is also computed below for comparison.

8(a) — Principle of cathodic protection

Cathodic protection works by deliberately polarizing the entire metal surface to a potential more negative than its freely-corroding $E_{corr}$, either by connecting it to a more active sacrificial metal (a galvanic anode, e.g. Zn or Mg on steel) or by forcing current onto it from an external DC power supply through an inert or consumable auxiliary anode (impressed current). Driving the potential down the metal's own cathodic polarization curve suppresses the metal's anodic (dissolution) partial-current reaction — per mixed-potential theory, moving the whole surface's potential below $E_{corr}$ pushes the net reaction at every point on the surface toward reduction, cutting the anodic dissolution current toward zero. At full protection the applied/galvanic current supplies essentially all of the electrons the cathodic reaction (e.g. O$_2$ or H$^+$ reduction) demands, so the metal's own oxidation is no longer needed to supply them, and it stops corroding.

8(b) — Cathodic overpotential for the reduced corrosion rate

Given. Anodic Tafel slope $\beta_a=0.100$ V/decade; bare (unprotected) corrosion current density $i_1=500\ \mu\text{A/cm}^2$; target protected current density $i_2=0.5\ \mu\text{A/cm}^2$ (see Verify note above).

Approach. As the surface is polarized cathodically from $E_{corr}$ to a more negative protection potential $E_p$, the metal's own anodic dissolution current continues to follow its same anodic Tafel line (extrapolated below $E_{corr}$): $i_a(E_p)=i_1\,10^{(E_p-E_{corr})/\beta_a}$. Setting $i_a(E_p)=i_2$ and solving for the (negative) potential shift gives the required overpotential magnitude.

Solve for the overpotential. $$\eta = |E_p-E_{corr}| = \beta_a\log_{10}\!\left(\frac{i_1}{i_2}\right) = (0.100)\log_{10}\!\left(\frac{500}{0.5}\right) = (0.100)(3)$$ $$\boxed{\eta \approx 0.300\ \text{V} \ (300\ \text{mV cathodic shift})}$$ This is exactly a 3-decade current reduction, so the shift is exactly 3 anodic Tafel decades — the clean-number tell that $i_2=0.5\ \mu\text{A/cm}^2$ is the intended target.

8(c) — Actual (bare) corrosion current density

Reading the value the question itself supplies as the metal's kinetic starting point (and per the Verify note, treating this as the intended "current density" ask): $$\boxed{i_{corr} \approx 500\ \mu\text{A/cm}^2}$$

8(d) — Applied current required for full protection

Approach. At the protection potential $E_p$ (found in part (b)), the total cathodic partial current follows the cathodic Tafel line through the same $(E_{corr},i_1)$ point, extrapolated by the same overpotential but scaled by $\beta_c$: $i_c(E_p)=i_1\,10^{\eta/\beta_c}$. The externally applied current must supply the difference between what the cathodic reaction now demands and what the metal's own (now much-reduced) anodic dissolution still contributes.

Cathodic demand at $E_p$. $$i_c(E_p) = i_1\,10^{\eta/\beta_c} = (500)\,10^{0.300/0.200} = (500)(10^{1.5})$$ $$\boxed{i_c(E_p) \approx 15{,}811\ \mu\text{A/cm}^2}$$
Applied protection current. $$i_{applied} = i_c(E_p) - i_2 = 15{,}811 - 0.5$$ $$\boxed{i_{applied} \approx 15{,}810\ \mu\text{A/cm}^2 \approx 15.8\ \text{mA/cm}^2}$$

Final results — Question 8
Quantity	Result
Bare corrosion current density, $i_{corr}$	500 μA/cm²
Target protected current density, $i_2$	0.5 μA/cm²
Cathodic overpotential, $\eta$	0.300 V
Applied protection current density	≈15.8 mA/cm²

Verify (alternate reading)

If instead $i_2=1.5\ \mu\text{A/cm}^2$: $\eta=\beta_a\log_{10}(500/1.5)=0.100\times2.523\approx0.252$ V, and $i_c(E_p)=500\times10^{0.252/0.200}\approx 9127\ \mu\text{A/cm}^2$, giving $i_{applied}\approx9125\ \mu\text{A/cm}^2\approx9.1$ mA/cm$^2$ — the same method applies regardless of which reading of the ambiguous target is correct; only the numeric decade count changes.

Concept & Theory

Topic. Cathodic protection quantified by Tafel extrapolation from a known corrosion current density and Tafel slopes.

Key relations. Anodic Tafel $i_a=i_{corr}\,10^{(E-E_{corr})/\beta_a}$; cathodic Tafel $i_c=i_{corr}\,10^{-(E-E_{corr})/\beta_c}$ (both referenced through the same $(E_{corr},i_{corr})$ point); applied current at a protection potential $i_{applied}=i_c(E_p)-i_a(E_p)$.

Why this works. Both Tafel lines are anchored at the single measured point $(E_{corr},i_{corr})$ — that is what a Tafel/polarization-resistance test measures in the first place — so extrapolating each line by the same potential shift $\eta$ but through its own slope correctly separates "how much has the metal's own corrosion rate fallen" (anodic slope) from "how much current must the protection system now supply" (cathodic slope). Because $\beta_c=0.200$ is twice $\beta_a=0.100$ here, the same $\eta$ moves the cathodic branch only half as many Tafel decades ($\eta/\beta_c=1.5$) as it moves the anodic branch ($\eta/\beta_a=3$), so while the metal's own dissolution current falls by three decades, the cathodic demand at that same potential has already risen by 1.5 decades above $i_{corr}$ — which is why the applied protection current ends up far larger than $i_{corr}$ itself, not smaller.

Common pitfall. Using the cathodic Tafel slope to compute the overpotential in part (b) instead of the anodic slope — the quantity being suppressed (corrosion rate) is fundamentally the anodic partial current, so it is the anodic slope that connects $\eta$ to the target reduced current density.

Source. Fontana, Corrosion Engineering, 3rd ed., Ch. 4 (cathodic protection) and Ch. 6 (Tafel/polarization diagrams); Jones, Principles and Prevention of Corrosion, 2nd ed., Ch. 13.