Corrosion rate of iron from its corrosion current density: 22-Mec-B8 Engineering Materials, Undated paper, Q1

Question 1: Corrosion rate of iron from its corrosion current density (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.

Given. Iron dissolving uniformly under free corrosion in a dilute salt electrolyte, the whole surface acting as a mosaic of anodic and cathodic sites:

Given data
Quantity	Symbol	Value
Corrosion current density	i	1.1 µA/cm² = 1.1 × 10⁻⁶ A/cm²
Atomic mass of iron	M	55.85 g/mol
Electrochemical valence of iron	n	2 (Fe → Fe²⁺ + 2e⁻)
Density of iron	ρ	7.87 g/cm³
Faraday constant	F	96 485 C/mol
Target thickness loss for part (b)	r_b	0.01 mm/year

Find. (a) the depth of iron removed from the surface in one year at the measured current density, expressed in millimetres per year, and (b) the current density that would correspond to a loss of exactly 0.01 mm/year.

Uniform corrosion of iron under free-corrosion conditions. Anodic and cathodic sites are distributed over the whole surface and continually change places, so the metal thins evenly rather than pitting. Because every step from current to penetration rate is linear, the two parts of the question are a single straight line read in opposite directions: (a) reads the rate off a known current density, (b) reads the current density off a required rate.

Approach. Faraday's law converts a current into a mass of metal dissolved; dividing by the density and by the exposed area turns that mass into a depth of penetration, and because every step is linear in current, part (b) is a single ratio taken from part (a).

Write the anodic reaction and fix the valence. In a near-neutral salt solution iron dissolves to the ferrous ion, $$\mathrm{Fe} \rightarrow \mathrm{Fe}^{2+} + 2e^{-}$$ so $n = 2$ electrons are released per atom of iron removed. The balancing cathodic reaction on the same surface is oxygen reduction, $\mathrm{O}_2 + 2\mathrm{H_2O} + 4e^{-} \rightarrow 4\mathrm{OH}^{-}$, but it does not enter the arithmetic: the measured current density already tells us how fast electrons are leaving the metal.
Apply Faraday's law per unit area. The mass of iron dissolved per unit area per unit time is $$\frac{m}{At} = \frac{i\,M}{n\,F}$$ where $i$ is the current per unit area. Substituting, $$\frac{m}{At} = \frac{(1.1\times10^{-6})(55.85)}{2(96\,485)} = 3.184\times10^{-10}\ \text{g}/(\text{cm}^2\!\cdot\!\text{s})$$ Over a 365-day year, $t = 3.1536\times10^{7}$ s, this amounts to $1.004\times10^{-2}$ g of iron lost from every square centimetre.
Part (a) — convert the mass loss into a penetration depth. A mass lost per unit area, divided by the density, is a thickness: $$r = \frac{i\,M}{n\,F\,\rho} = \frac{(1.1\times10^{-6})(55.85)}{2(96\,485)(7.87)} = 4.045\times10^{-11}\ \text{cm/s}$$ and multiplying by the seconds in a year and by ten to reach millimetres, $$r = 4.045\times10^{-11} \times 3.1536\times10^{7} \times 10$$ $$\boxed{\ r = 1.28\times10^{-2}\ \text{mm/year} = 0.0128\ \text{mm/year}\ }$$ That is roughly 13 micrometres a year, or about 1.3 mm in a century — a slow, benign general attack, which is exactly what a microamp-level current density implies.
Cross-check with the standard corrosion-penetration-rate formula. Handbooks package the same physics as $\mathrm{CPR}\,[\text{mpy}] = 0.129\,i\,a/(n\rho)$ with $i$ in µA/cm², $a$ the atomic mass and $\rho$ in g/cm³. Substituting, $$\mathrm{CPR} = \frac{0.129(1.1)(55.85)}{2(7.87)} = 0.503\ \text{mils per year}$$ and since one mil is 0.0254 mm, that is 0.0128 mm/year — the same answer by an independent route, which confirms both the constant and the arithmetic.
Part (b) — scale the current density to the target rate. Every factor relating $i$ to $r$ is a constant, so the two are strictly proportional and no re-derivation is needed: $$\frac{i_b}{i_a} = \frac{r_b}{r_a} \quad\Longrightarrow\quad i_b = 1.1 \times \frac{0.0100}{0.012757}$$ $$\boxed{\ i_b = 0.862\ \mu\text{A/cm}^{2} = 8.62\times10^{-7}\ \text{A/cm}^{2}\ }$$ A thickness loss of 0.01 mm/year is slower than the measured one, so the required current density is correspondingly lower — about 78 per cent of the measured value. A candidate who obtains a number larger than 1.1 µA/cm² has inverted the ratio.

Results for the corroding iron
Quantity	Relation	Value
Anodic reaction	—	Fe → Fe²⁺ + 2e⁻, n = 2
Mass loss per unit area	iM/(nF)	1.004 × 10⁻² g/(cm²·yr)
Penetration rate	r = iM/(nFρ)	4.045 × 10⁻¹¹ cm/s
(a) Thickness lost in one year	r	0.0128 mm/year (0.503 mils/year)
(b) Current density for 0.01 mm/year	i_b = i_ar_b/r_a	0.862 µA/cm² (8.62 × 10⁻⁷ A/cm²)

Check: the question does not state the valence or the properties of iron, so the standard values are assumed: n = 2, M = 55.85 g/mol and ρ = 7.87 g/cm³. Ferrous dissolution is the correct choice for a dilute, near-neutral salt solution; if the iron dissolved directly to Fe³⁺ (n = 3), the rate would fall by one third to 0.0085 mm/year, so the assumption should be stated in the answer book. A 365.25-day year changes the result by 0.07 per cent and is immaterial. The calculation also assumes the attack is uniform: the same total current concentrated into a few pits would penetrate hundreds of times faster, which is why a corrosion current density alone never establishes that a component is safe.

Topic. Faraday's law as the bridge between corrosion current and metal loss, and the conversion of a current density into an engineering penetration rate.

Key relations. Faraday's law m = ItM/(nF); corrosion rate per unit area r = iM/(nFρ); the handbook form CPR[mpy] = 0.129 i a/(nρ) with i in µA/cm², and CPR[mm/yr] = 0.00327 i a/(nρ); one mil per year = 0.0254 mm/year; the mixed-potential principle, that at the freely corroding potential the anodic and cathodic currents are equal and their common value is i_corr.

Why this works. Corrosion is not a chemical process that happens to be accompanied by a current — it is the current. Every iron atom that leaves the lattice as Fe²⁺ leaves two electrons behind, and those electrons must be consumed somewhere on the same piece of metal, usually by reducing dissolved oxygen. So counting coulombs is exactly the same as counting atoms, with Faraday's constant as the exchange rate: 96 485 coulombs per mole of electrons. Dividing the resulting mass loss by density and by area converts it into the quantity a designer actually needs, a rate of wall thinning that can be multiplied by a design life to give a corrosion allowance. That is why electrochemical measurements — linear polarisation resistance, potentiodynamic scans, corrosion coupons instrumented for current — can predict wall loss in weeks rather than waiting years for a mass-loss coupon.

Common pitfall. Forgetting the density and reporting a mass loss where a thickness was asked for; forgetting to convert years to seconds, which throws the answer out by seven orders of magnitude; and mixing units inside the handbook formula, which is written for µA/cm² and g/cm³ and returns mils per year, not millimetres. The conceptual pitfall is treating a low uniform-corrosion rate as proof of safety: localised attack — pitting, crevice corrosion, stress-corrosion cracking — carries the same average current density into a far smaller area and can perforate a wall while the average rate still looks negligible.

Source. Fontana, Corrosion Engineering, 3rd ed., Ch. 9 (modern theory: mixed potentials and corrosion rate from current); Callister & Rethwisch, Ch. 17.3 (corrosion rates, CPR); Askeland & Wright, Ch. 22.

Question 1 of 8: Corrosion rate of iron from its corrosion current density

Question 1: Corrosion rate of iron from its corrosion current density (20 marks)