Basic Corrosion Theory: 21-Mat-A7 Environmental Degradation of Materials, May 2015, Q1

Question 1: Basic 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.

1.1 — (a) Nernst-corrected electrode potentials at pH 7, $p_{O_2}=0.2$ atm

Given. Two reduction half-reactions with their standard (1 M, 1 atm, 25 °C) potentials, and the actual service condition of a neutral aerated solution.

Given data — Question 1(a)
Quantity	Symbol	Value
Oxygen reduction half-cell	$O_2+4e^-+4H^+\rightarrow 2H_2O$	$E^\circ=1.229$ V, $n=4$
Ferrous-hydroxide reduction half-cell	$Fe(OH)_2+2e^-+2H^+\rightarrow Fe+2H_2O$	$E^\circ=-0.048$ V, $n=2$
Solution pH	$pH$	7 (so $[H^+]=10^{-7}$ M)
Oxygen partial pressure	$p_{O_2}$	0.2 atm
Nernst constant at 25 °C	$2.303RT/F$	0.0592 V (as quoted on the paper)

Find. The actual (non-standard) electrode potential of each half-cell at pH 7 and $p_{O_2}=0.2$ atm.

Approach. Apply the Nernst equation $E=E^\circ-\dfrac{0.0592}{n}\log Q$ to each half-cell as written (a reduction), with $Q$ built from the activities of the species that appear explicitly: $O_2$ as its partial pressure, $H^+$ as $10^{-pH}$, and the two solids ($Fe$, $Fe(OH)_2$) at unit activity.

Oxygen half-cell. For $O_2+4H^++4e^-\rightarrow 2H_2O$ the reaction quotient for the reduction is $Q=1/\big(p_{O_2}[H^+]^4\big)$ ($H_2O$ at unit activity), so $$E_{O_2/H_2O} \;=\; E^\circ_{O_2/H_2O} - \frac{0.0592}{4}\log\!\left(\frac{1}{p_{O_2}[H^+]^4}\right) \;=\; E^\circ_{O_2/H_2O} + \frac{0.0592}{4}\log p_{O_2} - 0.0592\,pH$$ Substituting $p_{O_2}=0.2$ and $pH=7$, $$E_{O_2/H_2O} \;=\; 1.229 + \frac{0.0592}{4}\log(0.2) - 0.0592(7) \;=\; 1.229 - 0.0103 - 0.4144$$ $$\boxed{E_{O_2/H_2O} \;\approx\; 0.804\ \text{V}}$$ Dropping the pH from 0 (standard, acidic) to 7 costs the oxygen electrode $0.0592\times7=0.414$ V — by far the larger of the two corrections — while the depressed oxygen activity ($p_{O_2}=0.2$ instead of 1 atm) costs only about 10 mV.
Ferrous-hydroxide half-cell. For $Fe(OH)_2+2H^++2e^-\rightarrow Fe+2H_2O$, $Fe$ and $Fe(OH)_2$ are both solids at unit activity, so $Q=1/[H^+]^2$ and $$E_{Fe(OH)_2/Fe} \;=\; E^\circ_{Fe(OH)_2/Fe} - \frac{0.0592}{2}\log\!\left(\frac{1}{[H^+]^2}\right) \;=\; E^\circ_{Fe(OH)_2/Fe} - 0.0592\,pH$$ $$E_{Fe(OH)_2/Fe} \;=\; -0.048 - 0.0592(7) \;=\; -0.048-0.4144$$ $$\boxed{E_{Fe(OH)_2/Fe} \;\approx\; -0.462\ \text{V}}$$ This half-cell has no $O_2$ term at all, so its entire shift from standard conditions is the $-0.0592\,pH$ term — the same 414 mV correction as above, because both half-reactions consume the same number of protons per electron (1:1).

Final results — Question 1(a)
Half-cell	Symbol	Result
Oxygen reduction	$E_{O_2/H_2O}$	$+0.804$ V
Ferrous hydroxide reduction	$E_{Fe(OH)_2/Fe}$	$-0.462$ V

1.2 — (b) The overall corrosion reaction, $E_{corr}$, and spontaneity

Find. The overall (rusting) reaction obtained by combining the two half-cells, the resulting corrosion cell potential $E_{corr}$, and whether the reaction proceeds spontaneously as written.

Approach. In the corrosion cell, iron is oxidised (the anodic reaction, the reverse of the printed $Fe(OH)_2/Fe$ reduction) and oxygen is reduced (the cathodic reaction, exactly as printed). Balance electrons between the two half-reactions, add them, and evaluate $E_{corr}=E_{cathode}-E_{anode}$ using the two Nernst-corrected potentials from part (a) — the anode's potential is still taken from its own reduction half-reaction, not reversed in sign, because that is how the cell-potential formula is defined.

Balance the electrons and add the half-reactions. The oxygen reduction needs 4 electrons; the iron oxidation (reversed ferrous-hydroxide half-cell) supplies only 2, so the anodic half-reaction is doubled: $$2Fe + 4H_2O \;\longrightarrow\; 2Fe(OH)_2 + 4H^+ + 4e^- \qquad \text{(anodic, oxidation, }\times 2\text{)}$$ $$O_2 + 4H^+ + 4e^- \;\longrightarrow\; 2H_2O \qquad \text{(cathodic, reduction)}$$ Adding and cancelling the $4H^+$ and $2H_2O$ common to both sides gives the overall corrosion reaction $$\boxed{2Fe + O_2 + 2H_2O \;\longrightarrow\; 2Fe(OH)_2}$$ which is mass-balanced (2 Fe, 4 O, 4 H on each side) and is exactly the familiar first step of the rusting of iron: ferrous hydroxide, which then oxidises further in air to the hydrated ferric-oxide "rust" the question refers to.
Evaluate the corrosion cell potential. With oxygen reduction as the cathode and the ferrous-hydroxide couple as the anode, $$E_{corr} \;=\; E_{cathode}-E_{anode} \;=\; E_{O_2/H_2O}-E_{Fe(OH)_2/Fe} \;=\; 0.804-(-0.462)$$ $$\boxed{E_{corr} \;\approx\; +1.267\ \text{V}}$$
Test spontaneity. The free-energy change for the reaction as balanced (4 electrons transferred) is $$\Delta G \;=\; -nFE_{corr} \;=\; -(4)(96\,490\ \text{C/mol})(1.267\ \text{V}) \;\approx\; -489\ \text{kJ}$$ Because $E_{corr}>0$, $\Delta G<0$: the reaction is spontaneous as written. Physically this says the oxygen half-cell's potential remains well above the iron half-cell's potential even after both are corrected to the actual pH and oxygen activity of the service solution, so iron sitting in neutral aerated water has a large thermodynamic driving force to rust — consistent with everyday experience, and with the fact that neither half-cell needed to be reversed in sign to build a positive-$E_{corr}$ cell.

Final results — Question 1(b)
Quantity	Symbol	Result
Overall corrosion reaction	—	$2Fe+O_2+2H_2O\rightarrow 2Fe(OH)_2$
Corrosion cell potential	$E_{corr}$	$+1.267$ V
Free energy change (4 mol e$^-$ basis)	$\Delta G$	$-489$ kJ
Spontaneity	—	Spontaneous ($E_{corr}>0$, $\Delta G<0$)

Concept & Theory

Topic. The Nernst equation applied to a corrosion half-cell, and the sign convention that turns two half-cell potentials into a corrosion driving force.

Key relations. Nernst equation $E=E^\circ-\dfrac{0.0592}{n}\log Q$ (25 °C); $[H^+]=10^{-pH}$; cell potential $E_{corr}=E_{cathode}-E_{anode}$ (both taken as reduction potentials); spontaneity criterion $\Delta G=-nFE_{corr}$, spontaneous iff $E_{corr}>0$.

Why this works. A half-cell's potential is only "standard" at unit activity of every species that appears in $Q$; the moment the solution departs from pH 0 or the gas from 1 atm, the Nernst correction must be applied before the two half-cells can be compared. Here both half-reactions consume protons at the same 1-electron:1-proton ratio, so both shift by the identical $-0.0592\,pH$ term and the pH dependence very nearly cancels out of $E_{corr}$ — the corrosion driving force for rusting is almost pH-independent in the near-neutral range, which is why iron rusts about as readily in slightly acidic rain water as in slightly alkaline tap water. What does not cancel is the oxygen availability: $E_{O_2/H_2O}$ falls (and the driving force with it) as $p_{O_2}$ falls, which is the thermodynamic basis for oxygen-starvation corrosion control (deaeration, coatings that exclude oxygen) treated further in Questions 4 and 6.

Common pitfall. Reversing the sign of $E^\circ_{Fe(OH)_2/Fe}$ before subtracting because iron is "the one that oxidises" — the $E_{cathode}-E_{anode}$ formula already accounts for the reversal; reversing it a second time gives a negative, non-spontaneous-looking $E_{corr}$ for a reaction that is known to proceed. A second common error is to apply the $-0.0592\,pH$ correction only to one half-cell.

Source. Jones, Principles and Prevention of Corrosion, 2nd ed., Ch. 2 (thermodynamics of corrosion, Pourbaix-diagram construction) and Ch. 3; Fontana, Corrosion Engineering, 3rd ed., Ch. 2.

Question 1 of 6: Basic Corrosion Theory

Question 1: Basic Corrosion Theory (20 marks)

1.1 — (a) Nernst-corrected electrode potentials at pH 7, $p_{O_2}=0.2$ atm

1.2 — (b) The overall corrosion reaction, $E_{corr}$, and spontaneity