Pourbaix (E-pH) Diagrams

Corrosion engineering remains the diagrams' home turf: choosing a cathodic-protection potential for a buried steel pipeline, or predicting whether a stainless-steel implant will passivate in physiological fluid (pH ≈ 7.4, mildly oxidizing), starts with locating the operating point on the relevant Pourbaix diagram before any kinetic (Tafel, EIS) data are collected. Nuclear-plant primary-circuit chemistry uses temperature-corrected Pourbaix diagrams (rescaling the $0.0592$ V factor for $T\approx300\,^\circ$C) to keep zirconium alloy cladding and steel piping inside immunity or tightly protective-oxide domains, since even brief excursions into corrosion domains can seed stress-corrosion cracking. Hydrometallurgy leans on the same tool from the opposite direction: extracting copper, uranium, or gold from ore relies on deliberately pushing pH and potential (via oxidants like O₂, Fe³⁺, or Mn⁴⁺, and acids/bases) into the region where the metal ion is soluble, then later precipitating it by shifting the same coordinates back into a solid-phase domain — leach-then-precipitate flowsheets are literally paths traced across a Pourbaix diagram. Environmental geochemistry uses element-specific diagrams (arsenic, chromium, uranium, selenium) to predict the mobile versus immobile speciation of contaminants in groundwater as a function of redox condition and pH, directly informing remediation strategy (e.g., whether reducing conditions will immobilize chromium as insoluble Cr(OH)₃ rather than mobile, carcinogenic CrO₄²⁻). Battery and electrolyzer chemistry uses the water-stability lines specifically: any aqueous battery or electrolyzer operating outside the narrow ~1.23 V window between the O₂/H₂O and H₂O/H₂ lines risks parasitic water electrolysis, which is why aqueous zinc, lead-acid, and nickel-metal-hydride chemistries are engineered to keep working potentials inside (or only briefly, kinetically, outside) that window — the same water lines that bound every metal's own Pourbaix diagram.

Ask the tutor: How would the iron Pourbaix diagram change qualitatively if chloride ions were present at high concentration, and why does this explain pitting corrosion of stainless steel?

Start from what you already know: the pH Dependence of Redox Potentials tells you that when a half-reaction consumes or produces H⁺, its potential shifts linearly with pH via the Nernst equation. A Pourbaix diagram simply performs this calculation for every relevant couple of an element simultaneously, then finds where each pair of species has equal free energy — i.e., equal activity under the Nernst equation — and plots that condition as a line in the $(pH, E)$ plane. The regions between lines are named for whichever species is thermodynamically dominant there, exactly as in a phase diagram, where a species is "stable" in the coexistence sense used throughout Redox Stability in Water. Three families of reactions generate three families of lines. A pure electron-transfer reaction, such as

involves no protons, so its Nernst potential does not depend on pH: it is drawn as a horizontal line at $E = E^\circ$ (for unit activities). A pure acid–base reaction, such as

involves no electron transfer, so the equilibrium is fixed by pH alone (via $K_{sp}$ and the proton stoichiometry) and independent of $E$: a vertical line. A mixed reaction, such as

couples electron and proton transfer, so the Nernst potential depends on both; the line has a nonzero slope in $E$–pH space, fixed by the ratio of protons to electrons in the balanced equation. Layered underneath every metal diagram are two universal dashed reference lines for water itself, because water can act as both oxidant and reductant:

Both have slope $-0.0592$ V/pH (four electrons/four protons and two electrons/two protons respectively, giving the same $m/n=1$ ratio). Above the upper line, water oxidizes to O₂; below the lower line, water reduces to H₂; between them, water is thermodynamically stable — the "water stability window" that frames every aqueous Pourbaix diagram.

Reading such a diagram is an exercise in localization: pick a point $(pH, E)$ representing your solution and applied potential (or open-circuit corrosion potential), and whichever region contains that point names the thermodynamically stable species. Everything else — corrosion rate, kinetics of passivation, whether a film is protective — lies outside the diagram's remit and must come from kinetic data, exactly as Predicting Redox Behavior from Diagrams cautions when interpreting any predominance map.

The entire construction rests on the Nernst equation, applied systematically and combined with mass-balance (solubility) equilibria. For a general half-reaction written as the reduction

the potential is

where:

• $E^\circ$ is the standard potential (all activities unity, pH 0),
• $n$ is the number of electrons transferred,
• $m$ is the number of protons transferred,
• $[\text{Ox}], [\text{Red}]$ are activities (approximated by concentrations) of the oxidized/reduced species,
• the factor $0.0592$ V is $RT\ln10/F$ at $T = 298.15$ K.

Reading it: the coefficient of pH, $-0.0592\,m/n$, is the slope of the boundary line in $E$–pH space. When $m=0$ (no protons), the slope is zero — a horizontal line. When $n=0$ is not possible for an electron-transfer reaction but is the case for a pure acid–base equilibrium, the reaction has no $E$-dependence at all and the line is vertical, governed instead by

which is just the mass-action law for the solubility/hydrolysis equilibrium, entirely independent of potential. For mixed lines, the slope in mV per pH unit follows directly:

where: $m/n$ is the proton-to-electron ratio fixed by how you balance the half-reaction. A 1e⁻/1H⁺ process (like Fe(OH)₃/Fe²⁺ shown above, once fully balanced) gives slope $-0.0592\times3 = -0.1776$ V/pH for that particular 1-electron/3-proton case; a 2e⁻/2H⁺ process gives the same slope as water's lines, $-0.0592$ V/pH; a 1e⁻/2H⁺ process gives $-0.1184$ V/pH. The slope is a direct fingerprint of stoichiometry — a chemist can read the electron:proton ratio of a reaction straight off the graph. Finally, every diagram is drawn for a chosen total dissolved-metal activity, conventionally $10^{-6}$ M for corrosion diagrams (the operational definition of "negligible corrosion"). Changing this reference concentration slides the vertical/sloped solid-line boundaries between soluble ion and solid phases, because

where: lowering $[\text{ion}]$ (a more demanding corrosion criterion) shifts the intercept and hence the whole boundary, enlarging the "immunity"/passivation domains at the expense of the "corrosion" domain — this activity-dependence is itself diagnostic and is exploited deliberately in corrosion engineering.

Example 1 — Foundational: classify three lines by inspection. Consider three half-reactions relevant to a hypothetical metal M: (a) $\text{M}^{3+} + e^- \rightleftharpoons \text{M}^{2+}$ (b) $\text{M(OH)}_2 + 2\text{H}^+ \rightleftharpoons \text{M}^{2+} + 2\text{H}_2\text{O}$ (c) $\text{MO}_2 + 4\text{H}^+ + 2e^- \rightleftharpoons \text{M}^{2+} + 2\text{H}_2\text{O}$ For (a): no H⁺ appears, so $m=0$ — a horizontal line at $E = E^\circ_{a} - 0.0592\log([\text{M}^{2+}]/[\text{M}^{3+}])$. For (b): no electrons appear — this is a pure solubility/acid–base equilibrium, so it is a vertical line located at the pH fixed by $K_{sp}$: from $K_{sp} = [\text{M}^{2+}]/[\text{H}^+]^2$ (schematically), $\text{pH} = \tfrac12(\log K_{sp}-\log[\text{M}^{2+}])$, independent of $E$. For (c): $m=4$, $n=2$, so slope $= -0.0592\times(4/2) = -0.1184$ V/pH — a sloped line with half the slope of a $4e^-/4H^+$ process (like O₂/H₂O) at the same $m$, because here $n$ is twice as large per proton transferred. This triage — horizontal/vertical/sloped by simply counting H⁺ and $e^-$ in the balanced equation — is the first and most important reading skill for any Pourbaix diagram, and it requires no numerical data at all.

Example 2 — Applied: does iron corrode or passivate at pH 3, $E = +0.40$ V? Using the boundary derived above, $E_{\text{Fe(OH)}_3/\text{Fe}^{2+}} = 1.412 - 0.1776\,\text{pH}$, evaluate at pH = 3.00:

The operating point $E=0.40$ V lies below 0.879 V, so at pH 3 the system sits in the Fe²⁺ (soluble) region relative to this boundary — active corrosion is thermodynamically favored, not passivation, at this potential and pH. Cross-checking against the Fe²⁺/Fe (immunity) boundary (roughly $E \approx -0.44 - 0.0592\log[\text{Fe}^{2+}]$, pH-independent for this couple since no protons participate) confirms $E=0.40$ V is well above that immunity line too, so Fe²⁺ — not metallic Fe — is the stable species: the point (pH 3, +0.40 V) lands squarely in the corrosion domain, consistent with the known aggressive dissolution of iron in dilute, mildly oxidizing acid.

Example 3 — Exam/competition level: constructing a two-line intersection (triple point) and predicting a buffer's fate. A student is given two boundary lines for element M (all at the standard $10^{-6}$ M convention): Line 1 (M²⁺/M, pure electron transfer, $m=0,n=2$): $E_1 = -0.250$ V (horizontal, no pH dependence). Line 2 (M(OH)₂/M²⁺, pure acid–base): vertical at pH $= 8.50$. Line 3 (MO₂·H₂O/M²⁺, mixed, $m=2,n=2$, from $\text{MO}_2\cdot\text{H}_2\text{O} + 2\text{H}^+ + 2e^- \rightleftharpoons \text{M}^{2+}+2\text{H}_2\text{O}$, with intercept $E^\circ=0.700$ V and reference $[\text{M}^{2+}]=10^{-6}$ M): using the general formula,

Task: find the triple point where all three phases (M, M²⁺, and either M(OH)₂ or MO₂·H₂O) could in principle meet, and determine which solid phase is actually stable at pH 8.50, $E = -0.10$ V. Step 1. At pH = 8.50 (the vertical line's location), evaluate Line 3:

This is well above Line 1's value of $-0.250$ V, so at pH 8.50 the ordering from low to high $E$ is: M (below $-0.250$ V) → M²⁺ ($-0.250$ to $+0.374$ V) → MO₂·H₂O (above $+0.374$ V). The vertical line (M(OH)₂/M²⁺) only matters below the potential where MO₂·H�2O is stable, i.e., it competes with M²⁺ in the same window; since M(OH)₂ is the lower oxidation-state solid, its true domain lies between Line 1 and Line 3, replacing part of the naive M²⁺ region wherever it is thermodynamically favored — determining this requires comparing $\Delta G_f$ of M(OH)₂ against dissolved M²⁺ directly, which is exactly the vertical-line construction. Step 2. At the test point $(pH, E) = (8.50, -0.10\ \text{V})$: compare $-0.10$ V against Line 1 ($-0.250$ V) and Line 3 ($+0.3744$ V). Since $-0.250 < -0.10 < 0.3744$, the point lies in the intermediate band — the M²⁺/M(OH)₂ competition zone. Because we are exactly at the vertical line's pH (8.50), this point sits precisely on the boundary between soluble M²⁺ and solid M(OH)₂: an infinitesimal decrease in pH would place the point in the "M²⁺ stable" (corroding) region, while an infinitesimal increase in pH would place it in the "M(OH)₂ stable" (passivating) region. This razor's-edge sensitivity is a completely general and practically important feature of Pourbaix diagrams: near a nearly-vertical boundary, small pH shifts (a local acidification in a crevice, for instance) can flip a system from passivation to active corrosion, which is precisely the mechanism behind crevice corrosion and pitting in engineering alloys.