CHEM3120 · Environmental and Analytical Chemistry
Metal Ions: Hydrolysis, Redox & pE-pH Diagrams
Lectures 21-24 cover the environmental chemistry of metal ions in CHEM3120: stepwise hydrolysis and speciation, redox in natural waters described by pE, the construction and reading of pE-pH (Pourbaix) diagrams, and metal transport and toxicity (mercury, lead). This is the most quantitatively demanding part of Prof. Kepert's examinable Block 3 list, with pE calculations and Pourbaix-line derivations as Part A short-answer.
What this chapter covers
- 01Stepwise metal-ion hydrolysis M^n+ → M(OH) → ... with successive pKa values
- 02Dominant hydroxide species between successive pKa; amphoteric hydroxides redissolve at high pH
- 03pE = -log[e-] as an electron-availability scale; pE° = E°/0.0592 at 25 °C
- 04Nernst-style pE lines pE = pE° - (m/n)pH from combining half-reactions with H+/e-
- 05pE-pH (Pourbaix) diagrams: lines separating oxidised/reduced and soluble/solid regions
- 06The iron example: Fe2+, Fe3+, Fe(OH)3 and Fe(s) regions
- 07Metal transport (Stokes sedimentation) and heavy-metal toxicity (Hg vapour, methylmercury, Pb-EDTA chelation)
From a standard potential to pE
- +1(a) Convert the standard potential to pE°: pE° = E°/0.0592 = 0.77/0.0592 = 13.0.
- +1(b) For Fe3+ + e- ⇌ Fe2+, pE = pE° - log([Fe2+]/[Fe3+]). With [Fe3+] = [Fe2+] the ratio is 1 and log 1 = 0, so pE = pE° = 13.0.
- +1(c) pE = 13.0 is a high (positive) value, meaning very low electron availability — strongly OXIDISING conditions, consistent with the Fe3+/Fe2+ couple sitting in the oxic zone of a natural water.
Key terms
- Metal-ion hydrolysis
- Stepwise loss of protons from aquated metal ions, M^n+ → M(OH) → ..., governed by successive pKa values; the dominant hydroxo species changes between consecutive pKa.
- Amphoteric hydroxide
- A metal hydroxide (e.g. Al(OH)3, Fe(OH)3) that dissolves both in acid (as the metal ion) and in strong base (as a hydroxo-anion), so it is least soluble at intermediate pH.
- pE
- pE = -log[e-], the electron-availability analogue of pH; pE° = E°/0.0592 at 25 °C. High pE means oxidising conditions, low pE reducing.
- pE-pH (Pourbaix) diagram
- A map of the thermodynamically stable species over pE and pH, with lines separating oxidised/reduced and soluble/solid regions; each line comes from a half-reaction involving H+ and/or e-.
- Pourbaix line pE = pE° - (m/n)pH
- A boundary whose slope -(m/n) is set by the ratio of protons to electrons in the balanced half-reaction; horizontal lines are pH-independent redox couples, vertical lines pure acid-base boundaries.
- Methylmercury (CH3Hg+)
- A highly toxic, bioaccumulative mercury species formed in the environment (CH4 + Hg2+ → CH3Hg+ + H+); it crosses the blood-brain barrier and biomagnifies in aquatic food chains.
Metal Ions: Hydrolysis, Redox & pE-pH Diagrams FAQ
What is pE and how does it relate to E°?
pE = -log[e-] is to electrons what pH is to protons: a measure of how oxidising or reducing a solution is. It links directly to electrode potential through pE° = E°/0.0592 at 25 °C, so a couple with a high standard potential has a high pE° and describes oxidising conditions. High pE means electrons are scarce (oxidising); low pE means electrons are abundant (reducing).
How do I read a pE-pH (Pourbaix) diagram?
Each region is the species that is thermodynamically stable at that combination of redox intensity (pE, vertical axis) and acidity (pH, horizontal axis). Boundaries come from half-reactions: a horizontal line is a redox couple with no H+ involved, a vertical line is an acid-base or solubility boundary with no electrons, and a sloped line pE = pE° - (m/n)pH involves both. Locate your water's pE and pH, and the region you land in tells you the dominant form (e.g. Fe2+, Fe3+ or solid Fe(OH)3).
Why do amphoteric metal hydroxides redissolve at high pH?
Because they can react as both bases and acids. In acid, the hydroxide dissolves by taking up protons to give the free metal ion; in strong base, it dissolves by losing protons to form a soluble hydroxo-anion. Between these extremes it is least soluble and precipitates. This U-shaped solubility is why metals like aluminium and iron are mobilised at both low and very high pH but immobilised near neutral.
Can AI help me with the metal-ion and pE-pH material in CHEM3120?
Yes. Sia can convert an E° to pE with you, derive a Pourbaix line from a balanced half-reaction step by step, and talk through metal hydrolysis, transport and toxicity examples like methylmercury and lead chelation. It explains the method and checks your reasoning; it does not do graded assessment for you, and University of Sydney academic-integrity rules apply.
Exam move
This is the most calculation-heavy Block 3 chapter, so build it in layers. First lock the conversion pE° = E°/0.0592 and the equal-concentration anchor (pE = pE° when the redox partners are equal). Then practise deriving a Pourbaix line pE = pE° - (m/n)pH by balancing a half-reaction for both H+ and e- and reading off the slope; the iron system (Fe2+/Fe3+/Fe(OH)3/Fe(s)) is the template to rehearse. Layer on the qualitative pieces — stepwise hydrolysis and amphoteric solubility, Stokes-law transport, and the mercury and lead toxicity chemistry. Because these methods are supplied on the formula sheet but demand practised execution, drill them for speed under closed-book conditions and keep them warm across the semester. Confirm the exam date, room and permitted materials on Canvas and the exam timetable.
Working through Metal Ions: Hydrolysis, Redox & pE-pH Diagrams in CHEM3120? Sia is AskSia’s AI Chemistry tutor — ask any CHEM3120 Metal Ions: Hydrolysis, Redox & pE-pH Diagrams question and get a clear, step-by-step explanation grounded in how CHEM3120 is taught and assessed. Read this chapter free, then take your hardest questions to Sia.
