CHEM20018 · Chemistry: Reactions And Synthesis
Inorganic Redox: Ellingham, Latimer, Frost & Pourbaix Diagrams
This section is about reading four redox diagrams rather than memorising tables of potentials. An Ellingham diagram plots ΔG° vs T for metal-oxide formation (slope = −ΔS°) and tells you which metal can reduce another's oxide — the basis of smelting. A Latimer diagram strings species in order of oxidation state with E° on each arrow, and lets you combine couples by an electron-weighted average. A Frost diagram plots nE° (volt-equivalents) against oxidation number so that the steepest descent is the strongest oxidant and a convex 'bump' marks a species that disproportionates. A Pourbaix diagram maps stability against E and pH, including the water stability window.
What this chapter covers
- 01Ellingham diagram: ΔG° vs T for oxide formation; slope = −ΔS°; a lower line reduces an oxide above it
- 02Carbon line crossing metal-oxide lines as the thermodynamic basis of smelting
- 03Latimer diagram: species ordered by oxidation state with E° on each step
- 04Combining Latimer steps: E°overall = Σ(niE°i) ÷ Σni (electron-weighted average)
- 05Disproportionation vs comproportionation and the E°(right) > E°(left) test
- 06Frost diagram: nE° vs oxidation number; steepest descent = strongest oxidant; convex point disproportionates, concave is stable
- 07Pourbaix diagram: E vs pH stability map; diagonal Nernst lines (slope −0.0592 × H⁺/e⁻ ratio)
- 08Water stability window between the H₂ and O₂ lines; effect of complexation on E°
Latimer diagram: combine couples and test for disproportionation
- 1 mark — recognise you weight by electron count, not arithmetic mean(a) Combine the two steps by an electron-weighted average. The M³⁺ → M change is 3 electrons total: 1 e⁻ at +1.50 V and 2 e⁻ at +0.40 V.
- 2 marks — correct weighted average and valueE°(M³⁺/M) = [ (1)(+1.50) + (2)(+0.40) ] ÷ (1 + 2) = (1.50 + 0.80) ÷ 3 = 2.30 ÷ 3 = +0.77 V.
- 1 mark — set up the correct cell from the two flanking potentials(b) Disproportionation of M²⁺ is 3 M²⁺ → 2 M³⁺ + M (M²⁺ is both oxidised to M³⁺ and reduced to M). Build the cell from the two flanking couples: cathode = reduction M²⁺/M (+0.40 V), anode = oxidation M²⁺ → M³⁺ (reverse of M³⁺/M²⁺, +1.50 V).
- 2 marks — negative E°cell + correct conclusionE°cell = E°(reduction, M²⁺/M) − E°(M³⁺/M²⁺) = 0.40 − 1.50 = −1.10 V < 0, so disproportionation is non-spontaneous. Equivalently, E°(left, M³⁺/M²⁺ = +1.50) > E°(right, M²⁺/M = +0.40), the condition that prevents disproportionation; the reverse (comproportionation) is favoured.
Key terms
- Ellingham diagram
- A plot of ΔG° against temperature for the formation of metal oxides; the slope equals −ΔS° and a metal whose line lies below another's can reduce that metal's oxide.
- Latimer diagram
- A linear scheme listing a species' oxidation states in order with the standard reduction potential written above each connecting arrow.
- Frost diagram
- A plot of nE° (volt-equivalents) against oxidation number; steepest downward slope is the strongest oxidising couple, a convex point disproportionates, a concave point is stable.
- Disproportionation
- A reaction in which one species is simultaneously oxidised and reduced to higher and lower oxidation states; favoured when E° to its right exceeds E° to its left.
- Pourbaix diagram
- A map of the thermodynamically stable species as a function of electrode potential E and pH, including the stability window of water.
Inorganic Redox: Ellingham, Latimer, Frost & Pourbaix Diagrams FAQ
How do I combine two Latimer potentials into one?
Use an electron-weighted average: E°overall = Σ(niE°i) ÷ Σni, where each ni is the number of electrons in that step. You cannot just average the two voltages unless both steps involve the same number of electrons.
What is the quickest way to spot disproportionation?
On a Latimer diagram, a middle species disproportionates when the potential to its right is more positive than the potential to its left. On a Frost diagram the same species sits on a convex 'bump' above the line joining its neighbours. Either picture gives the answer without a full calculation.
What does the water stability window on a Pourbaix diagram mean?
It is the band of E and pH between the line for H⁺/H₂ reduction and the line for O₂/H₂O oxidation. A species whose stability region falls inside this band is stable in water; one outside it will oxidise or reduce water (e.g. a strong enough oxidant above the O₂ line liberates O₂).
Exam move
This topic is graphical literacy, so practise interpreting diagrams, not memorising numbers. For Ellingham, learn that slope = −ΔS° and that a lower line reduces an oxide above it (with carbon's crossing line explaining smelting). For Latimer, drill the electron-weighted average and the right-greater-than-left disproportionation test. For Frost, be able to convert a Latimer set into a Frost plot and read slope = E°, convex = disproportionation, concave = stable. For Pourbaix, know horizontal (pH-independent), vertical (redox-independent) and diagonal (both, Nernst slope) lines and the water window. Work several past redox questions purely by reading the picture and stating the conclusion in one or two lines — that is exactly how the 30-mark short-answer section is graded.
