CSIR-NET Redox Diagrams — Latimer, Frost and Pourbaix
A single element can exist in half a dozen oxidation states, and the question "which of these survives in acid, and which falls apart?" is answered not by memory but by three diagrams. They contain the same thermodynamic information in three different presentations: Latimer diagrams are compact, Frost diagrams are visual, and Pourbaix diagrams add pH. This article shows how to compute with each one — including the combination rule that most candidates get wrong — using manganese and copper worked in full.
Latimer diagrams: the compact form
A Latimer diagram lists the species of one element in decreasing oxidation state, with the standard reduction potential written over each arrow. For manganese in acidic solution (pH 0) the standard series is:
Each arrow is a couple, and the number of electrons transferred is the difference in oxidation state between the two species it joins.
The combination rule — why you cannot simply average potentials
Standard potentials are not additive, because they are intensive. Gibbs energies are additive, and ΔG° = −nFE°. Adding the ΔG° values and dividing by the total nF gives the rule you must use:
It is an electron-weighted average, not a plain one. A one-electron step and a two-electron step do not count equally.
Worked example 1 — E°(MnO4−/MnO2) in acid.
The path is MnO4− → MnO42− (1 electron, +0.56 V) then MnO42− → MnO2 (2 electrons, +2.27 V). Total 3 electrons.
E° = (1 × 0.56 + 2 × 2.27) ÷ 3 = (0.56 + 4.54) ÷ 3 = 5.10 ÷ 3 = +1.70 V.
A plain average would have given (0.56 + 2.27)/2 = 1.42 V, which is simply wrong. The tabulated value for the permanganate–manganese dioxide couple in acid is +1.70 V, so the weighted rule reproduces reality and the plain average does not.
Worked example 2 — E°(Cu2+/Cu) from the copper Latimer diagram.
Cu2+ —(+0.153)→ Cu+ —(+0.521)→ Cu, both one-electron steps.
E° = (1 × 0.153 + 1 × 0.521) ÷ 2 = 0.674 ÷ 2 = +0.337 V, matching the tabulated +0.34 V for the Cu2+/Cu couple. With equal electron counts the weighted average happens to coincide with the plain one — which is exactly why this example is a poor test and example 1 is a good one.
Predicting disproportionation from a Latimer diagram
Take any intermediate species with a potential on its left (reduction to it) and one on its right (reduction of it).
The cell emf for the disproportionation is E°(right) − E°(left).
The logic is that the species is a better oxidant than it is a reductant, so two of it can react together, one going up and one going down.
Worked example 3 — is Cu+ stable in water?
Left of Cu+: E°(Cu2+/Cu+) = +0.153 V.
Right of Cu+: E°(Cu+/Cu) = +0.521 V.
Since 0.521 > 0.153, Cu+ disproportionates:
2 Cu+ → Cu2+ + Cu.
E°cell = 0.521 − 0.153 = +0.368 V, with n = 1.
ΔG° = −nFE° = −(1)(96 485)(0.368) = −35 506 J mol−1 =
−35.5 kJ mol−1.
Equilibrium constant, by the exponential route:
ln K = nFE°/RT = 35 506 ÷ (8.314 × 298.15) = 35 506 ÷ 2478.8 = 14.324, so
K = e14.324 = 1.66 × 106.
Cross-check by the log shortcut: log K = nE°/0.05916 = 0.368 ÷ 0.05916 = 6.221, giving K = 106.221 = 1.66 × 106. The two routes agree, so the arithmetic is safe. This is why simple Cu(I) salts are unstable in water, while CuCl and [Cu(NH3)2]+ survive — insolubility or complexation removes Cu+ from solution and changes the effective potentials.
Worked example 4 — Mn3+ in acid.
Left: E°(MnO2/Mn3+) = +0.95 V. Right: E°(Mn3+/Mn2+) = +1.51 V. Since 1.51 > 0.95, Mn3+ disproportionates to MnO2 and Mn2+, with E°cell = 1.51 − 0.95 = +0.56 V. This is the reason Mn(III) is not a stable aqueous species in acid unless it is complexed or precipitated.
Frost diagrams: the same data, made visual
A Frost diagram plots N·E° (which is −ΔG°/F for forming that species from the element) on the vertical axis against oxidation state N on the horizontal axis. Build it by cumulative addition: start the element at the origin and add n·E° for each step.
| Species | N | Step used | Cumulative N·E° (V) |
|---|---|---|---|
| Mn | 0 | — | 0 |
| Mn2+ | 2 | 2 × (−1.18) | −2.36 |
| Mn3+ | 3 | +1 × 1.51 | −0.85 |
| MnO2 | 4 | +1 × 0.95 | +0.10 |
| MnO42− | 6 | +2 × 2.27 | +4.64 |
| MnO4− | 7 | +1 × 0.56 | +5.20 |
Four reading rules make the diagram do all the work:
- The slope of the line joining any two points is the E° of that couple. For MnO4−/Mn2+: (5.20 − (−2.36)) ÷ (7 − 2) = 7.56 ÷ 5 = +1.51 V, matching the tabulated value — an independent check that the table is correct.
- The lowest point is the most stable species. Here Mn2+ at −2.36, which is why manganese chemistry in acid ends up at Mn(II).
- A species lying above the line joining its neighbours disproportionates. Mn3+ sits at −0.85; the line from Mn2+ (−2.36) to MnO2 (+0.10) passes through (−2.36 + 0.10)/2 = −1.13 at N = 3. Since −0.85 is above −1.13, Mn3+ disproportionates — the same conclusion as worked example 4, reached a different way.
- A species below that line is stable against disproportionation, and two species that lie on a convex-downward pair will tend to comproportionate.
Worked example 5 — manganate. For MnO42− at N = 6,
the line from MnO2 (4, +0.10) to MnO4− (7, +5.20) sits at
0.10 + (5.20 − 0.10) × (6 − 4)/(7 − 4) = 0.10 + 5.10 × 0.667 = +3.50 at N = 6.
The actual point is +4.64, which is above +3.50, so manganate disproportionates in
acid to permanganate and manganese dioxide — the familiar green-to-purple change on
acidification. The diagram predicted it without any additional data.
Pourbaix diagrams: adding pH
Latimer and Frost diagrams are quoted at a fixed pH (usually 0 for acid, 14 for base). A Pourbaix diagram plots E against pH, so it shows how a species' stability field changes with acidity. Every boundary line comes from the Nernst equation, and its slope is fixed by the proton-to-electron ratio of the half-reaction.
| Half-reaction | m | n | Slope (V per pH unit) | Line type |
|---|---|---|---|---|
| Fe3+ + e− → Fe2+ | 0 | 1 | 0 — horizontal | Electron only |
| MnO2 + 4H+ + 2e− → Mn2+ + 2H2O | 4 | 2 | −0.1183 | Sloping |
| O2 + 4H+ + 4e− → 2H2O | 4 | 4 | −0.0592, intercept +1.229 V | Upper water line |
| 2H+ + 2e− → H2 | 2 | 2 | −0.0592, intercept 0 V | Lower water line |
| Fe(OH)3 + 3H+ → Fe3+ + 3H2O | 3 | 0 | Vertical | Acid–base only, no electrons |
The two water lines are the single most useful feature. A species whose stability field lies above the upper line will oxidise water and release O2; one lying below the lower line will reduce water and release H2. Anything outside the band between them is thermodynamically unstable in aqueous solution, which is why some paper-strong oxidants and reductants cannot be kept in water at all. Note that a Pourbaix diagram is thermodynamic only — a species outside the band may still persist for a long time if the reaction is kinetically slow. Permanganate is the standard example: it is a strong enough oxidant to oxidise water, yet its solutions are usable in the laboratory because that reaction is slow.
Mistakes that cost marks
- Averaging potentials without electron weighting. Always use Σ(niEi)/Σni. This is the single biggest source of wrong answers in the topic.
- Adding potentials for a series. E° values never add. ΔG° values do.
- Reading the disproportionation test backwards. It is E°(right) > E°(left), where "right" means reduction of the species in question.
- Plotting E° instead of N·E° on a Frost diagram. The vertical axis must be the free-energy-like quantity, or the slope rule stops working.
- Mixing acid and base tables. Latimer diagrams differ completely between pH 0 and pH 14; check which one the question supplies before you start.
- Treating a Pourbaix diagram as a kinetic map. It predicts what is thermodynamically possible, not what happens quickly.
- Forgetting that complexation and precipitation change everything. Cu+ is unstable as the free ion but perfectly stable as CuCl or in ammonia.
Where this appears in the paper
| Sub-topic | Typical question form |
|---|---|
| Latimer combination | Compute a non-adjacent couple's E° from a supplied diagram |
| Disproportionation | Which species is unstable in acid or base, and what does it give? |
| ΔG° and K | Convert a computed E° into ΔG° or an equilibrium constant |
| Frost reading | Identify the most stable oxidation state; use slope and convexity |
| Pourbaix | Slope of a boundary from m/n; can the species exist in water? |
| Applications | Why permanganate persists; why Cu(I) needs stabilising |
That is a map of the sub-topics, not a claim about the number of questions or the marks attached. The paper structure for your session is defined by the current official notification, and that is the only source to rely on for it.
Turn a potential into a number you can trust. The Nernst Equation calculator handles the E, E°, n and reaction-quotient relationship at any temperature — useful for checking the non-standard-condition part of these questions once you have combined the standard potentials by hand.
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