Calculating Nonadjacent Potentials from Latimer Diagrams

Weighted-average skip-step potentials are the computational backbone of every quantitative disproportionation and comproportionation argument in descriptive inorganic chemistry — deciding, for example, whether Cu⁺ disproportionates to Cu²⁺ and Cu(s) in water (it does, famously, because the Cu²⁺/Cu⁺ couple sits below the Cu⁺/Cu couple), or whether chlorine oxyanions like ClO⁻ are kinetically persistent despite being thermodynamically unstable toward disproportionation to ClO₃⁻ and Cl⁻. Analytical chemists use the same machinery to compute the effective potential of multi-electron titration end-points (permanganate titrations formally involve the 5-electron MnO₄⁻/Mn²⁺ couple computed exactly as in Example 3), and the identical formalism underlies constructing accurate Frost Diagrams, where the vertical axis plots $nE°$ (free energy in disguise) precisely so that straight-line segments between states can be read off, extended, and averaged using ordinary Cartesian slope arithmetic rather than potential arithmetic. In biochemistry, the same weighted-summation logic underlies multi-electron cofactor potentials — for instance relating the one-electron semiquinone couples of ubiquinone to the overall two-electron quinone/hydroquinone potential that drives mitochondrial electron transport, a calculation formally identical to Example 1. In environmental and geochemistry, skip-step potentials for iron, manganese, and sulfur species (e.g., relating Fe³⁺/Fe²⁺ and Fe²⁺/Fe(s) potentials to a hypothetical direct Fe³⁺/Fe(s) three-electron couple) feed directly into Eh–pH (Pourbaix) diagram construction, where stability-field boundaries are literally the loci of skip-step potentials as a function of pH. Battery and materials chemists performing multi-electron intercalation analyses (lithium- and sodium-ion cathode materials cycling through several formal oxidation states of a transition metal) use exactly this weighted free-energy summation to relate measured average discharge voltages to the individual redox couples resolved by operando spectroscopy.

Ask the tutor: In the acidic manganese Latimer diagram, use the weighted-average method to find E° for the direct MnO₄²⁻ → Mn²⁺ couple (skipping MnO₂ and Mn³⁺), and check whether MnO₂ is stable with respect to disproportionation into MnO₄²⁻ and Mn³⁺.

A Latimer diagram is a chain of oxidation states of one element, connected by arrows, each arrow labeled with the standard reduction potential $E°$ for that specific one-step (or few-step) half-reaction. For manganese in acid solution, a typical span looks like:

Every number on that chain is an $E°$ for a specific electron count: Mn(VII)→Mn(VI) is a one-electron step, Mn(VI)→Mn(IV) is a two-electron step, Mn(IV)→Mn(III) is a one-electron step, Mn(III)→Mn(II) is a one-electron step. The diagram never tells you, for instance, the potential for Mn(VI) going straight to Mn(III) (a three-electron span crossing the Mn(IV) node) or for Mn(VII) going all the way to Mn(II) (a five-electron span). Those numbers are not printed because they are not independent thermodynamic data — they are derivable from what's already there, provided you route the calculation through Gibbs energy rather than through potential. The physical reason potentials don't add is dimensional: $E°$ is an intensive quantity (a per-electron, per-mole free-energy density in some sense), while $ΔG°$ is extensive. Two reactions in series only add cleanly in the extensive variable. If reaction 1 releases $ΔG°_1$ for $n_1$ electrons and reaction 2 releases $ΔG°_2$ for $n_2$ electrons, gluing them into one overall reaction that moves $n_1+n_2$ electrons releases $ΔG°_1+ΔG°_2$ total — but the potential of that combined reaction is this total energy divided by the total charge moved, not the sum or the plain average of the two original potentials.

Notice the boxed comparison in the figure: the naive average of 2.26 V and 0.95 V is 1.605 V, but the correct electron-weighted value is 1.823 V — a 218 mV discrepancy, easily enough to change a stability conclusion. The weighting matters because the two-electron step contributes twice as much free energy per volt as the one-electron step; ignoring that is exactly the error this article exists to prevent.

The master relation converting a reduction potential into a free energy is

where: $n$ is the number of electrons transferred in that specific half-reaction, $F$ is the Faraday constant ($96{,}485\ \text{C mol}^{-1}$), and $E°$ is the standard reduction potential of that step in volts. Reading: this is simply the definition of electrical work done reversibly by $n$ moles of electrons falling through a potential difference $E°$; it is the bridge between the extensive quantity ($ΔG°$) and the intensive one ($E°$). For a path made of $m$ consecutive adjacent steps connecting a starting oxidation state to an ending oxidation state, additivity of state functions gives

where: $n_i$ and $E°_i$ are the electron count and potential of the $i$-th adjacent arrow along the path, and the sum runs over every arrow between the two states of interest (never skipping any — every intermediate node's arrow must be included exactly once). Reading: because $ΔG°$ is a state function, the overall free-energy drop from start to end is path-independent, so summing consecutive steps down the diagram gives the true, single overall $ΔG°$. Converting that total back into a single potential for the combined (nonadjacent) couple requires dividing by the total number of electrons transferred over the whole span, $n_{\text{total}} = \sum_i n_i$:

where: $E°_{\text{skip}}$ is the potential for the direct, one-step (in electron-bookkeeping terms, $n_{\text{total}}$-electron) half-reaction from the initial to the final oxidation state, and the sums are over the same intermediate arrows as before. Reading: this is a weighted average of the component potentials, with weights equal to each step's electron count — heavier (more-electron) steps pull the answer more strongly toward themselves, exactly as a two-electron step should outweigh a one-electron step when both feed the same overall couple. A companion identity, useful as a sanity check, restates the whole thing purely in terms of $ΔG°$ without ever introducing $F$ explicitly if you keep $nE°$ as a single "weighted potential unit":

where: the left side is the "$nE°$ product" for the combined couple and the right side is the sum of the $nE°$ products of every intervening step. Reading: this is the form you actually compute with on paper — never divide until the very last line, and never touch $F$ numerically at all since it cancels.

Example 1 — Foundational: a clean two-step skip with unequal $n$. Consider a (constructed, pedagogically clean) three-state Latimer fragment:

Find $E°$ for the direct $X^{5+}\rightarrow X^{1+}$ couple (a 4-electron skip). Apply the weighted-average formula directly:

Check against the naive (wrong) unweighted average: $(1.10+0.42)/2 = 0.760\ \text{V}$ — off by 170 mV, because the three-electron step (which dominates the free-energy budget) was under-weighted. As a limiting-case sanity check, if instead $n_2$ had also been $1$ (equal electron counts), the formula would correctly reduce to the plain average $(1.10+0.42)/2 = 0.760$ V, confirming the formula behaves as expected in that special case even though it is not the right answer for the actual $n_2=3$ problem.

Example 2 — Applied: nitrogen chemistry, a three-electron skip crossing two intermediate nodes. Nitrogen's Latimer diagram in acid solution includes the fragment

Find the potential for the direct $\text{NO}_3^-\rightarrow\text{N}_2\text{O}$ couple, a 4-electron overall process ($n_{\text{total}} = 2+1+1 = 4$). Sum the $nE°$ products:

Divide by the total electron count:

The unweighted average of the three printed potentials would be $(0.94+1.00+1.59)/3 = 1.177$ V — a 55–60 mV error, smaller here because the electron counts are less lopsided than in Example 1, but still wrong in principle and non-negligible for a quantity that decides whether disproportionation is thermodynamically favorable at the tens-of-millivolt level. This example illustrates that the size of the weighting error scales with how unequal the $n_i$ are, not with the number of steps.

Example 3 — Exam/competition level: testing manganese(III) for disproportionation using a 5-electron skip. Using the acidic manganese Latimer diagram given at the start of this article,

Part (a): find $E°$ for the direct MnO₄⁻ → Mn²⁺ couple (a 5-electron overall reduction). Total electrons: $n_{\text{total}} = 1+2+1+1 = 5$. Sum of $nE°$ products:

This matches the well-known experimental value for the permanganate/Mn²⁺ couple in acid ($\approx +1.51$ V), a good consistency check on the whole calculation and on the individual adjacent potentials fed into it. Part (b): is Mn³⁺ thermodynamically stable with respect to disproportionation into MnO₂ (Mn⁴⁺-containing solid) and Mn²⁺ in acid? This requires comparing the couple feeding into Mn³⁺ (the reduction of MnO₂ to Mn³�⁺, $E°=+0.95$ V, one electron) against the couple leaving Mn³⁺ (the reduction of Mn³⁺ to Mn²⁺, $E°=+1.51$ V, one electron). Disproportionation of Mn³⁺ means:

built from one Mn³⁺ being oxidized back to MnO₂ (reverse of the $+0.95$ V couple, so oxidation potential $-0.95$ V) and one Mn³⁺ being reduced to Mn²⁺ (the $+1.51$ V couple, reduction). The cell potential for this comproportionation/disproportionation combination is

Since $E°_{\text{cell}}>0$, disproportionation of Mn³⁺ is thermodynamically favorable ($ΔG°<0$) — this is the standard "is the middle species higher or lower on a Frost plot than the line joining its neighbors" test, and it is exactly the comparison that motivates building Frost Diagrams: a species disproportionates whenever the potential of the couple below it exceeds the potential of the couple above it, which is visually the statement that the point sits above the chord joining its neighbors on a Frost plot. The nonadjacent-potential machinery of this article is what supplies the numbers for that comparison whenever the two flanking couples aren't already adjacent single steps in the printed diagram.

The name permanently attached to this diagrammatic style is Wendell Mitchell Latimer, an American physical chemist at the University of California, Berkeley, whose 1938 book The Oxidation States of the Elements and Their Potentials in Aqueous Solutions first systematized the compact chain-of-arrows notation now used worldwide. Latimer's project was fundamentally an act of data compression: rather than tabulating every conceivable pairwise couple for an element with seven or eight oxidation states — a combinatorial explosion — he recognized that only the adjacent one-electron (or minimal-electron) potentials needed to be tabulated, because every other couple was thermodynamically recoverable from them via exactly the free-energy additivity argument developed here. In that sense, the weighted-average method is not an add-on to the Latimer diagram; it is the reason the diagram is allowed to be so short in the first place. Latimer also contributed foundational work on entropies of aqueous ions and on hydrogen bonding (the Latimer–Rodebush hydrogen-bonding hypothesis, developed with Worth Rodebush in 1920, is one of the earliest explicit statements of hydrogen bonding as a distinct interaction), but it is the oxidation-potential compendium that bears his name in every modern inorganic chemistry course.