Latimer Diagram
A Latimer diagram of a chemical element is a summary of the standard electrode
potential data of that element. This type of diagram is named after Wendell Mitchell Latimer,
an American chemist.
Construction
In a Latimer diagram, the most highly oxidized form of the element is on the left, with
successively lower oxidation states to the right. The species are connected by arrows, and the
numerical value of the standard potential (in volts) for the reduction is written at each arrow.
For example, for oxygen, the species would be in the order O2 (0), H2O2 (–1), H2O (-2))
The arrow between O2and H2O2 has a value +0.68 V over it, it indicates that the
standard electrode potential for the reaction:
O2(g) + 2 H+ + 2 e− ⇄ H2O2(aq)
is 0.68 volts.
Applications
Latimer diagrams can be used in the construction of Frost diagrams, as a concise
summary of the standard electrode potentials relative to the element. Since ΔrGo = -nFEo, the
electrode potential is a representation of the Gibbs energy change for the given reduction.
The sum of the Gibbs energy changes for subsequent reductions (e.g. from O2 to H2O2, then
from H2O2 to H2O) is the same as the Gibbs energy change for the overall reduction (i.e.
from O2 to H2O), in accordance with Hess's law. This can be used to find the electrode
potential for non-adjacent steps, which gives all the information necessary for the Frost
diagram.
A simple examination of a Latimer diagram can also indicate if a species will
disproportionate in solution under the conditions for which the electrode potentials are given:
if the potential to the right of the species is higher than the potential on the left, it will
disproportionate. Therefore, hydrogen peroxide is unstable and will disproportionate (see
diagram above).
Latimer Diagrams
written with the most oxidized species on the left, and the most reduced species on the
right We can determine whether a species in an intermediate oxidation state will
disproportionate, ie, convert to the more oxidized species to its right and the more reduced
species to its left, by comparing the redox potential for its formation vs. the redox potential
for its reduction.
We can determine whether a species in an intermediate oxidation state will
disproportionate, ie, convert to the more oxidized species to its left and the more reduced
species to its right, by comparing the redox potential for its formation vs. the redox potential
for its reduction.
Latimer Diagrams
Because many elements can have more than one oxidation state, Latimer devised a
simplified way of summarizing the potentials involved in the chemical interrelations among them.
These so-called Latimer diagrams (or linear potential diagrams) are drawn by sequentially writing
each chemical species in its different oxidation states in the form of a reduction progression. The
standard potential for each reaction is written on the uniting line. The potentials for any equilibria
among non-sequential oxidation states are represented with a uniting segment on which the
corresponding potential is shown. The steps involved in the hydrogen peroxide decomposition
(Example 2.11) can thus be drawn as follows:
The equilibrium between the non-sequential oxidation states (i.e., 0 and −2 in O2 and H2O,
respectively) cannot be written on the same line as those already drawn, and another line is needed
to depict it. In order to calculate the potential involved in this reaction, its stoichiometry must be
found first. This can be done by combining the O2 → H2O2 and H2O2 → H2O equations discussed in
the Example 2.11 in such a way that the peroxide is eliminated from the overall reaction; the oxygen
appears as a reactant, and water as the product:
O2(g) → H2O(l)
Redox Processes
The balanced equation is
O2(g) + 4H+ + 4e− → 2H2O(l)
Using the same rationale as in the previous example for calculating the G0 value for this reaction,
and adding the O2 → H2O2 and H2O2 → H2O equations, one obtains G0 3 = G0 1 + G0 2 = −2F(0.68)
+ (−2F × 1.78) (2.62a) = −2F(2.46) = −4F E0 3 From here, the E0 value is calculated: E0 3 = 2.46 2 =
1.23 V (2.62b) This value is written on the line that unites these two non-sequential oxidation states,
and the Latimer diagram for oxygen is completed:
