The whole basis of Frost diagrams is to have a graphical representation of the Gibbs free energy of formation ($\Delta{G}_{f}$) of different oxidation states of an element using the proportional relation between Gibbs free energy and standard reduction potential:

$\Delta$$G$ = $-nFE^{o}$

From which we can derive:

$\frac{{-}\Delta{G}}{F}$ = $nE^{o}$

Thus, if we graph $nE^{o}$ against the oxidation number (N), we can accurately describe the relationship between oxidation state and ($\Delta{G}_{f}$).

I encountered a problem in Shriver's Inorganic Chemistry that asks to draw the Frost diagram of oxygen from its Latimer diagram.

Construct a Frost diagram for oxygen from the Latimer diagram below.

enter image description here

I initially though that we would use the standard reduction equations below, which give $nE^{o}$ as $-1.36V$ ($-2$ times $0.68V$) for $\ce{H2O2}$ and $-4.92V$ ($-4$ times $1.23V$) for $\ce{H2O}$.

$\ce{O2(g) + 2H+ (aq) + 2e- -> H2O2 (aq)}$ $E^{o}$=$+0.68V$

$\ce{O2(g) + 4H+(aq) + 4e- -> 2H2O(l)}$ $E^{o}$=$+1.23V$

However, the book seems to be using the change in the oxidation number of oxygen to calculate $nE^{o}$ for $\ce{H2O2}$ and $\ce{H2O}$.

enter image description here

Online images of Frost diagrams corroborate this calculation.

enter image description here

Why is the change in oxidation state used as the standard to calculate $nE^{o}$ instead of the number of electrons transferred given that $n$ in $nE^{o}$ is the number of electrons transferred?


Shriver, D. F., Weller, M. T., Overton, T., Rourke, J., & Armstrong, F. A. (2014). Inorganic chemistry 6th Edition.

  • $\begingroup$ You have just multiplied the given equation by 2. In your case $\Delta G^0$ will change and yes it is multiplied by no. of electrons transferred only. Think properly. $\endgroup$ – Soumik Das Dec 9 '18 at 17:29
  • $\begingroup$ I've made what I'm asking more clear with a specific example. $\endgroup$ – Ethiopius Dec 9 '18 at 18:37
  • 1
    $\begingroup$ I have to agree with you, it's confusing. To calculate $\Delta G^0$ you'd do as you suggest. Presumably the point here is to provide a measure of the potential on a per atom basis. It seems a matter of convention. In real life, reporting how you calculated a result is often enough, if a general convention is not evident. $\endgroup$ – Try Hard Dec 13 '18 at 23:00
  • $\begingroup$ The reaction used to calculate the nE values is the formation of the zero oxidations state (designates as nE=0) from the oxidation state in question, correct? If so, then measuring the potential on a per atom basis would be flawed and would not provide a plot that reflect accurately upon the formation of the oxidation state from the zero oxidation state (opposite of reaction used to calculate nE)? $\endgroup$ – Ethiopius Dec 14 '18 at 21:46

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