Why does standard reduction potential get multiplied by the change in oxidation state and not by the number of electrons transferred?

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:

$$\DeltaG$$ = $$-nFE^{o}$$

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. 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 $$\ce{H2O}$$. Online images of Frost diagrams corroborate this calculation. 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?

References

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

• 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. – Soumik Das Dec 9 '18 at 17:29
• I've made what I'm asking more clear with a specific example. – Ethiopius Dec 9 '18 at 18:37
• 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. – Buck Thorn Dec 13 '18 at 23:00
• 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)? – Ethiopius Dec 14 '18 at 21:46