How do I estimate ΔG⁰f for Fe2+?

How would one use the equation ΔG° = −nFE° to estimate the ΔG⁰f of Fe2+(aq)? Since Fe2+ appears in multiple entries in the following table, I'm not sure which value to use. 


[![data](https://i.stack.imgur.com/eTBup.png)](https://i.stack.imgur.com/eTBup.png)




This is not possible. The potentials are given only as references to the standard reduction potential for hydrogen. As such, they are not absolute.


You can convert between potential and free energy for a full redox reaction.


Actually, your statement has a deeper conceptual flaw. The standard formation free energy of the Fe(II) ion does not make any sense. The standard free energy of formation requires making something from its constituent elements in their standard states, but you cannot make an ion from its constituent neutral atoms. You'll need to account for the energy of the electrons and their binding energy to the ion. Therefore, the cation alone cannot have a standard free energy of formation.


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The Gibbs free energy of formation of $\ce{Fe^2+ (aq)}$ would be given by the equation


$$\ce{Fe (s) -> Fe^2+ (aq) + 2e-}$$


since iron in its reference state is a solid. Therefore, this should already point you towards the correct half-equation to use.


It is true that the electrode potentials given are relative, i.e. it is not possible to measure an absolute Gibbs energy of formation, since it is not possible to prepare a solution of $\ce{Fe^2+}$ without any corresponding counterions. Therefore, for the half-reaction


$$\ce{Fe^2+ (aq) + 2e- -> Fe (s)} \qquad E^\circ = -0.44~\mathrm{V}$$


the value of $\Delta G^\circ = -nFE^\circ$ ($n = 2$) actually refers to the value of $\Delta G^\circ$ for the reaction


$$\ce{Fe^2+ (aq) + H2 (g) -> Fe (s) + 2H+ (aq)} \qquad \Delta\_\mathrm{r} G^\circ = +84.9~\mathrm{kJ~mol^{-1}}$$


One way of proceeding is to note that the Gibbs energy of reaction can be expressed in terms of the Gibbs energies of formation of all the constituent species, in an exactly analogous manner to enthalpy.


$$\Delta\_\mathrm{r}G^\circ = 2G\_\mathrm{f}^\circ(\ce{H+(aq)}) + G\_\mathrm{f}^\circ(\ce{Fe (s)}) - G\_\mathrm{f}^\circ(\ce{H2(g)}) - G\_\mathrm{f}^\circ(\ce{Fe^2+(aq)})$$


Since the Gibbs energies of formation of $\ce{Fe(s)}$ and $\ce{H2(g)}$ are zero this leaves us with


$$84.9~\mathrm{kJ~mol^{-1}} = 2G\_\mathrm{f}^\circ(\ce{H+(aq)}) - G\_\mathrm{f}^\circ(\ce{Fe^2+(aq)})$$


In order to assign a Gibbs energy of formation to ions, it is conventional to set


$$G\_\mathrm{f}^\circ(\ce{H+(aq)}) = 0$$


Atkins' *Physical Chemistry*, 10th ed. (p 136) writes



> 
> In essence, this definition adjusts the actual values of the Gibbs energies of formation of ions by a fixed amount, which is chosen so that the standard value for one of them, $\ce{H+(aq)}$, has the value zero.
> 
> 
> 


**Under this convention**, $G\_\mathrm{f}^\circ(\ce{Fe^2+(aq)})$ would be calculated to be $-84.9~\mathrm{kJ~mol^{-1}}$. Note that by virtue of setting $G\_\mathrm{f}^\circ(\ce{H+(aq)}) = 0$, this is still only a **relative** value.


(Disclaimer: In Table 2C.5 of Atkins (p 976), $G\_\mathrm{f}^\circ(\ce{Fe^2+(aq)})$ is given as $-78.9~\mathrm{kJ~mol^{-1}}$. I am still not entirely sure where this discrepancy arises from.)


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