# If the ferric ion is more stable than the ferrous ion, then why is it readily reduced to the latter?

The standard reduction potential value for the Ferric ion: \begin{align} \ce{Fe^3+ + e- &<=> Fe^2+} & E^\circ &= 0.77 \end{align}

The value is positive, hence it means that ferric ion is readily reduced to ferrous ion, i.e. ferrous ion must be more stable than ferric ion.

But in a book I have also seen that ferric ion is more stable than ferrous ion as it has stable $d^5$ configuration. I am little bit confused that if ferric is more stable, then why ferric is readily reduced to ferrous. Please explain it in detail.

• – Linear Christmas Feb 5 '17 at 12:25
• The problem with nearly every question about "stability" is that it is unclear what exactly you mean by "stable". This is no exception. For example, do you think the stability of ferric ion aqueous solution would be different if it were under an atomsphere of pure hydrogen, or if it were under an atmosphere of pure oxygen? – Curt F. Feb 5 '17 at 15:48
• The short answer is that ‘stable $\mathrm{d^5}$ configuration’ is little more than a myth. There are a lot more factors determining whether something is more stable than something else. – Jan Feb 5 '17 at 23:04

As already pointed out, the notion of stability needs to be defined much more accurately.

The fact that the standard reduction potential is positive has no direct bearing on the gas-phase stabilities of the two ions. The only thing that the standard reduction potential actually reflects, is that the standard Gibbs free energy of the following reaction

$$\ce{Fe^3+ (aq) + 1/2H2 (g) -> Fe^2+ (aq) + H+ (aq)}$$

is negative.

So, for example, Curt already hinted in the comments that under an atmosphere of oxygen, $\ce{Fe^3+}$ would be more stable. This is simply because the reaction

$$\ce{4Fe^2+(aq) + O2 (g) + 4H+ (aq) -> 4Fe^3+ (aq) + H2O (l)}$$

(or some variation of it) has a negative Gibbs free energy. That is very easy to prove, since we know that $E^\circ(\ce{O2}/\ce{H2O}) = +1.23~\mathrm{V}$ and $E^\circ(\ce{Fe^3+}/\ce{Fe^2+}) = +0.77~\mathrm{V}$.

In addition if you were to carry out the reaction in base, the stability of $\ce{Fe(III)}$ relative to $\ce{Fe(II)}$ will be increased greatly, since any $\ce{Fe^3+}$ will be precipitated as $\ce{Fe(OH)3}$.

Remember that a standard reduction potential is measured against an arbitrarily determined zero. If, instead of the standard hydrogen electrode, we chose to use the $\ce{O2/H2O}$ reaction as the standard, then $E^\circ(\ce{Fe^3+}/\ce{Fe^2+})$ would have a negative value.

Let's also get this myth out of the way. $\ce{Fe^2+}$ does not, under any circumstances, have a $\mathrm{(3d)^5(4s)^1}$ configuration. Period.

Ok, with all of that settled, let's answer the question you were actually trying to ask. Yes, $\ce{Fe^3+}$ is a decently good oxidant. There are plenty of half-reactions that have $E^\circ$ below the $\pu{0.77 V}$ of $\ce{Fe^3+/Fe^2+}$. If you want to use $\ce{H+/H2O}$ as a reference point (which is perfectly fine, but you have to make it very clear that this is a relative stability, not an absolute stability), then yes, $\ce{Fe^3+}$ is a better oxidant than $\ce{H+}$.

The reason behind that has nothing to do with $\mathrm{d}^5$ stability. It is simply because $\ce{Fe^3+}$ has a pretty high charge and would quite like having another electron so that it doesn't have quite so high a charge.

More formally, you could construct a cycle for the half-reaction $\ce{Fe^3+ + e- -> Fe^2+}$, neglecting entropy for the time being.

$$\require{AMScd} \begin{CD} \ce{Fe^3+ (aq) + e-} @>\large\Delta H^\circ>> \ce{Fe^2+ (aq)} \\ @V\large-\Delta_\mathrm{hyd}H(\ce{Fe^3+})VV & @AA\large\Delta_\mathrm{hyd}H(\ce{Fe^2+})A \\ \ce{Fe^3+ (g) + e-} @>>\large -I_3(\ce{Fe})> \ce{Fe^2+ (g)} \\ \end{CD}$$

The quantity $\Delta_\mathrm{hyd}H(\ce{Fe^2+}) - \Delta_\mathrm{hyd}H(\ce{Fe^3+})$ is positive since $\Delta_\mathrm{hyd}H(\ce{Fe^3+})$ is more negative than $\Delta_\mathrm{hyd}H(\ce{Fe^2+})$. So, the hydration enthalpy terms disfavour the reduction of $\ce{Fe^3+}$ to $\ce{Fe^2+}$.

However, the third ionisation energy of Fe is pretty large, and it turns out that in the 3d block, the third ionisation energy is the major factor that controls the reduction potential of $\ce{M^3+}$ to $\ce{M^2+}$. In particular, it is quite large because Fe is the 6th element in the 3d-block; you should know that $Z_\mathrm{eff}$, as well as ionisation energies, steadily increase across the 3d block due to the poor shielding provided by 3d electrons.

Put simply, the large value of $I_3$ makes it difficult to oxidise $\ce{Fe^2+}$ to $\ce{Fe^3+}$.

The $\mathrm{d^5}$ configuration of $\ce{Fe^3+}$ does have an impact. It can explain why $E^\circ(\ce{Fe^3+}/\ce{Fe^2+})$ is relatively small compared to its neighbours: $E^\circ(\ce{Mn^3+}/\ce{Mn^2+})$ and $E^\circ(\ce{Co^3+}/\ce{Co^2+})$ are both much larger than $\pu{0.77 V}$. However, this "half-filled shell stabilisation" (more properly phrased in terms of exchange energy) is not quite a large enough factor to make $E^\circ(\ce{Fe^3+}/\ce{Fe^2+})$ negative.