The standard reduction potential of Fe3+ is 0.77V vs SHE, and the standard reduction potential of [Fe(CN)6]3- is 0.37V vs SHE.
Prussian Blue, or Iron Hexacyanoferrate, is a coordination polymer made of interconnecting Fe cations and [Fe(CN)6] anions. Cations such as K+ or Na+ may fill the interstitial sites, along with zeolitic water. Additionally, due to the strong ligand field (CN)-, the C-coordinated Fe of [Fe(CN)6] groups are in the low-spin state, while N-coordinated Fe remains in the high-spin state.
It has been proven that the redox potential of low-spin [Fe(CN)6] is higher than that of high-spin Fe in Prussian blue.
example: Article Rhombohedral Prussian White as Cathode for Rechargeable Sodi...
If the standard reduction potential of high-spin Fe3+ in water is higher than low-spin [Fe(CN)6]3- in water, why is it the opposite in Prussian Blue? My explanation has always been that the oxidized electron in low-spin [Fe(CN)6] is at lower energy than high-spin Fe, but that also assumes the barycenter is the same. However, this explanation doesn't work for the redox potential of aqueous ions.
Edit: For clarification, this is an example of the intercalation and reduction of both Fe. Redox potential is for aqueous Na+ battery from this source:
https://pubs.rsc.org/en/content/articlehtml/2017/cc/c7cc02516e
Stable complex compound formation plays its significant role in the redox potential. Better saying, if a complex compound to be formed during the redox reaction is stable enough, the redox potential will be alterred.
I think the key here, is aqueous ions. To my knowledge, Prussian blue isn't very soluble, so it wouldn't be a fair comparison.
Based on the article you've attached, the insoluble compounds (Prussian White and Prussian Blue) are used as a catalytic surface, where as In the case of Fe3+ and [Fe(CN)6]3-/4-, these are solution based reactions.
Further to this, one should also consider the thermodynamics of the system. How does Prussian Blue's chemical potential differ to that of the [Fe(CN)6]3- and Fe3+. Remember, Prussian Blue is a mixed-valency compound, so there's an interplay between the Fe2+/Fe3+ couple.
Apologies if I completely misunderstood your question.
Adrian Fortuin , thank you for your reply. I am not sure if catalytic surface is the right word to describe Prussian blue in this paper, but I understand that you are saying it is different than ions in solution.
Yes, Prussian blue is mixed valence, but in its fully oxidized state (Berlin Green) there is only Fe3+, FeIII[FeIII(CN)6]. The low-spin [Fe(CN)6] is always reduced first. I have added some clarification to the discussion description.
I agree with your response about thermodynamics, the chemical potential of the system must change which changes the redox potential. I think Volodymyr Tkach is correct in saying the stable complex formed will affect the redox potential, but specifically why or how does it change. The compound may be more stable than its ions in solution, which lowers its chemical energy (higher voltage) for low-spin [Fe(CN)6], but actually high-spin Fe has a lower redox potential than when it is an ion in solution.
Correct me if I am wrong.
Thinking along the threads further, in thermodynamic aspects: the stronger the bonding/ the more insoluble the oxidized products the higher the potential.
From the aqueous phase to solid phase, the formation of insoluble NaFeIII[FeII(CN)6], FeIII[FeIII(CN)6] and Na2FeII[FeII(CN)6] introduce Gibbs free energy change=-RTln(1/Ksp). After substitution of aqueous ions with the final insoluble products, the change of redox potential equals to RT/nF ln(Ksp of oxidized product/Ksp2 of reduced product), which also implies Ksp(FeIII[FeIII(CN)6])
- Badges
- Science topic
- Similar topics
- Chemistry