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Iron oxidation states

CHRIS25 - 3-12-2014 at 06:05

Don't know why but I am always having problems with Iron when it comes to working out oxidation states

These first two I hope I have understood correctly:
(CN has a charge of -1)
Potassium Ferricyanide:
K₃[Fe(CN)₆] has an oxidation state of +3, this makes sense because the K has +3 (3x +1) and the Fe-CN complex has an overall 3-

Ferrous Sulphate:
FeSO₄ Fe has an oxidation state of +2 due to the Sulphate cation

But the following:

Prussian Blue:
Fe₃[Fe(CN6)]₂ has an oxidation state of +3 since the Fe is also Fe₃ but the Fe-CN complex has an overall 2- lost here?

Turnballs Blue:
Fe₄[Fe(CN6)]₃ according to what I read has an oxidation state of +2 yet the Fe is in a 4x plus the Fe-CN complex is in 3- (3x 1-) lost here also?

So how do I work out the oxidation state of the Fe in the CN complex and then its overall charge?

[Edited on 3-12-2014 by CHRIS25]

gdflp - 3-12-2014 at 07:26

This is just a very general rule, but if you have a species with one anion and one cation, in this case Fe(CN)6 and Fe, then the number present of each ion is the charge of the opposing ion. This rule doesn't apply if the compound is multivalent, or if the structure dictates that the coefficients be multiplied by a factor (e.g. P4O10 is P5+ and O2- not P10+ and O4-). In Prussian Blue, this dictates that the Fe cation has a charge of 2+ and the anion has a charge of 3-. This gives the Fe in the cyanide complex an overall charge of 3+. In Turnballs Blue a similar case occurs, the iron cation has a charge of 3+ and the ferricyanide ion has a charge of 4-, giving that iron an overall charge of 2+.

Ozone - 3-12-2014 at 07:41

Unless there is an excess of either 3+ or 2+...In this case, the so-called insoluble form results, and Fe is present both at 3+ and 2+ in the complex. This is what Wiki says, anyway.

There is further substantiation in the literature, however, as Hansen, et al. describe it as KFe(III)[Fe(II)(CN)6].

http://pubs.acs.org/doi/abs/10.1021/ed046p46

O3

DraconicAcid - 3-12-2014 at 09:24

Quote: Originally posted by gdflp

This rule doesn't apply if the compound is multivalent, or if the structure dictates that the coefficients be multiplied by a factor (e.g. P4O10 is P5+ and O2- not P10+ and O4-).

Phosphorus oxides are molecular, and do not contain ions.

Just remember that the charges have to add up to zero, and that the anion charges are seldom variable. If you have something like Fe₃O₄, the oxygens are each -2, so you have a total of -8. The irons must add up to +8 to balance them, so you have to have two Fe(III) and one Fe(II).

CHRIS25 - 3-12-2014 at 09:39

So as I understand everyone's very helpful posts here (thanking you all) You can actually have, in the same solution when combined, a mixture of Fe ions co-existing harmoniously in both the 2+ and 3+ oxidation states, if so then this solves my confusion.
@Ozone, I see I have to purchase the article, It's almost christmas, scrooge is on the prowl, pity.

[Edited on 3-12-2014 by CHRIS25]

blogfast25 - 3-12-2014 at 09:59

Quote: Originally posted by CHRIS25

Prussian Blue:
Fe₃[Fe(CN6)]₂ has an oxidation state of +3 since the Fe is also Fe₃ but the Fe-CN complex has an overall 2- lost here?

Turnballs Blue:
Fe₄[Fe(CN6)]₃ according to what I read has an oxidation state of +2 yet the Fe is in a 4x plus the Fe-CN complex is in 3- (3x 1-) lost here also?

Well, you managed to choose an interesting case there, with those mixed oxidation states.

But we don't consider the structures to be Fe₃[Fe(CN6)]₂ and Fe₄[Fe(CN6)]₃ anymore. Further research showed both compounds are one and the same, written as Fe7(CN)18·xH2O. I.e. PB = TB.

[Edited on 3-12-2014 by blogfast25]

DraconicAcid - 3-12-2014 at 10:11

Or, you could have something like Cu[Fe(bipy)(CN)₄] (where bipy is the neutral bipyridine ligand). This could be Cu(I) and Fe(III) or Cu(II) and Fe(II).

CHRIS25 - 3-12-2014 at 10:11

Quick reply because I have to dash - I just looked at the journal of Art chemistry, you are right. Plus a few other links say the same thing, though countless chemistry papers (not blogs or such like) do not even mention this, why?

blogfast25 - 3-12-2014 at 11:25

Quote: Originally posted by CHRIS25

why?

Mainly age of papers, I think.

aga - 3-12-2014 at 12:39

Have a shuftie at Chevreul's Salt.

Copper in two oxidation states.

Ozone - 3-12-2014 at 13:14

Well, you learn something new everyday. Damn, I must be getting old...They used to teach that the iron was present in multiple states.

O3

Edit-I suppose not, I misread Blogfast's post. Oops. This edit is just to avoid confusion for anyone reading the whole thread from the beginning.

[Edited on 3-12-2014 by Ozone]

DraconicAcid - 3-12-2014 at 13:18

Quote: Originally posted by Ozone

Well, you learn something new everyday. Damn, I must be getting old...They used to teach that the iron was present in multiple states.

It is. Fe₇(CN)₁₈·xH2O would have four Fe(II) and three Fe(III).

Ozone - 3-12-2014 at 13:29

Ah, yes. I misread the following posts (busy at work), oops. And, here is some confirmation via Mossbauer spectroscopy (at least up to 1984):

http://deepblue.lib.umich.edu/bitstream/handle/2027.42/24978...

Although, it looks as if there may be some room for speculative nit-picking on data interpretation:

"The asymmetry on the r.h.s. of the more intense peak indicates that the observed envelope is the sum of
two unresolved peaks. An investigation using an 56Fe labeled Prussian Blue39 substantiates this type of fitting, where both an Fe(III) (non-octahedral symmetry) and Fe(II) (octahedral symmetry are present, as do our data for the analogues."

But, the plots looked pretty straight, to me.

O3

CHRIS25 - 3-12-2014 at 14:37

Ok, so this something new to me, copper and iron co-existing in two oxidation states, I had a look at Chevreul's Salt, will be doing that - on my list now, looks interesting. Was this the same Michel Eugène Chevreul born in 1789? He seems to be famous for fatty acids and many other things, but no mention about his 'salt'?

Tests to produce precise amounts of Ferrous and Ferric

CHRIS25 - 6-12-2014 at 05:35

Object here is to precipitate known ratios of Fe₂⁺ and Fe₃⁺

0.026m FeSO₄ + 0.006m K₃[Fe(CN)6] gave 11.1 grams Fe₇(CN)18.14-16H₂O precipitate

The stoichiometric ratios of reactants are 4 Ferrous ions to 1 Ferric ion, producing within the Fe₇ part of the precipitate a 2Fe₂⁺ and 1 Fe₃⁺ -- is this logic correct? so that now I have a good understanding from which to experiment further.

blogfast25 - 6-12-2014 at 06:56

Have a look at that paper Ozone linked too.

In the older notation PB is written as Fe4[Fe(CN)6]3, which implies 4 Fe3+ + 3 Fe2+ which is +18 charges (not "four Fe(II) and three Fe(III)" as Draconic wrote: that only gives 17+ charges and cannot accommodate 18 CN-

4 x (3+) + 3 x (2+) = 18+

Chris, mixing exactly stoichiometric amounts isn't really all that necessary here. If there was a little FeSO4 excess it would not react and simply stay in solution. Similarly if there was a bit of K3Fe(CN)6 in excess. You need to filter and wash there precipitates anyway, so any excess reagent will be washed out.

[Edited on 6-12-2014 by blogfast25]

blogfast25 - 6-12-2014 at 07:03

Another common mixed oxidation state iron compound is Magnetite:

Fe3O4 = Fe2O3.FeO

2 Fe(III) for each Fe(II).

If you precipitate from a mixed Fe(III),Fe(II) solution with sodium hydroxide (or ammonia solution), you obtain a black precipitate, i.e. Magnetite, which as the name implies responds to magnets (see also 'ferrofluid' magnetic liquids).

[Edited on 6-12-2014 by blogfast25]

CHRIS25 - 6-12-2014 at 08:52

Quote: Originally posted by blogfast25

Have a look at that paper Ozone linked too.

In the older notation PB is written as Fe4[Fe(CN)6]3, which implies 4 Fe3+ + 3 Fe2+

[Edited on 6-12-2014 by blogfast25]

Sorry as far as Ozone's paper is concerned, First I have to weed out the information that is a total foreign language to me and then I am left with nothing to read that makes any sense anyway, I will never ever reach that point in chemistry.

Secondly this is getting annoying why the hell are there two different methods/formulas for explaining the same formula for Potassium Hexocyanoferrate? They don't even appear to be conveying the same data. Sorry. The Fe₄ "version" clearly tells me that there are 4 ions bonded to each Fe(CN) complex with the charges as you explained, 4 x 3+ and 3 x 2+ balancing out the -18 charge on that Fe₄. BUT, The Fe₇ "version" blows my mind away. this means 7 Fe ions attached to the Fe(CN) complex and so to equal -18 we now have what? I an=know I am an awkward learner in chemistry, sorry, but I am trying so hard to get to grips with this.

[Edited on 6-12-2014 by CHRIS25]

blogfast25 - 6-12-2014 at 09:39

Quote: Originally posted by CHRIS25

Secondly this is getting annoying why the hell are there two different methods/formulas for explaining the same formula for Potassium Hexocyanoferrate?

I'm not sure whether you realise this but there are indeed 2 versions of 'Potassium Hexacyanoferrate'. namely Potassium Hexacyanoferrate(II) and Potassium Hexacyanoferrate(III).

In the (II) version the iron is in oxidation state (II) and the anion is Fe(CN)64- and it requires 4 K ions for neutrality, so its formula is: K4Fe(CN)6. It is yellow in colour.

In the (III) version the iron is in oxidation state (III) and the anion is Fe(CN)63- and it requires 3 K ions for neutrality, so its formula is: K3Fe(CN)6. It is red in colour.

Prussian Blue or Turnbull's Blue can be prepared in two ways:

1. add a ferrous salt solution to a solution of Potassium Hexacyanoferrate(III)

2. add a ferric salt solution to a solution of Potassium Hexacyanoferrate(II)

Both create deeply blue precipitates which for a long time were considered similar yet separate compounds. Further research showed them to be identical and of the formula Fe7(CN)18.xH2O, with an Fe(II)/Fe(III) ratio as explained above.

I accept that it is confusing, additionally because as the paper explains:

"In the preparation of the compound which has
been called “Turnbull’s blue” but which is actually
Prussian Blue, it is clear that linkage isomerism or
oxidation-reduction has occurred."

So initially the two precipitates are not identical but one of them converts immediately to the same chemical structure as the other, because of this internal oxidation-reduction. Both precipitates thus end up being the same compound, despite different preparation methods.

Not easy.

[Edited on 6-12-2014 by blogfast25]

bismuthate - 6-12-2014 at 10:56

The iron in the ferrocyanide ion has an oxidation state different from the iron ion is what I always remember.
My question is why ferrate/perferrate are used interchangeably.
And ferrite at that. Geeze we need more studies done on ferrates (IV,V, and VI)

[Edited on 6-12-2014 by bismuthate]

blogfast25 - 6-12-2014 at 11:24

Potassium ferrocyanide and ferricyanide are probably more correct names than hexacyanoferrate. Ferrate should probably be reserved for Fe(VI), like FeO42-.

Squabbling over nomenclature isn't very interesting though.

Stoichiometrically the transition from Turnbull's Blue to Prussian Blue is probably like this:

2 Fe3[Fe(CN)6]2 === > Fe7(CN)18 + 3 Fe(CN)2

That 'ejection' of Fe(II) restores the Fe(III)/Fe(II) balance to Prussian Blue's.

Quote: Originally posted by bismuthate

The iron in the ferrocyanide ion has an oxidation state different from the iron ion is what I always remember.

No, not really. The complex is a fairly straightforward coordination complex of Fe(III). It shows all the characteristics of a strong complex.

[Edited on 6-12-2014 by blogfast25]

DraconicAcid - 12-12-2014 at 18:58

Quote:

Potassium ferrocyanide and ferricyanide are probably more correct names than hexacyanoferrate. Ferrate should probably be reserved for Fe(VI), like FeO42-.

I strongly disagree.

Quote:

Squabbling over nomenclature isn't very interesting though.

Well, okay, that, I can't argue with...

blogfast25 - 13-12-2014 at 06:15

Quote: Originally posted by DraconicAcid

I strongly disagree.

On what grounds?

Wiki lists potassium ferrocyanide, then goes on to state the IUPAC name is "Potassium hexacyanidoferrate(II)". It's the 'do' that makes me laugh...