Sciencemadness Discussion Board - Very cool reaction -- explanation???

Very cool reaction -- explanation???

woelen - 17-12-2006 at 14:12

I stumbled across a really cool reaction, which is totally unexpected to me.

I was experimenting with making H4Fe(CN)6 and oxidizing this to H3Fe(CN)6. By accident, trying to reduce one of the compounds I made, using Na2S, I found a reaction sequence, which is really neat. I did some trial and error experimenting and found the following sequence to be reliable, allowing reproduction easily by others:

Dissolve some yellow prussiate of potash, K4Fe(CN)6.3H2O, in water. Make the solution as concentrated as possible. Prepare 1 ml of this liquid.
Take another 1 ml of appr. 40% HNO3 and add this to the 1 ml of the concentrated solution of K4Fe(CN)6.

When this is done, then the solution becomes somewhat green with blue and it becomes turbid. I expect a mix of H3Fe(CN)6 and H4Fe(CN)6 is formed.
Heat the liquid to 70 C or something like that (not critical, it just needs to be hot, but not boiling). The liquid at first seems not to change, but at once, it becomes very dark, almost black. Also some bubbles of gas are produced (NO, HCN, (CN)2 ???). Just to err on the safe side, do this with good ventilation, I don't know what gases are formed. Keep the liquid hot, until formation of gases ceases, or if you are not sure about the end-point of the reaction, just keep it hot for 5 minutes.

After this step let it cool down and take a break. After a few hours, add this 2 ml of liquid to a lot of water (200 ml or so). This makes a dilute pale yellow/brown solution.
Next, neutralize this liquid by adding slight excess of NaOH (I also tried with Na2CO3 and that is suitable also).

And now the surprising thing. If Na2S is added, or H2S is bubbled through the alkaline liquid, then it becomes beautifully deep purple, like a solution of KMnO4. On shaking for a time, the liquid even becomes deep indigo blue/purple, a very beautiful color.
When this liquid is acidified, then the purple/indigo color disappears. This color also is not long-lasting. On shaking for some time with air-contact, it also disappears.

I'm really surprised to see this special reaction with such common reagents. The effect is very striking. I intend to make a web-page on this subject, but I already post it here.

The use of nitric acid as oxidizer is essential. I also tried with H2O2 as oxidizer and then no purple color can be obtained after neutralization and adding Na2S.
I also tried adding Na2S to a solution of K3Fe(CN)6 and that does not give the nice color. It just leads to lightening of the solution, due to reduction of the ferricyanide to ferrocyanide (that also was my initial intent with my experiments).

[Edited on 17-12-06 by woelen]

The_Davster - 17-12-2006 at 14:18

Could excess nitrate be being reduced to nitrite, which could then coordinate to the metal centre making nitrito or nitro complex?

EDIT: I might just be thinking of this because I have been reading some preps for nitrito and nitro cobalt complexes which I intend to try soon...They are very colourfull.

[Edited on 17-12-2006 by The_Davster]

garage chemist - 17-12-2006 at 15:25

The reaction of hexacyanoferrate with nitric acid is the first step of the synthesis of sodium nitroprusside.
Look here: http://www.versuchschemie.de/topic,6061,0,-Synthese+von+Nitr...

Further reading: Brauer (can be downloaded in the library here).

[Edited on 17-12-2006 by garage chemist]

guy - 17-12-2006 at 15:38

Sulfide is able to reduce Fe(III) to an Fe(II) state, but with the stabilizing CN- ligands, I suspect it just forms a very strong charge transfer complex. Shaking it with air might have oxidized the sulfide.

not_important - 17-12-2006 at 20:26

I'm going along with garage chemist - HNO3 makes nitroprusside, which gives a violet complex with alkaline sulfide.

LawnMaster1 - 18-12-2006 at 07:46

Would you mind posting any pictures?

woelen - 18-12-2006 at 13:39

Thanks for the info on the nitroprusside/sulfide complex. I indeed found more information about this and this is quite a nice reaction. I even happened to have information about it myself in one of my old German books, but I had never read that section on nitroprusside, its synthesis and its uses in analytical chemistry.
I could not find, however, information on the chemical structure of this purple/indigo compound. Can this purple compound be isolated? And what is its chemical formula?

Quote:

Would you mind posting any pictures?

I'll make a page about this subject anyway, even now the riddle (at least partially) is solved. The effect is really striking and very beautiful, so you will get your pictures in due time (I hope next weekend).

Quote:

from link of garage chemist

Ein Arbeiten im Freien oder unter dem Abzug ist NICHT notwendig, evtl . am Fenster arbeiten !

I strongly disagree with this. For everyone, who wants to try this himself, you definitely have to use good ventilation. I read in my book that in this reaction (CN)2 can be formed, together with HCN. Both gases are VERY toxic and very fast acting. I only did the experiment with 200 mg or so, but the scale, suggested in the link is with tens of grams. The amount of toxic gases, formed from this, can be lethal! I think that more than just CO2 is produced in the reaction.

[Edited on 18-12-06 by woelen]

not_important - 18-12-2006 at 19:16

Sorry, I don't recall reading of an isolation of the complex, or even anything on its stability. BTW, acetone and some other ketones also form a violet complex with nitroprusside when used in place of H2S/S(2-).

guy - 3-1-2007 at 00:11

Hey woelen,

Try this experiment. It has very similar results to your's.

Put NaNO2 or KNO2 in some FeSO4. Then add some HCl. You should have a dark solution. Dilute until a very intense permangate purple is obtained.

I wonder what it can be?

I think it is a nitrosyl chloro iron complex. I dont have any other strong acids at hand, but I would try without chloride. I think that just forms a brown solution (brown ring test).

[Edited on 1/3/2007 by guy]

Sauron - 3-1-2007 at 01:51

Why use archaic nomenclature? "Yellow prussiate of potash" -- hell, you might as well use alchemical symbology.

Any of the modern naming systems will do, the 18th and 19th centuries are toast.

YT2095 - 3-1-2007 at 03:13

well I`ve never been one to rub muriat of natrium into the wound, but I actualy like such terms myself, and don`t think they should ever be lost, I`m even guilty of using them myself sometimes.
have no doubt, Woelen also knows the IUPAC names too

woelen - 3-1-2007 at 13:08

Quote:

Originally posted by guy
Hey woelen,

Try this experiment. It has very similar results to your's.

Put NaNO2 or KNO2 in some FeSO4. Then add some HCl. You should have a dark solution. Dilute until a very intense permangate purple is obtained.

I wonder what it can be?

I think it is a nitrosyl chloro iron complex. I dont have any other strong acids at hand, but I would try without chloride. I think that just forms a brown solution (brown ring test).

[Edited on 1/3/2007 by guy]

Could you be a little more precise about what you did (especially relative amounts of the chemicals used)? I did the experiment, but I could not reproduce the purple color.

I did two experiments, one with 10% H2SO4 and one with 10% HCl. Both experiments result in formation of a dark brown/green solution, but the liquid with HCl very quickly looses its color and it becomes bright yellow, the well-known color of the iron(III)/chloro complex, FeCl4(-). The liquid with H2SO4 remains dark for a much longer time.
I could not obtain the purple color you described. I used lab reagent grade chemicals.

I also modified the experiment by changing the order in which the chems were mixed, but also that did not give me the purple compound. It looks quite interesting what you did and I realluy would like to reproduce it.

I certainly can imagine that such nitrosyl/chloro/iron complex exists. I discovered a similar copper(II) complex, which only is formed by the combination of nitrite, chloride and copper (II) in acidic environment. This copper complex is very deep blue/indigo.

[Edited on 3-1-07 by woelen]

guy - 3-1-2007 at 13:46

Quote:

Originally posted by woelen

Quote:

Hmm.. I just used KNO2 from my KNO3/Pb experiments and HCl from a toilet cleaner (it had a blue dye and it was gelly, but I diluted it a lot so I thought that wouldnt matter).

This is what I did. To test my nitrite solution from the KNO3/Pb experiment, I added some FeNH4(SO4)2 to it. A precipitate was formed because NO2- is basic. Then I added a few drops of 10% HCl (from the toilet cleaner). It turned brownish with bubbling and a smell of NO2. Then I diluted it with water until it turned purple. Further dilution results in the familiar orange Fe(III) color.

[Edited on 1/3/2007 by guy]

Edit: I just found some good H2SO4 from my aquarium test kit. I tested it with NaCl, FeSO4 and KNO2 and did NOT get a purple. I guess the inpurities in the toilet cleaner made a cool purple color. The purple color, however, can ONLY be obtained once KNO2 is added...so thats interesting in itself.

[Edited on 1/3/2007 by guy]

Velzee - 3-5-2018 at 07:19

The violet solution appears to be an indicator of sulfides: https://youtu.be/NYBV5rI0a2I

[Edited on 5/3/2018 by Velzee]

Velzee - 3-5-2018 at 07:51

Quote: Originally posted by LawnMaster1

Would you mind posting any pictures?

I too, would like to see photos, mainly of the solution upon adding acid and after

AJKOER - 3-5-2018 at 18:34

Some possible chemistry which may help.

First, from "Fenton chemistry in biology and medicine" by Josef Prousek, to quote reaction (15) on page 2330:

"For Fe(II) and Cu(I), this situation can be generally depicted as follows [20,39],

Fe2+/Cu+ + HOX → Fe3+/Cu2+ + ·OH + X- (15)

where X = Cl, ONO, and SCN. "

In the current context:

Fe2+ + HONO2 → Fe3+ + ·OH + NO2-

From the hydroxyl radical acting on nitrite, some nitrogen oxides formation:

•OH + NO2- = OH- + NO2 (g)

NO2 + H2O = HNO2 + HNO3

2 HNO2 → NO2 (g) + NO (g) + H2O

H+ + CN- + OH• = •CONH2 (85–95%) or •HC(O)NH (see http://pubs.rsc.org/en/content/articlelanding/2000/p2/a90960... )
......
--------------------------------------------

With the introduction of H2S in aqueous conditions:

H2S + H2O = H3O+ + HS-

Fe(lll) + HS- → Fe(ll) + HS•

In the presence of oxygen:

HS• + O2 → HSO2•

HSO2• → H+ + SO2•-

SO2•- + O2 → SO2 + O2•-

SO2 + H2O = H+ + HSO3-

HS• + O2•- → S + HO2-

HO2- + H+ → H2O2

Fe(ll) + H2O2 + H+ --> Fe(lll) + OH• + H2O

OH• + H2S --> H2O + HS•

Approximate theoretical implied net reaction ignoring, not listed, problem side reactions (consuming radicals) and further oxidation of sulfite:

2 H2S + 2 O2 --Fe(ll)/Fe(lll)--> S + H2O + H2SO3

Reference: see comments and links at http://www.sciencemadness.org/talk/viewthread.php?tid=81401&...

[Edited on 4-5-2018 by AJKOER]

Melgar - 4-5-2018 at 01:06

Quote: Originally posted by AJKOER

H+ + CN- + OH• = •CONH2 (85–95%) or •HC(O)NH

I don't think it would usually work like that. In most cases, peroxide will oxidize cyanide to cyanate, unless it's under really specific, controlled conditions. Something like 2OH• + CN- = H2O + OCN-

Quote: Originally posted by AJKOER

With the introduction of H2S in aqueous conditions:

H2S + H2O = H3O+ + HS-

Fe(lll) + HS- → Fe(ll) + HS•

There would NEVER be a reaction that would go in the direction of the second equation. Generally free radicals don't last very long, are unpredictable in how they react, and are formed either from peroxides or halogens + light. I think you might be missing that these reactions often go in a direction that takes multiples of the reactants, and produces some combination of products. Like:

2Fe3+ + 2HS- = Fe2+ + Fe(II)S2 + H2

FeS2 is iron pyrite, aka fool's gold, and is insoluble and would then leave the solution.

Quote: Originally posted by AJKOER

In the presence of oxygen:

HS• + O2 → HSO2•

HSO2• → H+ + SO2•-

SO2•- + O2 → SO2 + O2•-

SO2 + H2O = H+ + HSO3-

HS• + O2•- → S + HO2-

HO2- + H+ → H2O2

Fe(ll) + H2O2 + H+ --> Fe(lll) + OH• + H2O

OH• + H2S --> H2O + HS•

This is more or less impossible. Except for the part about bisulfite forming from SO2 and water. That does happen. Really, you have to realize that most of these reactions you're proposing have other, thermodynamically favorable reaction possibilities that use multiples of one of the reactants. Like:

2Fe2+ + H2O2 + 2H+ = 2Fe3+ + 2H2O

AJKOER - 4-5-2018 at 05:03

Melgar:

On the first page of the SM thread I referenced there is this link: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4295652/ to a 1999 work 'Free Radicals and Chemiluminescence as Products of the Spontaneous Oxidation of Sulfide in Seawater, and Their Biological Implications', by DAVID W. TAPLEY, GARRY R. BUETTNER, and J. MALCOLM SHICK published in Biol Bull. 1999 Feb 1; 196(1): 52–56.; 196(1): 52–56, doi: 10.2307/1543166 .

All the sulfur radical related equations are extracted therefrom, none of my personal work. You may wish to pass your comments along to the authors of that work.
----------------------------------------

The reaction of the hydroxyl radical with cyanide is surmised from a 2000 article in Journal of the Chemical Society, Perkin,Transactions 2, titled 'Common intermediates in the OH-radical-induced oxidation of cyanide and formamide', by Florinella Muñoz, Man Nien Schuchmann, Gottfried Olbricha and Clemens von Sonntag. Here is the full abstract:

"OH radicals generated in the pulse radiolysis of N2O-saturated water react with formamide (by H-abstraction) and with cyanide (by addition and rearrangement) to give the same radicals, the main radical being the ·CONH2 radical 1 (85–95%), the other most likely being the HC(O)NH· radical 2 (5–15%). Quantum-chemical calculations support the preferred formation of 1 from formamide as well as from cyanide. Radical 1 reduces tetranitromethane (TNM) to the nitroform anion (2.8 × 108 dm3 mol−1 s−1). In the presence of oxygen, it rapidly adds oxygen (2.7 × 109 dm3 mol−1 s−1) to give the corresponding peroxyl radical ·OOCONH2 (3) which absorbs more strongly in the wavelength region 250–400 nm than radical 1 [ε(320 nm) ≈ 180 dm3 mol−1 cm−1]. Peroxyl radical 3 deprotonates in basic solution (pKa ≈ 9.6), and its anion rapidly eliminates O2·− (106 s−1) to give cyanic acid. Product studies under γ-radiolysis conditions show that in the absence of O2 less than half of radical 1 decays by disproportionation to produce cyanate and formamide. In basic solution and in the presence of O2, the G values of cyanate and H2O2 confirm that all of the peroxyl radical 3 decays by O2·−-elimination"

You are correct that eventual disproportionation in the presence of oxygen (or I would guess, an oxygen source) can lead to a cyanate product as well. However, upon addition of H2S to the system, I would wait for verification of a visibly distinct cyanate salt.

Link repeated: http://pubs.rsc.org/en/content/articlelanding/2000/p2/a90960...

[Edited on 4-5-2018 by AJKOER]

Tsjerk - 4-5-2018 at 05:31

@AJOEKER; I think there is no need to inform the author of that article about your misinterpretation of his work.

Have you looked at the circumstances, environments, concentrations that article is talking about? Have you heard Woelen talk about elemental sulfur in his experiments

All the reactions you pull from some published reference are quite likely possible somewhere at some time under some conditions. But not here, not as Woelen describes.

Edit: I would be suprised if there was a single radical in this whole reaction that actually participates. Why do you think oxidation by nitric acid would form a hydroxyl radical?

[Edited on 4-5-2018 by Tsjerk]

AJKOER - 4-5-2018 at 06:02

Quote: Originally posted by Tsjerk

@AJOEKER; I think there is no need to inform the author of that article about your misinterpretation of his work.

Have you looked at the circumstances, environments, concentrations that article is talking about? Have you heard Woelen talk about elemental sulfur in his experiments

All the reactions you pull from some published reference are quite likely possible somewhere at some time under some conditions. But not here, not as Woelen describes.

Edit: I would be suprised if there was a single radical in this whole reaction that actually participates. Why do you think oxidation by nitric acid would form a hydroxyl radical?

[Edited on 4-5-2018 by Tsjerk]

Well, you are entitled to your opinion, but please explain the likely formation of nitrogen oxides (per Woelen's comment: "Also some bubbles of gas are produced (NO,..") absence the generation of hydroxyl radicals. Note, NO and NO2 are sometimes presented by myself as •NO and •NO2, as they are examples of stable free radicals! If you accept the presence of radicals, adding H2S could further add sulfur based radicals.

I presented the homogeneous chemistry as a background contributing to a possible explanation as to products relating to the different experiment at hand. My first words are "Some possible chemistry which may help." However, my first reference, to quote: "Fenton chemistry in biology and medicine" by Josef Prousek, is hardly an example of claimed radical production in a homogeneous setting.

Others may comment or expand on my reactions, as did Melgar, suggesting a possible path to cyanates, especially if there is some suggestive visible evident of such salts.

[Edited on 5-5-2018 by AJKOER]

Tsjerk - 4-5-2018 at 06:36

Quote: Originally posted by AJKOER

I said "participate" in the reaction, I know NO and NO2 are radicals, but they don't participate. They happily bubble out.

Quote: Originally posted by AJKOER

I presented the homogeneous chemistry as a background contributing to a possible explanation as to products relating to the different experiment at hand. My first words are "Some possible chemistry which may help." [Edited on 4-5-2018 by AJKOER]

Well, there you go, you present the homogeneous chemistry... No you don't, your whole list of reactions is BS.

Quote: Originally posted by AJKOER

Others may comment or expand on my reactions, as did Melgar, suggesting a possible path to cyanates, especially if there is some suggestive visible evident of such salts.
[Edited on 4-5-2018 by AJKOER]

I will give you an example of a reaction that doesn't work in these circumstances/pressure/concentration to be relevant.

Have you ever heard of someone preparing H2O2 by bubbling O2 in H2S?

The article you pulled these reactions from is describing luminescence on the bottom of the deep sea near hydrothermal vents, while being focused on oxidative stress on organisms by radical oxygen species as a third overlooked mechanism of sulfuric toxicity. I can guess what the concentrations of these species are... I also guess Woelen didn't get a deep dark purple colour because of these.

AJKOER - 4-5-2018 at 08:15

Actually, H2S after conversion to HS• (from the same hydroxyl radical that produced the nitrogen oxides) does react with NO•, to quote a source:

"Simplifying depiction of chemical interaction between H2S and NO. The radical of H2S (HS•) reacts with that of NO (NO•) to generate thionitrous acid (HSNO)." Link: https://www.hindawi.com/journals/omcl/2016/6904327/ .
-------------------------

And, actually one can pass oxygen into a transition metal solution with say ascorbic acid (which I have done with an air pump and works even better so with light) to in situ create sufficient H2O2 and associated hydroxyl radicals to cause problems. So why not with H2S after conversion to HS• ?

Reference: See https://www.sciencemadness.org/whisper/viewthread.php?tid=77... , and here is an extract:

"Generation of Hydroxyl Radicals from Dissolved Transition Metals in Surrogate Lung Fluid Solutions" by Edgar Vidrio, et al at http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2626252/ . Cited reactions :

Cu(l)/Fe(II) + O2(aq) → Cu(ll)/Fe(III) + .O2-

As an alternate reference for the above reaction (which I have personally performed on Cuprous citrate using an air pump from an old fish tank), see for example, https://books.google.com/books?id=WjReuSXxl4YC&pg=PA17&a...

The reaction chain continues as:

Cu(l)/Fe(II) + .O2- +2 H+ → Cu(ll)/Fe(III) + HOOH

Cu(l)/Fe(II) + HOOH → Cu(ll)/Fe(III) + .OH + OH-

Net of the last three reactions:

3 Cu(l)/Fe(II) + O2(aq) +2 H+ → 3 Cu(ll)/Fe(III) + .OH + OH-

And, in the presence of sunlight (or a reductant like Citric or Ascorbic acid), a cyclic reaction could ensue in the case of sunlight:

Cu(ll)/Fe(lll) (aq) + hv → Cu(l)/Fe(ll) (aq) + HO• + H+

[Edited on 4-5-2018 by AJKOER]

Tsjerk - 4-5-2018 at 08:58

Quote: Originally posted by AJKOER

Actually, H2S after conversion to HS• (from the same hydroxyl radical that produced the nitrogen oxides) does react with NO•, to quote a source:

"Simplifying depiction of chemical interaction between H2S and NO. The radical of H2S (HS•) reacts with that of NO (NO•) to generate thionitrous acid (HSNO)." Link: https://www.hindawi.com/journals/omcl/2016/6904327/ .
-------------------------

In Woelen his reaction NO is actively removed by heating before the addition to Na2S.

Quote: Originally posted by AJKOER

And, actually one can pass oxygen into a transition metal solution with say ascorbic acid (which I have done with an air pump and works even better so with light) to in situ create sufficient H2O2 and associated hydroxyl radicals to cause problems. So why not with H2S after conversion to HS• ?

Reference: See https://www.sciencemadness.org/whisper/viewthread.php?tid=77... , and here is an extract:

"Generation of Hydroxyl Radicals from Dissolved Transition Metals in Surrogate Lung Fluid Solutions" by Edgar Vidrio, et al at http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2626252/ . Cited reactions :

Cu(l)/Fe(II) + O2(aq) → Cu(ll)/Fe(III) + .O2-

As an alternate reference for the above reaction (which I have personally performed on Cuprous citrate using an air pump from an old fish tank), see for example, https://books.google.com/books?id=WjReuSXxl4YC&pg=PA17&a...

How did you show you made radicals in your experiments? I don't see it in any of your posts. In your references they used HPLC, quite possibly you have one of those at work, but I don't believe you have one at home.

Quote: Originally posted by AJKOER

The reaction chain continues as:

Cu(l)/Fe(II) + .O2- +2 H+ → Cu(ll)/Fe(III) + HOOH

Cu(l)/Fe(II) + HOOH → Cu(ll)/Fe(III) + .OH + OH-

Net of the last three reactions:

3 Cu(l)/Fe(II) + O2(aq) +2 H+ → 3 Cu(ll)/Fe(III) + .OH + OH-

And, in the presence of sunlight (or a reductant like Citric or Ascorbic acid), a cyclic reaction could ensue in the case of sunlight:

Cu(ll)/Fe(lll) (aq) + hv → Cu(l)/Fe(ll) (aq) + HO• + H+

[Edited on 4-5-2018 by AJKOER]

All full-blown speculation as long as you don't show any prove of any reaction.

AJKOER - 4-5-2018 at 09:35

My point was that O2 with the proper support, can be a substitute for H2O2, and per the Fenton-type reference above, one can replace the H2O2 with HOCl, HONO2, ...In fact, I have on SM referenced a redox reaction with NO2.

The source of the hydroxyl radical is, as I claimed, via the HNO3 in place of H2O2 in the usual fenton reaction.

My cited experiment clearly produced a coloration change, per my recollection. This may only imply, albeit, a small amount of product.

My goal here is not to prove, but just suggest, possible reaction pathway(s) to match visible products.

Tsjerk - 4-5-2018 at 10:11

Quote: Originally posted by AJKOER

My point was that O2 with the proper support, can be a substitute for H2O2, and per the Fenton-type reference above, one can replace the H2O2 with HOCl, HONO2, ...In fact, I have on SM referenced a redox reaction with NO2.

There is no Fenton-type reaction here, there is no need for a Fenton theory explanation in this reaction as none of the oxidations here need a catalyst.

Quote: Originally posted by woelen

The use of nitric acid as oxidizer is essential. I also tried with H2O2 as oxidizer and then no purple color can be obtained after neutralization and adding Na2S.

[Edited on 17-12-06 by woelen]

Quote: Originally posted by AJKOER

The source of the hydroxyl radical is, as I claimed, via the HNO3 in place of H2O2 in the usual fenton reaction.

See the quote from Woelen, this does not work with straight peroxide.

Quote: Originally posted by AJKOER

My cited experiment clearly produced a coloration change, per my recollection. This may only imply, albeit, a small amount of product.

You mean the oxygen Cu(I) experiment? That will give you an coloration change anyway as Cu(I) is not stable in oxygen.

Quote: Originally posted by AJKOER

My goal here is not to prove, but just suggest, possible reaction pathway(s) to match visible products.

Sure.

AJKOER - 4-5-2018 at 13:16

Classic Fenton produces Fe(lll) and hydroxyl radicals.

The fenton-type reaction with a source of Fe(ll) and HONO2 forms Fe(lll), nitrite and hydroxyl radicals followed by nitrogen oxides.

So the observation that the products of these two reactions interacting with Na2S differently could suggest something other than just an hydroxyl radical exposure.
----------------------------------------------

Came across paper that may be relevant. See Equation (1) at https://www.academia.edu/1557498/Addition_and_Redox_Reactivi... with the formation of the ion Fe(CN)5N(O)SH]3- . Note, per prior discussion, HS• reacts with NO• to generate thionitrous acid (HSNO).

[Edited on 5-5-2018 by AJKOER]