Chemistry: crazy redox - WTF IS HAPPENING HERE?

Jimmy King

I'd be happy to provide the full document for this to anyone interested. Part of my current lab has some pre-lab problems. One of them is this:

What is the oxidation half reaction, reduction half reaction and the overall net reaction for sodium thiosulfate reaction with potassium ferricyanide.

... I have no fucking idea what is going on here.

Sodium thiosulfate: Na₂S₂O₃
Potassium ferricyanide: K₃Fe(CN)₆

So, the obvious reaction here is:
3Na₂S₂O₃ + 2K₃Fe(CN)₆ -> 2Na₃Fe(CN)₆ + 3K₂S₂O₃

I just did the balancing real quick, but it looks right. That part is not the problem anyway. So, the problem here is that I don't see any redox going on.

On the left side:
Na = +1
O = -2
S = +2

K = +1
CN = -1
Fe = +3

On the right side:
Na = +1
CN = -1
Fe = +3

K = +1
O = -2
S = +2

Now, based on a page of prior babbling about reacting Ferricyanide (3- oxidation) with Iodine and, separately, Thiosulfate with Iodine, to get Ferrocyanide (4- oxidation) and Tetrathionate... perhaps I am to infer (assume seems more accurate) that this different reaction, involving Sodium and Potassium and no iodine will also result in Ferrocyanide and Tetrathionate as products?

So instead becoming something similar to:

Na₂S₂O₃ + K₃Fe(CN)₆ -> Na₄Fe(CN)₆ + K₂S₄O₆

I am fairly confident this is never going to balance based on my attempts and have no idea what the correct half reactions are since I obviously do not have the correct product. That suggests to me that this is not a double replacement reaction... but I have nothing to suggest what it should be then. This reaction is far more complex than anything we have covered in class or that our textbook shows for redox reactions. Everything we have done with redox has been very straightforward single replacement, decomp, and combination.

Now, the end of the bit going on about Thiosulfate, Tetrathionate, Ferricyanide, Ferrocyanide, and Iodine ends with a straightforward balanced equation which just tosses out the various Iodine products with no explanation as to why, only stating "where the iodine, iodide and triiodide are not in the overall reaction. ".
That gives a simple

S₂O₃ + Fe(CN)₆ -> S₄O₆ + Fe(CN)₆

Which does have clear redox going on where Sulfur goes from a +2 to a +2.5 (oxidation) and Fe from +3 to +2 (reduction).

I have no background knowledge to have any reason to believe that I can just ignore the sodium and potassium and assume the same end product occurs in the reaction involving Sodium Thionate and Potassium Ferricyanideother than a wild guess based on this problem apparently being impossible without that assumption.

davidsdurions

I'm sad this hasn't been responded to and although the answer is well beyond my knowledge base, I want to know what is happening too. Is it a big boom or a little fizzle?

Figured the thread could use the bump to get some fresh eyes on the problem.

Kipling

Warning - the last time I actually had to do this was a long time ago. In the iodine reactions, one of them has to evolve I2, one consumes it. If it doesn't, this makes no sense.

From what you said...

"Thionate -> tetrathionate" oxidation which generates two electrons. 2x (2-) to 2-. That's two free electrons.

Ferri->ferro reduces from 3- to 4-, needing one electron. So for each thionate oxidation, 2 of this reaction are needed to balance electrons.

So...

4 Na + 2 SO3(2-) -> S2O6(2-) + 4 Na + 2 e

6 K + 2 FeCN6(3-) + 2e -> 6 K + 2 FeCN6(4-)

I'm guessing you get a sodium tetrathionate with ferrocyanide where some is sodium and some is potassium? That's only if you get the salts, of course. Otherwise, the Na+ and K+ are just ions in solution.

Jimmy King

The original substances are Na2S2O3 and KFe(CN)6. Now if those decomp down to Na + S3O4 and K + Fe(CN), that might also work. So:

2Na₂S₂O₃ -> 2Na + S₄O₆

The Na and K will not undergo any oxidation state changes. I don't have anything to suggest to me that that's the case vs the original double replacement (unless I've forgotten something from my previous class, which could be), aside from the fact that the double replacement doesn't cause a redox reaction and I am 100% definitely supposed to get a redox reaction.

Leaving me with a final

2Na₂S₂O₃ + K₃Fe(CN)₆-> S₄O₆ + Fe(CN)₆ + 2Na + 3K

Fuzzy Cumulonimbus Cloud

This might work better if you treated it as an aqueous reaction. What would happen to the K and Na if this were in solution? What ions would form then? Could you maybe write half reactions with the dissociated components? And maybe you could add water molecules to deal with all those oxygens?

Jimmy King

Reviewed electrolyte and solubility stuff, since you are right, this is going to be an aqeuous solution. That took care of what to do with the Na and K. Thanks for that. It's been quite awhile since I had to think about that and that was not my strongest section when I took it.

I don't think there's anything which needs done with the oxygens. They stay bonded so the sulfur as it goes from S2O4 to S4O6. I'm also pretty sure that part of redox is not dealt with for 7 more chapters in my textbook.

Fuzzy Cumulonimbus Cloud

Oh sorry. You will get to that if you take analytical or electrochemistry.
Either way: yay!

Jimmy King

Ha, nope. Last semester of Chemistry for me. As a CS major, I normally wouldn't be taking it at all, but the other option was physics and the physics teachers are notoriously super terrible.

We might hit that chapter with the fancier solubility stuff at the end of the semester, so you may be hearing from me again. Well, you'll definitely be hearing from me again... just maybe or maybe not about this in particular.

Fuzzy Cumulonimbus Cloud

Feel free! I used to teach chemistry. Will again once I finish my Ph.D.