The new forums will be named Coin Return (based on the most recent vote)! You can check on the status and timeline of the transition to the new forums here.
The Guiding Principles and New Rules document is now in effect.
Chemistry: crazy redox - WTF IS HAPPENING HERE?
Jimmy King
Registered User regular
Registered User regular
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.
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.
Jimmy King on
0
Posts
Figured the thread could use the bump to get some fresh eyes on the problem.
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.
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
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.
Either way: yay!
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.