Sciencemadness Discussion Board
Not logged in [Login ]
Go To Bottom

Printable Version  
Author: Subject: Zinc reduction of hydroxides ?
gboneu
Harmless
*




Posts: 30
Registered: 12-5-2015
Member Is Offline

Mood: No Mood

thumbup.gif posted on 25-9-2015 at 09:07
Zinc reduction of hydroxides ?


Hello guys !!
After looking at Nurdrage video in making potassium by magnesium catalyzed reduction i wander if other metals and catalist can be used --- Somthing like zinc, or aluminium. And catalyst like ketones, Etc.... :) :D
View user's profile View All Posts By User
violet sin
International Hazard
*****




Posts: 1482
Registered: 2-9-2012
Location: Daydreaming of uraninite...
Member Is Offline

Mood: Good

[*] posted on 25-9-2015 at 18:57


Do you have links to papers suggesting something along these lines, or are you blindly speculating here? It would be nice to see some one has tried it, or there was sound logic behind your assumption.
View user's profile View All Posts By User
gboneu
Harmless
*




Posts: 30
Registered: 12-5-2015
Member Is Offline

Mood: No Mood

[*] posted on 30-9-2015 at 19:47


i just wondered because like magnesium zinc is also reactive, and with some variations it may be possible, like high temperatures and other catalysts. If possible i will do some experiments but, its better to ask before xD :D
View user's profile View All Posts By User
UC235
National Hazard
****




Posts: 565
Registered: 28-12-2014
Member Is Offline

Mood: No Mood

[*] posted on 1-10-2015 at 08:50


If you'd made any attempt to read the giant abomination that is the sodium thread on this very subject (I'd skip to the middle and cut out most of the fighting and find where people start managing to reproduce the results), you'd know that the reaction is extremely fussy and hard to reproduce. It works....sometimes, and due to unknown factors like the phase of the moon. The first functional prep was reported by member Pok of Versuchschemie.de. We fought over it a whole lot and some members managed to reproduce the results. Nurdrage came in at the end, did some optimization of his own and made a video about it.

The reaction can't even be made to work for sodium hydroxide (yet) which normally would be a very minimal jump in reactivity and would be expected to work. So, I see no reason to think we could swap reducing metals to things that are quite different and expect results. Furthermore, we have a strongly suspected mechanism for the catalyst's functioning and your assumption that ketones would somehow work (probably because they work in one other unrelated high temp reaction) is completely unfounded.
View user's profile View All Posts By User
aga
Forum Drunkard
*****




Posts: 7030
Registered: 25-3-2014
Member Is Offline


[*] posted on 1-10-2015 at 09:10


It must be annoying that Nile Red and Nurdrage get the credit cos they did some videos, when the actual Work behind it all was done by extant others.

Can't see any way to redress this other than by you pointing to the Actual work (as you are doing) or by asking them to give credit where it's due in their videos.





View user's profile View All Posts By User
blogfast25
International Hazard
*****




Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline

Mood: No Mood

[*] posted on 1-10-2015 at 09:17


Quote: Originally posted by aga  
It must be annoying that Nile Red and Nurdrage get the credit cos they did some videos, when the actual Work behind it all was done by extant others.




Not really. Nurdrage did actually contribute a lot to that sticky thread, testing successfully various solvents.

As regards using zinc? Forget it. I even doubt if:

Zn + KOH === > K + ZnO + 1/2 H<sub>2</sub>

... is thermodynamically favourable. (Edit: a quick check shows that it isn't)

As regards catalysis by ketones, since we more or less understand how the catalysis with t-butanol works, we can be confident that ketones will do diddly squat here.

[Edited on 1-10-2015 by blogfast25]




View user's profile View All Posts By User
Bert
Super Administrator
Thread Moved
1-10-2015 at 10:08
Upsilon
Hazard to Others
***




Posts: 392
Registered: 6-10-2013
Member Is Offline

Mood: No Mood

[*] posted on 1-10-2015 at 11:00


You need a table of reduction potentials. You can easily find one on Google. Now these tables are for 1M concentration aqueous chemistry at 25C (or 20C, I don't remember exactly). Therefore they are not highly accurate for what you're trying to do. Find the half-reactions of the two metals involving their solid form. There is an E° voltage value associated with each half-reaction. You don't need to know what this means, but the general trend is that a metal with a lower (more negative) number will probably reduce a metal that has a higher (less negative) number.

When the numbers are close (like those of magnesium and potassium), it's hard to tell what will happen. However, if you're looking at zinc and potassium (-0.76 and -2.92, respectively), those numbers are so far off that it's pretty safe to assume that you'll never get zinc to reduce potassium.

[Edited on 1-10-2015 by Upsilon]
View user's profile View All Posts By User
aga
Forum Drunkard
*****




Posts: 7030
Registered: 25-3-2014
Member Is Offline


[*] posted on 1-10-2015 at 11:40


Quote: Originally posted by blogfast25  
As regards catalysis by ketones ...

Not that i know much myself, but there seems a really easy way to settle any chemistry argument : Experimentation !

I look forward to the Experiments gboneu.

Edit:

Not much experimentation going on these days, so we could use some inspiration.

[Edited on 1-10-2015 by aga]




View user's profile View All Posts By User
violet sin
International Hazard
*****




Posts: 1482
Registered: 2-9-2012
Location: Daydreaming of uraninite...
Member Is Offline

Mood: Good

[*] posted on 1-10-2015 at 12:00


I specifically din't mention my personal beliefs as to whether or not it would work, and simply asked if there were any references to support the theory, because I was really hoping that would spark gboneu to read, read, read :) maybe take a deeper look at what was required to make something like this even begin to look plausible, let alone understand in the mechanisms.

Just giving in to hunches and flights of fantasy can be rewarding, IF, you follow them up with research and experimentation. Without the latter, it is nothing more than a wild guess. Some great ideas were sparked by "woah, what if I...." moments. But they were worked at after curiocity sparked them.

View user's profile View All Posts By User
blogfast25
International Hazard
*****




Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline

Mood: No Mood

[*] posted on 1-10-2015 at 14:43


Quote: Originally posted by Upsilon  
You need a table of reduction potentials. You can easily find one on Google. Now these tables are for 1M concentration aqueous chemistry at 25C (or 20C, I don't remember exactly). Therefore they are not highly accurate for what you're trying to do. Find the half-reactions of the two metals involving their solid form. There is an E° voltage value associated with each half-reaction. You don't need to know what this means, but the general trend is that a metal with a lower (more negative) number will probably reduce a metal that has a higher (less negative) number.

When the numbers are close (like those of magnesium and potassium), it's hard to tell what will happen. However, if you're looking at zinc and potassium (-0.76 and -2.92, respectively), those numbers are so far off that it's pretty safe to assume that you'll never get zinc to reduce potassium.



This is quite wrong and really needs correcting.

The Standard Reduction Potentials are determined in watery solution and the Gibbs Free Energies involved are very different from those obtained in other conditions like in other solvents or in the absence of solvents.

For example, for the (hypothetical) reduction of K<sup>+</sup> by Zn, the reaction equation in aqueous medium would be:

2 K<sup>+</sup>(aq) + Zn(s) === > 2 K(s) + Zn<sup>2+</sup>(aq)

Taking into account the Gibbs Free Energies of these species the Standard Redox Potential then works out unfavourable.

But in the absence of solvents, the reaction equation would be (e.g.):

KOH(s) + Zn(s) ==== > K(s) + ZnO(s) + 1/2 H<sub>2</sub>(g)

To work out the overall change of Gibbs Free Energy very different numbers would have to be used.

Believe me, use the Standard Reduction Potential Tables for reactions carried out in aqueous media but not for any others.

[Edited on 1-10-2015 by blogfast25]




View user's profile View All Posts By User
Darkstar
Hazard to Others
***




Posts: 279
Registered: 23-11-2014
Member Is Offline

Mood: Sleepy

[*] posted on 1-10-2015 at 15:22


Quote: Originally posted by blogfast25  
Believe me, use the Standard Reduction Potential Tables for reactions carried out in aqueous media but not for any others.


Very true. For example, if you look at a reduction potential table, you might be given the impression that lithium is the strongest reducing agent of the alkali metals, when in reality it is actually the weakest under most non-aqueous circumstances because of its small size and how close its valence electron is to the nucleus. In water, however, lithium's small size and high charge density actually work to its advantage, making its valence electron extremely easy to remove, far easier than sodium or potassium or even cesium. But this is only because of the favorable interaction between lithium ions and water molecules, so you must take that into consideration before assuming that lithium is going to be anywhere near as strong of a reducing agent under non-aqueous conditions.
View user's profile View All Posts By User
Upsilon
Hazard to Others
***




Posts: 392
Registered: 6-10-2013
Member Is Offline

Mood: No Mood

[*] posted on 1-10-2015 at 15:39


I wasn't trying to imply that the table of reduction potentials is the appropriate tool to use here. I just feel like it can serve as a rough guideline for what he's trying to do. All of the 1A and 2A metals have very similar electrode potentials, so any non-aqueous redox-reactions involving a pair of any two of them will not be able to be predicted with the table. However, when you take two metals with drastically different potentials, a reaction is quite unlikely to occur, aqueous or not (like here, with potassium and zinc). I don't think you'll ever see iron reduce calcium, nickel reduce potassium, or zinc reduce lithium. If you find some bizarre reaction to prove me wrong, please don't hesitate to show me. I referenced the table not because of its ability to point out what is correct in non-aqueous chemistry, but instead to point out what is probably very obviously wrong (like the hypothesis that zinc can reduce potassium).
View user's profile View All Posts By User
blogfast25
International Hazard
*****




Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline

Mood: No Mood

[*] posted on 1-10-2015 at 15:45


Quote: Originally posted by Darkstar  


Very true. For example, if you look at a reduction potential table, you might be given the impression that lithium is the strongest reducing agent of the alkali metals, when in reality it is actually the weakest under most non-aqueous circumstances because of its small size and how close its valence electron is to the nucleus.


And despite that Li reduces CsCl at about 700 C under vacuum because in these conditions Cs is volatile and Li is not, which drives the equilibrium:

Li + CsCl < === > LiCl + Cs

... to the right by distilling off the Cs.

There are many factors that determine whether a chemical reaction is practical or not.

The SRPs really only tell the story of aqueous solutions and even then only the thermodynamic side of things, not the kinetics.




View user's profile View All Posts By User
blogfast25
International Hazard
*****




Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline

Mood: No Mood

[*] posted on 1-10-2015 at 15:55


Quote: Originally posted by Upsilon  
I don't think you'll ever see iron reduce calcium, nickel reduce potassium, or zinc reduce lithium. If you find some bizarre reaction to prove me wrong, please don't hesitate to show me.


I won't. Iron has been used to reduce KOH to K and H<sub>2</sub>, although the process is not in use any more. You'll find references to it in the sticky K thread.

Teaching newbies to use the SRPs when they shouldn't is just plain wrong. Already too many have invested 'magic powers' into SRPs. Please don't feed their general confusion/ignorance.

[Edited on 1-10-2015 by blogfast25]




View user's profile View All Posts By User
Upsilon
Hazard to Others
***




Posts: 392
Registered: 6-10-2013
Member Is Offline

Mood: No Mood

[*] posted on 1-10-2015 at 16:01


I see. I apologize for my assumptions.

It has been quite a while since I have learned about the nuances of redox reactions. If you could please show me where I can brush up on this type of thing, I would appreciate it.

[Edited on 2-10-2015 by Upsilon]
View user's profile View All Posts By User
Darkstar
Hazard to Others
***




Posts: 279
Registered: 23-11-2014
Member Is Offline

Mood: Sleepy

[*] posted on 1-10-2015 at 16:13


@blogfast25:

True, which is why I made sure to say "most" in order to leave room for those special or unusual circumstances. Just like the difference between lithium as a reducing agent in water and lithium as a reducing agent in ammonia, it ultimately comes down to the individual circumstances of the reaction conditions. Simply put, there are many factors that must be considered before you can say one way or another with any real degree of certainty whether or not a given reaction will occur.
View user's profile View All Posts By User
blogfast25
International Hazard
*****




Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline

Mood: No Mood

[*] posted on 1-10-2015 at 16:38


Quote: Originally posted by Upsilon  
If you could please show me where I can brush up on this type of thing, I would appreciate it.



That's such a broad field governed by such a large variety of considerations and theories that I would simply recommend the following: like everyone else, just keep learning and never assume you've finally understood it all.

Sapere Aude!

[Edited on 2-10-2015 by blogfast25]




View user's profile View All Posts By User
Upsilon
Hazard to Others
***




Posts: 392
Registered: 6-10-2013
Member Is Offline

Mood: No Mood

[*] posted on 1-10-2015 at 16:56


Quote: Originally posted by blogfast25  
Quote: Originally posted by Upsilon  
If you could please show me where I can brush up on this type of thing, I would appreciate it.



That's such a broad field governed by such a large variety of considerations and theories that I would simply recommend the following: like everyone else, just keep learning and never assume you've finally understood it all.

Sapere Aude!

[Edited on 2-10-2015 by blogfast25]


I was more referencing your ability to determine that the KOH + Zn reaction was not favorable. Did you just look it up or did you do a calculation?
View user's profile View All Posts By User
Darkstar
Hazard to Others
***




Posts: 279
Registered: 23-11-2014
Member Is Offline

Mood: Sleepy

[*] posted on 1-10-2015 at 17:09


Quote: Originally posted by Upsilon  
I don't think you'll ever see iron reduce calcium, nickel reduce potassium, or zinc reduce lithium. If you find some bizarre reaction to prove me wrong, please don't hesitate to show me. I referenced the table not because of its ability to point out what is correct in non-aqueous chemistry, but instead to point out what is probably very obviously wrong (like the hypothesis that zinc can reduce potassium).


As blogfast has already demonstrated, you can't make absolute claims like "X can't reduce Y" without first taking into consideration the specific conditions under which the reaction is to take place. For example, lithium reducing cesium or iron reducing potassium under certain conditions.

It's not so much that lithium or iron just flat-out can't give cesium or potassium an electron because they aren't strong enough reducing agents, it's just that the electron is very rapidly given back to them under most circumstances. Like in blogfast's example of lithium reacting with cesium chloride, the reaction proceeds because cesium is being removed before it has a chance to reduce lithium back. Under normal conditions the reaction wouldn't work because cesium's valence electron is so far from its nucleus (not to mention all of the repulsive electrons in between) and the lithium ion is extremely tiny, so the positive charge on it is distributed over a small area making it extremely strong--much stronger than the positive charge on a cesium ion. Because of this, cesium will just give the electron right back to lithium, returning us back to square one.
View user's profile View All Posts By User
blogfast25
International Hazard
*****




Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline

Mood: No Mood

[*] posted on 1-10-2015 at 17:22


Quote: Originally posted by Upsilon  


I was more referencing your ability to determine that the KOH + Zn reaction was not favorable. Did you just look it up or did you do a calculation?


When we say a reaction is thermodynamically favourable, that means that mathematically the change in Gibbs Free Energy ΔG is negative (< 0) in the given reaction conditions. ΔG = G<sub>reaction products</sub> - G<sub>reactants</sub> (very simply put).

Also, ΔG = ΔH - TΔS and in this case I was a little lazy and just calculated ΔH (and assumed Entropic effects would be small). I found ΔH > 0 at 298 K, which suggests this reaction is not thermodynamically favourable. But the devil is in the detail and at higher temperatures, as well as considering H<sub>2</sub> escapes, this reaction may yet prove to proceed. I didn't explore that scenario (because the reaction isn't of great interest to me).

A fairly decent guide to chemical energetics, here:

http://www.chemguide.co.uk/physical/energeticsmenu.html#top


[Edited on 2-10-2015 by blogfast25]




View user's profile View All Posts By User
Upsilon
Hazard to Others
***




Posts: 392
Registered: 6-10-2013
Member Is Offline

Mood: No Mood

[*] posted on 1-10-2015 at 17:26


Quote: Originally posted by blogfast25  

When we say a reaction is thermodynamically favourable, that means that mathematically the change in Gibbs Free Energy ΔG is negative (< 0) in the given reaction conditions. ΔG = G<sub>reaction products</sub> - G<sub>reactants</sub> (very simply put).

Also, ΔG = ΔH - TΔS and in this case I was a little lazy and just calculated ΔH (and assumed Entropic effects would be small). I found ΔH > 0 at 298 K, which suggests this reaction is not thermodynamically favourable. But the devil is in the detail and at higher temperatures, as well as considering H<sub>2</sub> escapes, this reaction may yet prove to proceed. I didn't explore that scenario (because the reaction isn't of great interest to me).

[Edited on 2-10-2015 by blogfast25]


Ah, I remember these formulas now. Thanks.
View user's profile View All Posts By User

  Go To Top