fluorescence
Hazard to Others
Posts: 285
Registered: 11-11-2013
Member Is Offline
Mood: So cold outside
|
|
Reduction of transition metals using sodium metal
Molybdenum blue, made by fusing elemental Sodium and Molybdenum(VI)oxide in a test tube till the molten sodium ignited by itself and then leaching out
with water over a week.
I tried several methods to reduce the MoO3 but none of them would work besides LiAlH4. Problem is that my LiAlH4 is not clean enough and a lot of grey
rests remain so you don't see the blue color until it has fully settled over a few days. Then I found this other method and tried it out.
According to literature heating sodium with V(>III) salts will form green solutions of V(III), Ti(IV) will form Ti(III) and MoO3 will form
Molybdenum blue. I tried the Vanadium one but it failed I guess.
I heated Ammoniummetavandate with Sodium where the Vanadium would turn black upon formation of the Pentaoxide. The product did not dissolve in water.
I guess if V(III) formed then it would be V(III)Oxide which is not that soluble in HCl. Addition of HCl however formed a yellow solution and I guess
this is the Vanadium(V) forming Polyvanadates rather then some sort of VOCl or VCl3 compound. Still searchin for a better acid that will dissolve
V(III) to show the green color but can't find much about it.
I also tried things I couldn't find in the literature but no success yet. Boric Acid does not seem to form any elemental Boron and Phosphates don't
form white phosphorus.... Still I am really satisfied with the Molybdenum one.
|
|
|
DraconicAcid
International Hazard
Posts: 4278
Registered: 1-2-2013
Location: The tiniest college campus ever....
Member Is Offline
Mood: Semi-victorious.
|
|
Quote: Originally posted by fluorescence  | I tried the Vanadium one but it failed I guess.
I heated Ammoniummetavandate with Sodium where the Vanadium would turn black upon formation of the Pentaoxide. The product did not dissolve in water.
I guess if V(III) formed then it would be V(III)Oxide which is not that soluble in HCl. Addition of HCl however formed a yellow solution and I guess
this is the Vanadium(V) forming Polyvanadates rather then some sort of VOCl or VCl3 compound. Still searchin for a better acid that will dissolve
V(III) to show the green color but can't find much about it. |
See my earlier posts in this thread about reduction of vanadium(V) oxide. Dissolve it in base and neutralize with sulphuric acid, reduce with
sulphite to give vanadium(IV) or with aluminum wire and hydrochloric acid to give vanadium(II).
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
|
|
|
fluorescence
Hazard to Others
Posts: 285
Registered: 11-11-2013
Member Is Offline
Mood: So cold outside
|
|
@DraconicAcid: Check out my YouTube Channel. I did this a while ago already. I'm rather looking for alternative methods.
|
|
|
fluorescence
Hazard to Others
Posts: 285
Registered: 11-11-2013
Member Is Offline
Mood: So cold outside
|
|
So I went into my lab today and saw that the Vanadium reaction had turned blue. So this is an overview of what I did together with a quick FE-Diagram
I made for the system at low pH values.
I guess the Sodium really reduced it to V(III), maybe V2O3 or something else. When I added HCl rests of the V(V) would react to
form the yellow compounds. But at lower pH values V(III) and V(V) should, according to my FE-Diagram synproportionate to V(IV) which is then the blue
compound.
Either this or the V2O3 reacted to form V2O4 which gives a blue solution...but I doubt that this will go
that fast under a layer of water and acid....
http://i.imgur.com/HvmGWMq.jpg
http://i.imgur.com/ZOpI567.jpg
[Edited on 5-1-2017 by fluorescence]
EDIT:
One thing I just though of woudl be the following:
(NH4)(VO3) -> (NH4)2(V8O21) : Thermal decomposition to the black product.
V5+ ----Na--> V2+ + V3+ + V4+
And according to Berzelius will V2O5 and V2O3 form VO2 as well....
Now we add HCl to it and I assume there are only Oxides present:
V(II): VO + HCl -> [V(H2O)6]2+ (purple) ----> [V(H2O)6]3+
Either by Oxidation with oxygen in the acid/water or by synproportionation:
V2+ + V5+ -> 2 V3+
V2+ + V4+ -> 2 V3+
So the V2+ should vanish quite fast from the system.
V(III): V2O3 is not really soluble in HCl but will after some time form the greeen [V(H2O)6]3+
as well and react once in solution with V5+ to form V4+ as well. So V3+ will leave the system, too.
Then we are left with V4+ which accumulates as V2O4, VO2 and already disolved
[VO(H2O)5]+ (blue). The Oxides also dissolve in HCl to form the blue solution and we are only left with bits of
V5+ which is either dissolved already or sticking as black mass to the bottom and V4+.
Might be one way to look at this but I am not sure whether I am not thinking too complicated about all of this.
[Edited on 5-1-2017 by fluorescence]
[Edited on 1-5-2017 by zts16]
|
|
|
Texium
|
Thread Moved
5-1-2017 at 10:40 |
fluorescence
Hazard to Others
Posts: 285
Registered: 11-11-2013
Member Is Offline
Mood: So cold outside
|
|
Many thanks to zts16 for moving this thread
Now that it is a single topic it might be better to summarize the facts again:
So the idea comes from Jander Blasius - Lehrbuch der analytischen und präparativen anorganischen Chemie (14. Auflage) and is mentioned there
as a test together with the famous Borax and Phosphate test for some metals.
-> pg 452: "Titanium: The sample is heated together with metallic sodium till the test tube becomes soft. After cooling down
the test tube is destroyed and the product is acidified. For titanium the solution should turn red (Ti3+) although I think they are
referring to the purple color."
-> pg 456: "Vanadium: H2S , SO2 and oxalic acid reduce V(V) to V(IV) in acidic conditions. Metals like
Zn, Cd, Al reduce to the purple V(II). If the sample is heated with metallic sodium V(V) is reduced to V(III). This can be seen as the product in
water turns green"
-> pg 463: "Molybdenum: " [...] gives molybdenum blue"
-> pg 466: "Tungsten" [...] tungsten blue"
Those were all I could find but I only have the book and no ebook so I might have missed something as I can't run a search for it.
Also I can't find much on the actual reaction but I believe it is similar to the thermite reaction as the sodium has to ignite according to
literature.
I think it would be quite interesting to test Mn, Cr, ... for it as well. I tried it with Cr(III) but nothing really happened and also I can't seem to
reduce compounds to their elements like Boron from Boric Acid for example which would be quite cool if it worked.
So if anyone has further suggestions or more literature on this reaction this might be quite interesting.
[Edited on 5-1-2017 by fluorescence]
|
|
|
![[*]](images/xpblue/default_icon.gif){kind=icon}
