Hello. Reading old book in the net over preparation manganese acetate (III) from Manganese acetate (II), acetic acid hot, and chlorine.
Someone have idea of mechanisms?????Thankblogfast25 - 28-5-2015 at 13:43
Go to library of this forum. Find 'Preparative Inorganic Chemistry' by Gregor Brauer. It has a procedure for Mn(III) acetate.GrayGhost - 28-5-2015 at 13:53
Only say manganese acetate is preparate by manganese permanganate or Chlorine in acetic acid hot. blogfast25 - 28-5-2015 at 14:30
Books like that are fairly lowly authorative.
I doubt if chlorine could oxidise Mn(II) to Mn(III) but I could be wrong on that.
Look up the Standard Reduction Potentials.
In water, Mn(III) will oxidise chloride to chlorine, first hand experience BTW. That strongly suggest the opposite
(what you want to do) is thermodynamically unfavourable.GrayGhost - 28-5-2015 at 14:36
I doubt if chlorine could oxidise Mn(II) to Mn(III) but I could be wrong on that.
Look up the Standard Reduction Potentials.
In water, Mn(III) will oxidise chloride to chlorine, first hand experience BTW. That strongly suggest the opposite
(what you want to do) is thermodynamically unfavourable.
That is the point, my dude. The book is wrong? Chlorine oxidise Mn (II) to Mn(III)? Apparently no.DraconicAcid - 28-5-2015 at 14:38
Standard reduction potentials only apply in aqueous solution. They may differ in glacial acetic acid.blogfast25 - 28-5-2015 at 14:40
The only way I could possibly see that happen (Mn(II) to Mn(III) by Cl(0)) is if Mn(III) acetate was insoluble in the solvent. Removal by
precipitation could drive the equilibrium to the right (Delta G is changed).
[Edited on 28-5-2015 by blogfast25]GrayGhost - 28-5-2015 at 14:49
Apparently the Mn(III) acetate should be soluble in acetic, but the chlorinated product of reaction if not is soluble , then equilibrium drive to the
right?
[Edited on 28-5-2015 by GrayGhost]blogfast25 - 28-5-2015 at 14:55
Apparently the Mn(III) acetate should be soluble in acetic, but the chlorinated product of reaction if not is soluble , then equilibrium drive to the
right?
That would suggest the insoluble reaction product is something other than Mn(III) acetate.
[Edited on 28-5-2015 by blogfast25]DraconicAcid - 28-5-2015 at 15:23
Classic example: oxidation of Fe(II) to Fe(III) is much 'easier' in neutral/alkaline than acid conditions because Fe(OH)<sub>3</sub> is so
insoluble.
An even better example- Cu(II) + Cu -> 2 Cu(I) is spontaneous in acetonitrile. The reverse is spontaneous in aqueous solution.GrayGhost - 28-5-2015 at 15:33
I thank you your valuable answers guys. blogfast25 - 28-5-2015 at 17:59
An even better example- Cu(II) + Cu -> 2 Cu(I) is spontaneous in acetonitrile. The reverse is spontaneous in aqueous solution.
Errmm. That works also in water, if there's chloride present:
Cu<sup>2+</sup>(aq) + Cu(s) + 2 Cl<sup>-</sup>(aq) === > 2 CuCl(s). Same reason though: Cu(I) being removed as solid
precipitate. A fairly standard way to prepare wet CuCl.
