CHRIS25
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e- affinity, ionic radius, radius density and ionization energy
Being reluctant to post more ignorance on this forum but I have to: I am trying to understand the Bigger picture, studying formulas and theories in
isolation is driving me nuts, they remain incongruent meaningless exercises unless I can combine them into a cohesive picture in this one simple
reaction. A bit like Gestalt psychology that I learned many years ago.
The Practical:
Aluminium metal in zinc sulphate solution, in a boiling tube at high temperature, hydrogen gas yielded reaction proceeds slowly.
2Al + 3ZnSO4 = 3Zn +Al2(SO4)3
Al reduces Zn, Al loses 3 e- while Zn needs 2 e- to become the displaced metal.
The Theory:
Zn has electronegativity of 1.65 / Al @ 1.61 zinc has a stronger affinity for e- than Al.
Question? e- affinity for Zn is 0 while for Al is 42.5 indicating the opposite tendency of the above electronegativity?
Ionization energy of Zn is 906 while Al is 577 meaning that Zn is more likely to give away e- than Al. Question? Is this
counter-productive at all to above reaction where Zn has 2+ charge and needs to gain 2 e-
Radius density of Zn is 142 pm (can't type A with 0 above it) Al has 118 pm; However I did read that this high radius density explains why in
solution ZnSO4 will hydrolize and produce a pH of around 4 - 5 though litmus tests indicate neutral still. Question?: The
ability of Zn to attract the e- from H2O and form hydronium and release H+ into solution was put down to this high density of the e- in Zn
radius, very difficult finding information about what is classed as high and what is classed as low. (Apart from this document http://www.saylor.org/site/wp-content/uploads/2011/06/Ionic-... where they are at odds with wikpedia anyway), excepting that low is 30 pm and high
is 200 pm, what is the relationship between the ability of Zinc to hydrolyse and another metal 'not' hydrolysing. Why does the density of the Zinc
cation in this instance determine whether the pH goes up or down in a solution?
Question? The Al needs to lose 3 e- to bond with sulphate, how much of this is determined by the pH of the solution, and how much of
this is determined by the fact that Al is higher on the activity series than Zn? And if pH is a strong factor then adding
H2SO4to the boiling tube should speed up the reaction?
[Edited on 13-2-2015 by CHRIS25]
[Edited on 13-2-2015 by CHRIS25]
‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some
Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)
Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)
The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by
precision and law. (me)
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blogfast25
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You are looking too far afield.
In watery medium ZnSO4 dissociates more or less completely:
ZnSO<sub>4</sub>(s) === > Zn<sup>2+</sup>(aq) + SO<sub>4</sub><sup>2-</sup>(aq)
The redox reaction then is:
3 Zn<sup>2+</sup>(aq) + 2 Al(s) ===> 3 Zn(s) + 2 Al<sup>3+</sup>(aq)
This reaction proceeds because the Gibbs Free Energy change ΔG is negative (< 0). This can be gleaned from the reduction/oxidation potentials of:
Zn<sup>2+</sup>(aq) + 2 e ===> Zn(s)
and:
Al(s) ===> Al<sup>3+</sup>(aq) + 3 e
Task: look up these potentials and work out the overal potential E for this redox reaction.
The factors you mention play a part but to determine ΔG ALL enthalpies and entropies need to be taken into account.
Electron affinity is really only used for non-metals.
[Edited on 13-2-2015 by blogfast25]
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CHRIS25
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What you ask is new ground still, not understanding really what these figures, in practical terms, really mean. However I arrived at +1.57v. And how
this relates to gibbs I have no idea, still trying to grapple with this. After 4 hours no zinc. Nevermind, back to the bloody internet again......
‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some
Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)
Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)
The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by
precision and law. (me)
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|
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blogfast25
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For the oxidation potential of Al I find + 1.66 V, for the reduction potential of Zn<sup>2+</sup> I find - 0.76 V.
http://hyperphysics.phy-astr.gsu.edu/hbase/tables/electpot.h...
For the overall redox potential both values have to be added up:
1.66 + (- 0.76) = + 0.9 V
This positive value indicates the reaction can proceed (ΔG < 0).
Unfortunately Zn itself can be oxidised by H<sub>3</sub>O<sup>+</sup> and this is the reason why you're not seeing any Zn
metal.
It's entirely normal: electropositive metals like Zn cannot be plated out in watery solution. The overall potential for the oxidation of Zn metal by
hydronium is also positive!
Try this reaction again with copper(II) sulphate instead of zinc sulphate. Contrast and report.
Incidentally, for an electrochemical redox reaction ΔG = - n F E
with n the number of electrons transferred, F Faraday's Constant and E the overall potential. So if E > 0, then ΔG < 0.
http://en.wikipedia.org/wiki/Nernst_equation
[Edited on 13-2-2015 by blogfast25]
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deltaH
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Kinda reminds me of the time blogfast sublimed AlCl3 by reacting Al with ZnCl2 anhydrous... I think the thread was named
chlorothermal reactions or some such nonsense 
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CHRIS25
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Ok, this is was is my big gripe....exactly what I was trying to understand ...the theoretical before the practical...Zinc can not be reduced from a
zinc sulphate solution due to hydrolysis, but I don't get at all what you mean by the fact that zinc is electropositive? and how can you tell that
zinc will be hydrolized anyway?
Is this "theory' correct, and I know it works because I have done it so many times, Ag is the oxidizer so the number here is +0.8 and the reducer
copper is +0.52 (if I am reading these darn charts correctly that is), therefore the reduction potential of Copper is + 1.32? We know that silver can
not be hydrolized....but how? How do we know that. I am reading about hydrolyisis all day and still do not know the bloody answer, bloody problem
with the internet is that you can not ask the teacher, darn bloody explanations leave you always with more bloody questions.
PS I can not do that copper sulphate reaction - I have no bloody zinc!
PPS Oh and appreciate that link, I already have one but this is better and the link is fantastic for getting me even more deeper into trouble and
confusion, but very much needed anyway.
PPPS You mean that the oxidation potential of H3O is +0.83?
[Edited on 13-2-2015 by CHRIS25]
[Edited on 13-2-2015 by CHRIS25]
[Edited on 13-2-2015 by CHRIS25]
‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some
Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)
Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)
The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by
precision and law. (me)
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blogfast25
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Quote: Originally posted by deltaH  | | Kinda reminds me of the time blogfast sublimed AlCl3 by reacting Al with ZnCl2 anhydrous... I think the thread was named
chlorothermal reactions or some such nonsense |
Huh? I'm sure it wasn't called that. Please explain.
http://www.sciencemadness.org/talk/viewthread.php?tid=30150
Oh, it was YOU! 
Chris25:
Electropositive elements in terms of electrochemistry are defined as having an oxidation potential for:
M(s) === > M<sup>z+</sup>(aq) + z e
that is larger than 0 V. So E<sub>Ox, M</sub> > 0
That is because for:
H<sup>+</sup> + e ==> 1/2 H<sub>2</sub>
... the reduction potential E<sub>red, H+</sub> = 0 V
For these metals the E for:
M + z H<sup>+</sup> ===> M<sup>z+</sup> + z/2 H<sub>2</sub>
is always positive: E = E<sub>Ox, M</sub> + E<sub>red, H+</sub> > 0
Such metals are always oxidisable by hydronium. But that is NOT a hydrolysis, BTW, it's a simple oxidation.
For silver E<sub>Ox, Ag</sub> = - 0.80 V and the overall E = - 0.80 + 0 = - 0.80 V
Silver (and other non-electropositive elements) cannot be oxidised by water or dilute non-oxidising acids.
You don't need any 'bloody zinc', only Al and copper sulphate:
3 CuSO4(aq) + 2 Al(s) === > 3 Cu(s) + Al2(SO4)3(aq)
Here metallic copper will precipitate because E<sub>Ox, Cu</sub> = - 0.34 V. Copper is not an electropositive element and is not oxidised
by hydronium ions.
DO IT!
[Edited on 13-2-2015 by blogfast25]
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deltaH
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The full title is "Chlorothermal reactions (chloride thermites) for preparing AlCl3 and KAlCl4" in the 'Beginnings' section.
Me? No, just my idea (10% inspiration), but you actually DID it (90% perspiration) 
[Edited on 13-2-2015 by deltaH]
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blogfast25
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Here's a few metals organised by E, electropositive ones on top, hydrogen separating the two groups (electro +/electro -):
http://www.chemguide.co.uk/physical/redoxeqia/ecs.html
[Edited on 13-2-2015 by blogfast25]
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CHRIS25
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Quick annoying question then, what is the point of a reactivity series of metals where, for example, aluminium metal (with no charge that is) can
displace zinc out of solution when it clearly can't? (noting of course that aluminium is not to blame but rather the zinc)
PS ".....Too often, you will find the equations involved written as one-way rather than reversible. That small mistake makes the whole topic quite
unnecessarily difficult to understand.".....totally agree! http://www.chemguide.co.uk/physical/redoxeqia/introduction.h...
[Edited on 13-2-2015 by CHRIS25]
‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some
Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)
Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)
The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by
precision and law. (me)
|
|
|
blogfast25
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| Quote: Originally posted by CHRIS25 | Quick annoying question then, what is the point of a reactivity series of metals where, for example, aluminium metal (with no charge that is) can
displace zinc out of solution when it clearly can't? (noting of course that aluminium is not to blame but rather the zinc)
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Because the 'reactivity series' (an annoyingly shitty term if there ever was one!) isn't there to determine which metal can be displaced by which
other one!
The electrochemical series, a collection of reduction half-potentials, finds all kinds of uses, that particular one only being one of them. And while
it tells you Al can displace Zn it also tells you that same Zn will be chewed up by water.
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CHRIS25
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I came back to delete this stupid question, getting tired, realised what I had asked duh...
‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some
Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)
Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)
The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by
precision and law. (me)
|
|
|
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