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Author: Subject: H2SO4 by displacement of CuSO4 with H2?
bluflame
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[*] posted on 5-9-2010 at 10:15
H2SO4 by displacement of CuSO4 with H2?


We know that more reactive metals will displace less reactive cations - such as when hydrogen gas is liberated when an acid attacks a metal. However, copper is less reactive then hydrogen itself - shouldn't it be replaced by hydrogen when they come in contact? The reaction H2 + CuSO4 -> H2SO4 + Cu seems to be a very easy route to sulfuric acid if that is the case. Is there a reason why this wouldn't work?
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Jor
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[*] posted on 5-9-2010 at 13:40


I have not tried this, but I think it would work, but the reaction would be very slow. Considering that you will probably waste a LOT of hydrogen (I think the absorption will be very inefficient due to low reaction speed and bubbling of H2 out of solution), this is not practical for making usable amounts of sulfuric acid.
You could however electrolyse a solution of copper sulfate, wich would form sulfuric acid, copper and oxygen. There is also a YouTube video where this is done.
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mr.crow
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[*] posted on 5-9-2010 at 13:51


You aren't replacing Cu with H2, you are replacing Cu2+ with 2 H+.

Where does the Cu2+ go? It needs to be reduced to Cu metal or precipitated as an insoluble compound.




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bluflame
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[*] posted on 5-9-2010 at 13:54


Quote: Originally posted by mr.crow  
You aren't replacing Cu with H2, you are replacing Cu2+ with 2 H+.

Where does the Cu2+ go? It needs to be reduced to Cu metal or precipitated as an insoluble compound.


It's a redox reaction: Cu2+ + H2 -> Cu + 2H+. The copper should precipitate out.
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[*] posted on 5-9-2010 at 15:23


If this is possible, we could use 19% HCl acid on iron to make hydrogen production slower.
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[*] posted on 5-9-2010 at 16:41


The fact is that hydrogen is displaced from most metal plus acid reactions as a very stable diatomic gas.
The reverse reaction is very unfavourable from a thermodynamic and equilibrium perspective.
Some reactions can be 'pushed backwards' compared with how they normally occur, this is not one of them!
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[*] posted on 5-9-2010 at 23:00


Hydrogen is not displaced by copper. I indeed can imagine that hydrogen could replace copper, but the conditions, needed for that probably are not very good. There are examples where hydrogen indeed replaces copper, e.g. if hydrogen gas is passed over hot CuO, then first Cu2O is formed and finally Cu. The oxygen then is bound to the hydrogen, giving water.

With red hot CuSO4 (anhydrous) this also could happen, but I doubt if the hydrogen only replaces copper. I can imagine that besides that, the sulfate also is reduced. Keep in mind that at the high temperatures required the sulfate ion also is a good oxidizer and it can be reduced to SO2, H2S and H2O by hydrogen. So, the final product to my opinion would be impure Cu-metal and a mix of SO2, S, H2S and H2O.




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kmno4
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[*] posted on 5-9-2010 at 23:59


Quote: Originally posted by bluflame  
We know that more reactive metals will displace less reactive cations - such as when hydrogen gas is liberated when an acid attacks a metal. However, copper is less reactive then hydrogen itself - shouldn't it be replaced by hydrogen when they come in contact? The reaction H2 + CuSO4 -> H2SO4 + Cu seems to be a very easy route to sulfuric acid if that is the case. Is there a reason why this wouldn't work?

This reaction is of course thermodynamically possible.
The only "but" is that H2 must be activated (presence of Pd, Pt....).
Without activation H2 will not reduce even Mn(VII) (=KMnO4), oxidizer much stronger than Cu(II).
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DDTea
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[*] posted on 6-9-2010 at 00:35


Quote: Originally posted by bluflame  
We know that more reactive metals will displace less reactive cations - such as when hydrogen gas is liberated when an acid attacks a metal. However, copper is less reactive then hydrogen itself - shouldn't it be replaced by hydrogen when they come in contact? The reaction H2 + CuSO4 -> H2SO4 + Cu seems to be a very easy route to sulfuric acid if that is the case. Is there a reason why this wouldn't work?


The term "reactive" has a very broad meaning and you should always specify what something is reactive *toward.* However, from the context, I understand what you mean.

The proposed reaction that you're speaking of is:

H2(g) + CuSO4(aq) --> H2SO4 (aq) + Cu(s)

Let's get rid of the spectator ion (SO4):

H2(g) + Cu(2+) (aq) --> 2 H(+) (aq) + Cu(s)

There are two principles that will determine whether this reaction is plausible: equilibrium and kinetics. Kinetics is a lot harder to measure, but equilibrium can easily be calculated by the change in free energy of the reaction. The relevant expression for the change in free energy is:

deltaG = -nF(deltaE)

where deltaG = Gibb's free energy
n = number of moles of electrons involved in redox reaction
F = Faraday's constant, ~ 96 485 C/mol
deltaE = cell potential

The relevant half-cell reactions here are:

Cu(2+) + 2e(-) --> Cu(s) , E = +0.340 V

2H(+) + 2e(-) --> H2 (g) , E = 0.000 V (by definition of standard cell potentials)

What's getting oxidized in our desired reaction? Hydrogen gas, right? That means, then, that Cu(2+) is getting reduced. So,

Cu(2+) + 2e(-) --> Cu(s), E = +0.340 V (cathode)

H2(g) ---> 2H(+) + 2e(-) , E = 0.000 V (anode)

become our half-cell reactions. So to use the Nernst equation, we need to calculated deltaE. To do that, we calculate: E(cathode) - E(anode) = (+0.340 V) - (0.000 V) = +0.340 V .

So, to calculate deltaG now:

deltaG = -nF(deltaE) = -(2 mols e)(96 485 C/mol e)(0.340 V) = -65 610 J = -65.6 kJ

The free energy change is large and negative, so in principle, this reaction should happen spontaneously. However, that says nothing about the kinetics of the reaction (well, it does, but that would get kind of obtuse). Also, all of this assumes the reaction is occurring under standard states (i.e.: 20 C, 100 kPa for H2, and 1 M concentration of CuSO4.

And I always screw up these calculations, so it wouldn't surprise me if I got something wrong here.

EDIT: Just kidding--it wasn't the Nernst equation!

[Edited on 9-6-10 by DDTea]




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Quantum_Dom
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[*] posted on 6-9-2010 at 07:45


Quote:


I'm going to do that with the Nernst equation:

deltaG = -nF(deltaE)

So to use the Nernst equation, we need to calculated deltaE. To do that, we calculate: E(cathode) - E(anode) = (+0.340 V) - (0.000 V) = +0.340 V .

So, to calculate deltaG now:

deltaG = -nF(deltaE) = -(2 mols e)(96 485 C/mol e)(0.340 V) = -65 610 J = -65.6 kJ




Nice DDTea ! Its very rare to see quality post that involves quantitative explanation based on thermodynamics in order to explain a chemistry concept. My compliments.

Sorry to get technical, but Im affraid the expression you using is simply the Gibb's free energy relation, not Nernst equation. The latter express the relation between the reduction potential of an half-cell, and the activities of the species involved, at equilibrium or the electromotive force of the full electrochemical cell.

QD

[Edited on 6-9-2010 by Quantum_Dom]




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DDTea
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[*] posted on 6-9-2010 at 08:16


Quote: Originally posted by Quantum_Dom  


Sorry to get technical, but Im affraid the expression you using is simply the Gibb's free energy relation, not Nernst equation. The latter express the relation between the reduction potential of an half-cell, and the activities of the species involved, at equilibrium or the electromotive force of the full electrochemical cell.

QD

[Edited on 6-9-2010 by Quantum_Dom]


Whoops, you're right. It *isn't* the Nernst equation. That's what I get for doing quantitative work in the dead of night :P

[Edited on 9-7-10 by DDTea]




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[*] posted on 7-9-2010 at 08:41


Yes, the reaction proceeds, but you need something to get the H2 into solution. It normally doesn't.

The hydrogen reference cell is done by bubbling (a mist, really) H2 gas at 1atm over a catalytic platinum mesh electrode.

In this case, you'd get copper plated onto said electrode, which would be a bit undesirable, and probably stop the reaction quickly. Copper may have some reactivity with hydrogen, but that may again change as it plates out (if levelers are used, the smooth surface will have less surface area).

The "hot" process mentioned above is probably the best. Disadvantages include H2SO4 distillation and dissociation. Of note, SO3 will be reduced, forming SO2 and more H2O. The resulting condensation product will be reducing, from the presence of H2SO3, and some gas will be left over (excess SO2).

Tim




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[*] posted on 29-4-2015 at 11:29


It actually is the nernst equation but it's for the ideal situation where all the concentrations are 1M and the activities are 1... in this case the logarithmic part cancels out.

I want to make the opposite CuSO4, SnSO4 and NiSO4 has anybody tried to make them using a battery like this https://www.youtube.com/watch?v=arlYPz3EP7A ? The reactions for Sn,Ni,Zn are favourable but I don't know about the kinetics of the reaction.


[Edited on 29-4-2015 by gatosgr]
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[*] posted on 29-4-2015 at 22:10


H2 is a very inert substance. It gets much more reactive at temperature of 200°C+, but I don't think you would want to mess around with hydrogen even at 200°C.
Also, solubility of hydrogen is a problem, so you need few MPa of pressure to make this reaction work with noticable rate.
Btw, hot CuO powder can be reduced with molecular hydrogen at 150-250° : CuO + H2 ==> Cu + H2O
You could try to use some platinum/palladium/nickel catalyst which is capable of absorbing the hydrogen efficiently, though I have no real examples to show.
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[*] posted on 30-4-2015 at 08:58


When I've done the reduction of CuO by H2 gas, I don't know what temperature the reaction starts at, but I can tell you that the pile of CuO "ignites" and a burn front moves through it until the pile is completely converted to Cu. It looks like the pile is "burning", but of course it is "unburning" (getting reduced).

With Fe2O3 there is no "burning"; the reaction is endothermic, but proceeds with the removal of the H2O by the gas stream; the iron metal produced is pyrophoric in my experiments (where I used fine rouge as for my iron oxide).




Any other SF Bay chemists?
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