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smaerd
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[*] posted on 7-9-2015 at 09:05
Electrochemistry Questions


So I am taking a course on electrochemistry. Unfortunately what I know about electrochemistry is more from the deeper end of the physical theory and I have never bothered to explore the fundamentals beyond general chemistry. So I could use some basic help!

Say I am trying to come up with an electrochemical cell that will allow for a reaction. Let's keep it incredibly simple. Say the reaction of interest is

H20 <-> H+ + OH-

Now this reaction can be easily broken into two pieces with standard reduction potentials, IE:
2H+ + 2e- <-> H2 & E* = 0V
2H2O + 2e- <-> H2 + 2OH- & E* = -0.828V

So the net rxn is

H2O <-> H+ + OH- & E* = -0.828V

No big deal. It's no surprise that the [s]autodissociation of water is spontaneous[/s] (Thanks DeltaH). Clearly no salt bridge is required as this is one solution.

Questions:
1. But how do I select proper electrodes to facilitate this reaction (or a more involved reaction)?

It seems like if a process is spontaneous/galvanic the best choice of electrode would be whatever salt is being reduced/oxidized (Ex: Sn2+ + 2e- -> Sn (s) thus Sn would be an electrode). In this case I'm don't think I need to even give an electrode pair at all?

When it is not spontaneous though how do I select electrodes?

2. Although water is a pure substance, it is involved in the redox process so wouldn't it need to be written in the cell notation?

Ex: Electrode 1 | H+ (ag), OH- (aq), H20 (L) | Electrode 2

Any help appreciated! Feel free to over explain things, I could use as much of a refresher as possible!

[Edited on 7-9-2015 by smaerd]




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Sulaiman
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[*] posted on 7-9-2015 at 10:10


mostly answered here https://en.wikipedia.org/wiki/Electrochemistry

from what I've read and experimented a little,
carbon rods from lantern batteries are useful for most cells,
due to the generally inert nature of graphite,
and the cost of platinum etc.
and sometimes due to thermal stability,
and in its original cell, pressure relief by diffusion through the rod.

I have read descriptions of a plethora of electrochemical cells,
each normally gives details of electrodes suitable for a particular reaction,
so I guess that you would need to find a reaction similar to your intended reaction, and go from there.

for water 316/A4/marine stainless steel seems electrode material of choice.
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smaerd
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[*] posted on 7-9-2015 at 10:20


Thanks sulaiman,

I'm asking more from a theoretical point of view. As is does the cell potential between two electrodes dictate feasbility of an electrochemical reaction or what.

edit- like for example say I have the following half cells

1) 2e- + PbSO4 <->Pb + SO42- & E* = -0.35V
2) PbSO4 + 2H2O <-> 2e- + 4H+ + PbO2 + SO42- & E* = -1.69V

I can discern the cell would be not spontaneous/electrolytic. The anode could be Lead. What would be a suitable cathode? How do I decide that?

The cell notation would be something like (tell me if I made a big error)
Pb|PbO2(S)|4H+(aq), 2SO42-(aq), H2O(L)|PbSO4(S)|Cathode?


[Edited on 7-9-2015 by smaerd]




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[*] posted on 7-9-2015 at 11:25


this is what I meant by
"I guess that you would need to find a reaction similar to your intended reaction, and go from there."
e.g. https://en.wikipedia.org/wiki/Lead%E2%80%93acid_battery

for the tin example, tin would be deposited at a tin electrode,
it would also be deposited at almost any electrode material,
often loosely adhered.
The electrode materials may add to the voltage required to cause current/ion flow reducing efficiency / heating the cell.

[Edited on 7-9-2015 by Sulaiman]
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[*] posted on 7-9-2015 at 11:33


Quote: Originally posted by smaerd  


H2O <-> H+ + OH- & E* = +0.828V

No big deal. It's no surprise that the autodissociation of water is spontaneous. Clearly no salt bridge is required as this is one solution.


NO! When you mix an acid and base you release a lot of energy, not the other way around.

Secondly, the autoionisation of water is thermodynamically very unfavourable, Kw = 10^-14!!!




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smaerd
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[*] posted on 7-9-2015 at 11:34


I appreciate you offering me experimental advice. Again though I need a more theoretical understanding.

For example, if I was given a test and was asked to create an electrolytic cell to perform a reaction and do a calculation on said cell. I could not say "Google a similar reaction to that reaction and pick a similar cathode/anode". Do you understand my concerns with your advice? I do appreciate it though.

I have done electrochemical reactions using the method you are suggesting though, and it works in practice. My coursework just won't allow for it.




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[*] posted on 7-9-2015 at 11:37


Delta H - Thanks for catching that mistake. That was supposed to be a negative sign!



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[*] posted on 7-9-2015 at 11:37


Quote: Originally posted by smaerd  
Thanks sulaiman,

I'm asking more from a theoretical point of view. As is does the cell potential between two electrodes dictate feasbility of an electrochemical reaction or what.

edit- like for example say I have the following half cells

1) 2e- + PbSO4 <->Pb + SO42- & E* = -0.35V
2) PbSO4 + 2H2O <-> 2e- + 4H+ + PbO2 + SO42- & E* = -1.69V

I can discern the cell would be not spontaneous/electrolytic. The anode could be Lead. What would be a suitable cathode? How do I decide that?

The cell notation would be something like (tell me if I made a big error)
Pb|PbO2(S)|4H+(aq), 2SO42-(aq), H2O(L)|PbSO4(S)|Cathode?


[Edited on 7-9-2015 by smaerd]


This is the lead acid battery, it works very well. The equations as you have written is in the charging direction, when discharging it's the reverse and the voltage is +0.35 + 1.69 = + 2.04V under standard conditions.




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[*] posted on 7-9-2015 at 11:39


'Standard Conditions'

Intriguing.

Do the voltages change for Non-STandard conditions ?




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[*] posted on 7-9-2015 at 11:41


yes, you use the Nernst equation to work out the non-standard conditions voltages, e.g. higher acid concentrations than 1M and temperature variation, etc.



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[*] posted on 7-9-2015 at 11:43


Thanks again DeltaH for the insight. Can you give me some advice on choosing electrodes for reactions. For example where is the cathode in that reaction? Or how can I determine a cathode or anode to make a reaction occur in general?

Is the PbSO4-2 the cathode? Or does PbSO4-2 coat a Pb cathode over time? How can I determine that by looking at two half cell reactions.

[Edited on 7-9-2015 by smaerd]




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[*] posted on 7-9-2015 at 11:45


Cathode is ALWAYS the one attracting the cations and anode ALWAYS the one attracting the anions, hence the name. I trust you can tell now?

BTW, the lead acid battery is interesting, because you end up with same species generated by both half reactions after discharge (PbSO4). In some distant chemically romantic way it can be viewed as a battery that reacts a practical hydrogen source with a practical oxygen source to make water, where the hydrogen came from acid reacting with lead and the oxygen came from acid reacting with lead dioxide (albeit very heavy and toxic sources for those two). The voltage is not quite the same, but then again it isn't quite hydrogen and oxygen either :P

The PbSO4 is mostly insoluble in the real lead acid battery, it coats the lead sponge as the battery discharges, but there is some equilibrium with the electrolyte so the reaction can be reversed.

Choosing of electrodes is mostly down to the practicality of having it survive. Lead resists sulfuric acid when a no potential exists, so that makes a good choice for the cathode. Lead dioxide is electrically conductive, so it works as an anode (pressed around a lead grid as current collector AFAIK).

Anodes are generally taking a beating in terms of corrosion, so they are usually the one to watch out for. You want the anode material not to be oxidised away under the conditions and chemistry of the battery in question (i.e. it is dependent on the battery).

Lead dioxide doesn't spontaneously react with sulfuric acid, so it works as the anode (same way lead metal doesn't spontaneously react with sulfuric acid and hence works as the cathode).

[Edited on 7-9-2015 by deltaH]




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[*] posted on 7-9-2015 at 12:15
Brainteaser


The astute may now be wondering WHY lead metal doesn't spontaneously react with the sulfuric acid electrolyte to make hydrogen gas AND why the lead dioxide doesn't spontaneously react to form water and oxygen?! After all, lead has a positive voltage referenced to SHE, so you would expect that to occur and similarly a PbO2 reaction with acid to make oxygen and water is also positive.

Care to take a stab at it?

CLUE: same reason when you charge it, it doesn't simply make hydrogen and oxygen (though a small amount usually forms, called out-gassing).

[Edited on 7-9-2015 by deltaH]




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[*] posted on 7-9-2015 at 12:16


That's really cool :). I knew about lead acid batteries but never thought of them in that way.

I know what a cathode is :). Thats good practical advice on selecting anodes!

Maybe I should give a better example...

Say I have these two half reactions (this is way more clear)
Reduction
4H+ + O2+ 4e- <-> 2H2O E* = -0.133V
Oxidation
2H2 <-> 4H+ + 4e- + E = 0.0V

So the net reaction is O2+ 2H2 <-> 2H2O

The reaction is clearly NOT spontaneous. So what two electrodes can I choose to do this reaction? Can it really be any two random pieces of metal?

I'm not talking about corrosion or degradation or pressure concerns. Is the standard potential of the electrode materials relevent in this decision?

I think that I need to select an anode with a more positive potential than the cell reaction and a cathode more negative. Is this right?


[Edited on 7-9-2015 by smaerd]




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[*] posted on 7-9-2015 at 12:17


Quote: Originally posted by deltaH  
The astute may now be wondering WHY lead metal doesn't spontaneously react with the sulfuric acid electrolyte to make hydrogen gas AND why the lead dioxide doesn't spontaneously react to form water and oxygen?! After all, lead has a positive voltage referenced to SHE, so you would expect that to occur and similarly a PbO2 reaction with acid to make oxygen and water is also positive.

Care to take a stab at it?

CLUE: same reason when you charge it, it doesn't simply make hydrogen and oxygen (though a small amount usually forms, called out-gassing).

[Edited on 7-9-2015 by deltaH]


Is it because of a kinetic effect? Or passivation of the electrode?




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[*] posted on 7-9-2015 at 12:18


Quote: Originally posted by smaerd  
That's really cool :). I knew about lead acid batteries but never thought of them in that way.

I know what a cathode is :). Thats good practical advice on selecting anodes!

Maybe I should give a better example...

Say I have these two half reactions (this is way more clear)
Reduction
H2O2 + 4H+ + O2+ 4e- <-> 2H2O E* = -0.133V
Oxidation
2H2 <-> 4H+ + 4e- E = 0.0V

So the net reaction is O2+ 2H+ <-> H2O

The reaction is clearly NOT spontaneous. So what two electrodes can I choose to do this reaction? Can it really be any two random pieces of metal?

I'm not talking about corrosion or degredation or pressure concerns. Is the standard potential of the electrode materials relevent in this decision?

I think that I need to select an anode with a more positive potential than the cell reaction and a cathode more negative. Is this right?



[Edited on 7-9-2015 by smaerd]


Your reactions aren't right again.




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[*] posted on 7-9-2015 at 12:21


Sorry I editted the original post! I added an extra H2O2 which was an intermediate step.



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[*] posted on 7-9-2015 at 12:23


Quote: Originally posted by smaerd  
Quote: Originally posted by deltaH  
The astute may now be wondering WHY lead metal doesn't spontaneously react with the sulfuric acid electrolyte to make hydrogen gas AND why the lead dioxide doesn't spontaneously react to form water and oxygen?! After all, lead has a positive voltage referenced to SHE, so you would expect that to occur and similarly a PbO2 reaction with acid to make oxygen and water is also positive.

Care to take a stab at it?

CLUE: same reason when you charge it, it doesn't simply make hydrogen and oxygen (though a small amount usually forms, called out-gassing).

[Edited on 7-9-2015 by deltaH]


Is it because of a kinetic effect? Or passivation of the electrode?
It is indeed a kinetic effect yes! Overpotential. The standard hydrogen electrode to which standard electrode potential table is quoted is using platinum (a fantastic hydrogenation catalyst) so that the kinetic effect is minimised. But lead is a piss poor hydrogenation catalyst, so some activation potential has to be sacrificed to get it over the activation energy on the lead surface, this results in a lead overpotential that significantly raises the hydrogen producing half reaction's potential. The same principle holds for lead oxide forming oxygen, large overpotential for that too.

Add activated platinum to both sides of a lead acid battery and it would fail miserably, it works so well because it doesn't work so well [as a catalyst for those reactions] :P

[Edited on 7-9-2015 by deltaH]




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[*] posted on 7-9-2015 at 12:24


Quote: Originally posted by deltaH  
The astute may now be wondering WHY lead metal doesn't spontaneously react with the sulfuric acid electrolyte to make hydrogen gas AND why the lead dioxide doesn't spontaneously react to form water and oxygen?!

6 weeks please to formulate an answer.

Still reeling from blogfast25's QM introduction.

So much to learn.

Chucking in BM (Battery Mechanics) so soon would cause overloadings of the brain (such as it is).

Edit:

I was too slow.

Reading the explanation would have Lobotomised my brain had i not been, as ever, very well prepared.

[Edited on 7-9-2015 by aga]




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[*] posted on 7-9-2015 at 12:25


Quote:
So the net reaction is O2+ 2H+ <-> H2O
This is still not right though.



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[*] posted on 7-9-2015 at 12:27


Damn it lol Its supposed to be 2H2! I'll fix it.

Now SciMad knows why my posts are always editted 10 times. :D

[Edited on 7-9-2015 by smaerd]




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[*] posted on 7-9-2015 at 12:30


Quote: Originally posted by smaerd  
Damn it lol Its supposed to be 2H2! I'll fix it.

Now SciMad knows why my posts are always editted 10 times. :D

[Edited on 7-9-2015 by smaerd]


Mine too, I'm notorious for this! You can re-read this thread, mine have all changed lol.




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[*] posted on 7-9-2015 at 12:32


Quote:
So the net reaction is O2+ 2H2 <-> H2O


Still not right :P

...and how can you say that the burning of hydrogen is clearly not spontaneous, come on :o:o:o

Remember it's over platinum and there's a sign error there!!! Blow hydrogen and oxygen at 1bar over platinum gauze and see what happens :D


[Edited on 7-9-2015 by deltaH]




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[*] posted on 7-9-2015 at 12:41


Wow thanks, yea its 2H2O wowww what is wrong with me. Okay yea I had the wrong half cell potential for one of my reactions. It's +2.46V for the entire cell. So it is quite spontaneous! Thus no electrodes are needed.

Well, in reality that reaction doesn't readily occur. Then again that could be due to other effects as you elucidated before.


Hmm so every half reaction should include the electrode of interest? Is that the idea I am missing by making little errors?

[Edited on 7-9-2015 by smaerd]




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[*] posted on 7-9-2015 at 12:53


Quote:
Thus no electrodes are needed.


I don't really understand where you going with this (or coming from). All reactions used for cells could be carried out without electrodes if you remove the physical separation between the oxidant and reductant and you will spontaneously release heat, usually a hell of a lot. That's kinda chemically shorting the cell.

By physically separating them and running an electrical wire between the two halves, you force the electrons to go via the external pathway. If there's negligible resistance, you still electrically short the cell and get lots of heat evolved anyway, same as chemically shorting the cell.

As for the electrodes, they're just a means to collect current since often the reactants are not very electrically conductive, however, some half reactions, often those that involve gases, need a catalyst, hence why platinum is used for their standard potentials, the catalytic case is a special case.




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