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smaerd
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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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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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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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Sulaiman
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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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deltaH
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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.
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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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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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smaerd
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Delta H - Thanks for catching that mistake. That was supposed to be a negative sign!
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deltaH
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| 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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aga
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'Standard Conditions'
Intriguing.
Do the voltages change for Non-STandard conditions ?
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deltaH
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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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smaerd
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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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deltaH
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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 
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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deltaH
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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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smaerd
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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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smaerd
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| 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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deltaH
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| 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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smaerd
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Sorry I editted the original post! I added an extra H2O2 which was an intermediate step.
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deltaH
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| 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]
[Edited on 7-9-2015 by deltaH]
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aga
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| 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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deltaH
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| Quote: | | So the net reaction is O2+ 2H+ <-> H2O |
This is still not right though.
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smaerd
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Damn it lol Its supposed to be 2H2! I'll fix it.
Now SciMad knows why my posts are always editted 10 times. 
[Edited on 7-9-2015 by smaerd]
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deltaH
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| 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.
[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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deltaH
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| Quote: | | So the net reaction is O2+ 2H2 <-> H2O |
Still not right
...and how can you say that the burning of hydrogen is clearly not spontaneous, come on 
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
[Edited on 7-9-2015 by deltaH]
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smaerd
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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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deltaH
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| 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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