Sciencemadness Discussion Board - Electrochemistry Questions

Electrochemistry Questions

smaerd - 7-9-2015 at 09:05

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]

Sulaiman - 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.

smaerd - 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]

Sulaiman - 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]

deltaH - 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!!!

smaerd - 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.

smaerd - 7-9-2015 at 11:37

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

deltaH - 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.

aga - 7-9-2015 at 11:39

'Standard Conditions'

Intriguing.

Do the voltages change for Non-STandard conditions ?

deltaH - 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.

smaerd - 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]

deltaH - 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

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]

Brainteaser

deltaH - 7-9-2015 at 12:15

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]

smaerd - 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]

smaerd - 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?

deltaH - 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.

smaerd - 7-9-2015 at 12:21

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

deltaH - 7-9-2015 at 12:23

Quote: Originally posted by smaerd

Quote: Originally posted 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]

aga - 7-9-2015 at 12:24

Quote: Originally posted by deltaH

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]

deltaH - 7-9-2015 at 12:25

Quote:

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

This is still not right though.

smaerd - 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.

[Edited on 7-9-2015 by smaerd]

deltaH - 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.

[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.

deltaH - 7-9-2015 at 12:32

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]

smaerd - 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]

deltaH - 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.

Brainteaser #2

deltaH - 7-9-2015 at 12:58

I made the statement that gases used in a cell generally require a catalyst, again... why?! This time a simple "for kinetic reasons" is too simple, I want a deeper answer.

Clue: Ideal gas law.

[Edited on 7-9-2015 by deltaH]

deltaH - 7-9-2015 at 13:15

OK, I'm tired and going to bed, so here's the answer not to keep you hanging...

We all know that chemical kinetics is a function of the concentration of reagents [to various powers]. For gases, one can re-arrange the ideal gas law such:

PV = NRT

<=> N/V = P/RT

Now N/V is nothing else than a concentration... like moles per litre

so for a gas, the concentration is P/RT, so at atmospheric pressure and room temperature, this amounts too:

(1atm)/(0.08205736L atm K−1 mol−1)/298K = 0.04 mol/l

Pretty f@#$%ing dilute!

Now on top of this, sometimes the gases are non-polar, e.g. oxygen and hydrogen, and so they are also extremely insoluble in the liquid phase (oxygen saturates in water in the ppm range).

No wonder the kinetics are slow with reagent concentrations like that and that's not even looking at the activation energies for dissociating them(which are case specific).

smaerd - 7-9-2015 at 14:20

Sorry DeltaH I got lost reading stuff about electrochemistry.

Yes I know the concentration of gasses in water is quite low, henry's law, etc.

I know platnium is catalytic for many hydrogen processes, but no clue about oxygen. Inface oxygen likely poison's platinums catalytical influence on those hydrogen proccesses.

Anyways, I am still completely stuck with my original question. How do I decide what electrodes can be used?

violet sin - 7-9-2015 at 20:54

my original question. How do I decide what electrodes can be.
______________________________

Will it carry enough current?
Will it withstand carrying that current in that solution?( without failing or polluting)
Will it be able to assist your purpose? (If it needs something special)
(don't forget,).. is it a valve metal placed in the right polarity?

Then it works

^^^ super simplified
______________________________
vvv less simplified, it really depends on what your doing.

- are you enacting a chemical reaction in one product by oxidation or reduction to make something new? like NaCl -> NaClO in a swimming pool sanitation system.
- are you destructively oxidizing the media directly like some of the papers on ammonia, NO(x), SO2 abatement. Attachment: CLEAN NOx & SO2.pdf (639kB)
This file has been downloaded 309 times.
- or are you simply plating out metal.

lots a ways to go with electrochemical cells, hence the oversimplified answer or the detailed oriented need for set-up constraints.
------------------------------------------

I had intended to throw more time at this answer, but family fist, sorry if it seems half-assed. I love electrochem. recently been reading some patents on the shor gold refining systems using ammonium chloride with nascent oxygen catalyst using gold scrap and a graphite rod as electrodes. ceramic diaphragm pot as a separator. no nitrates or fumes was a nice draw. http://www.google.com/patents/US5009755

[Edited on 8-9-2015 by violet sin]

this paper is somewhat informative about MMO Vs. Pt & graphite ( top of pg 4) Attachment: water_star.pdf (1.1MB)
This file has been downloaded 1277 times. only 9pg's long with some decent reading like common mode of failure for MMO. it also had this, among other things to say.

"in order to provide the most suitable MMO for a given application the electrode manufacturer needs specific information about the intended use. For example:- ... ..."

leading back to the simple/complicated way of answering will this metal work. special function or does it just have to not fail?

[Edited on 8-9-2015 by violet sin]

deltaH - 7-9-2015 at 22:29

Quote: Originally posted by smaerd

Sorry DeltaH I got lost reading stuff about electrochemistry.

Yes I know the concentration of gasses in water is quite low, henry's law, etc.

I know platnium is catalytic for many hydrogen processes, but no clue about oxygen. Inface oxygen likely poison's platinums catalytical influence on those hydrogen proccesses.

Anyways, I am still completely stuck with my original question. How do I decide what electrodes can be used?

Platinum is one of the better oxygen catalysts (relatively speaking) when also considering durability, so it's used in fuel cells on both sides. The oxygen overpotential is quite high though compared to hydrogen's, so there's much research into improving that and also in making non-precious metal catalysts with similar performance.

The wiki on 'overpotential' has a nifty table:
overpotentials.JPG - 31kB

Source: https://en.wikipedia.org/wiki/Overpotential

I don't know much about oxygen poisoning the surface of platinum, I would have imagined that in a hydrogen atmosphere, this is very easily converted to water and 'unpoisoned'

smaerd - 8-9-2015 at 02:41

Quote: Originally posted by deltaH

I don't know much about oxygen poisoning the surface of platinum, I would have imagined that in a hydrogen atmosphere, this is very easily converted to water and 'unpoisoned'

Show's what I know about catalysis

. Thanks for putting up with me DeltaH. I learned a good deal here.

I also found the answer to my question! So I read through 3 of my electrosynthesis books in their relevent sections. Turns out electrode selection is based on basically what sulaiman and DeltaH had said. It's typically experimentally determined based on corrosion, cost, etc. Although they do mention that some electrodes favor or disfavor reactions at certain potential differences. So there's no clear cut way of doing this. Like any good question it lead me to maybe 2 or 2000 more.

I'm still trying to wrap my head around the difference between the Cell potential verses the Electrode potential, and Open Circuit Potential (for trickier systems).

It seems like the cell itself has an potential based on a given redox reaction pair (something I am trying to do). Then depending on what voltage I apply to the physical electrodes I can either provide not enough electrode potential to instigate that reaction, enough to initiate the reaction, then if I go over board the excess potential results in more current/faster rate of the reaction.

What I'm trying to figure out now is how can I tell if the electrodes I pick have interfering reactions. Like if I pick a silver electrode with KBr then I have a rxn at 0.07V, but if I want to do a reaction at something like -0.10V the previous reaction would interfere with that happening(assuming I am driving electrode potential negatively).

[Edited on 8-9-2015 by smaerd]

deltaH - 8-9-2015 at 09:24

Quote:

Thanks for putting up with me DeltaH.

No putting up at all, I'm no expert in this, just sharing what I think I know

Quote:

What I'm trying to figure out now is how can I tell if the electrodes I pick have interfering reactions. Like if I pick a silver electrode with KBr then I have a rxn at 0.07V, but if I want to do a reaction at something like -0.10V the previous reaction would interfere with that happening(assuming I am driving electrode potential negatively).

Electrochemistry needs the details to be well defined or you may run into confusion.

Firstly, are these the half reactions you are referring to?

AgBr(s) + e- <=> Ag(s) + Br-(aq)...................E = +0.07133V
SnO(s) + 2 H+ + 2 e- <=> Sn(s) + H2O..........E = -0.10V

Now you say you're "driving the electrode potential negatively", I am not sure what you mean by this. Do you mean you are using it in this direction as written above and your other electrode is an oxidation reaction with more positive electrode potential, e.g.

Fe(s) <=> Fe(2+)(aq) + 2e-.........................E = +0.44V

smaerd - 13-9-2015 at 04:40

What I meant was applying an external voltage after the cell has reached it's open circuit potential. I think I understand these concepts now. Still don't know exactly how to go about calculating the open circuit potential. Sorry for the delay, I've been completely swamped!

The cell values I gave were completely arbitrary. My current understanding of deciding whether electrodes will have side reactions is literally looking up all the reactions I can think of in a table.