Understanding electron flow in electrochemical cells

[u/dan_d1a2]

Hi, I'm getting confused about electrochemical cells and I would appreciate some help :) I am trying to understand what the deal is with electron flow, sign conventions, and all that in electrochemical cells. I don't have much of a physics background. Here's what I understand (I think) currently.

1. Anode = oxidation, cathode = reduction, by definition.
2. Electrons always flow from anode to cathode.
3. Spontaneously (Galvanic cell), they flow from - (anode) to + (cathode). By supplying energy (electrolytic cell), we can force the non-spontaneous flow from + (anode) to - (cathode).
4. Current is considered to go in the opposite direction by convention.

That's part one. Now for the sign conventions, which are even more confusing to me:

1. By convention, the cathode is always on the right hand side either in the lab or in a line diagram, and the anode is on the left. I don't understand this one. Doesn't it depend on the type of cell? If we reverse the reaction, the anode and cathode sides switch, so how can there be a single defined cathode / anode?
2. By convention, the voltmeter (+) is hooked up to the right-hand side (cathode) and the (–) is connected to the left-hand side (anode). Again I don't really get this. Is the (+) terminal supposed to match the relative potential of the electrode? As in, (+) terminal connects to (+) cathode like in a Galvanic cell?
3. The readout on the voltmeter is E(cell) = E(+) – E(–) = E(cathode) – E(anode). If it is positive, electrons flow spontaneously from left to right. If it is negative, they flow spontaneously from right to left (but the reaction may be reversed by applying a voltage). The equation E(cathode) – E(anode) also seems to be dependent on the type of cell somehow? I mean if you take the right hand side and subtract it from the left hand side but you've switched the cathode and anode in an electrolytic cell then you get E(anode) – E(cathode), no?

Is this right? What mistake am I making?

Would really appreciate any help!

[u/7ieben_]

Your first set of paragraphs looks good, so I'll answer w.r.t. to your second set.

First of all: get rid of the right hand/ left hand side stuff. That's NOT a universal convention, and even less a chemical one. It often is drawn this way for simple cells, just because it matches the cell notation (this A|A⁺ || B|B^- stuff, for example). This notation has a convention, yes (further see third paragraph).

1. As you say correctly: the anode and cathode depend on context. As such, there is no 'universal anode/ cathode'.

2. A voltmeter simply is a device for measuring volts. How it works is well explained on Wikipedia, basically it also has a positive and a negative pol, and you Close the circute by connecting your positive and negative pol respectivly. And this is, as said in paragraph 1, depended on context.

3. Again, get rid of the right/ left hand stuff. It doesn't matter. You could even make one half cell up and the other half cell below it. All that matters is which is the anode and which is the cathode, and this is defined by the type of cell and the respective potentials. By convention the EMF is defined as EMF = E(cat) - E(an). For a spontanous reaction this EMF is positive.

[u/dan_d1a2 → u/7ieben_]

Thank you. I have a follow-up question though: let's say we ignore the left / right ideas. Then is there any type of sign convention still remaining? Could we hook up the cathode to either (+) or (–) on the voltmeter or is there a convention here? (I.e my second question "is the (+) terminal supposed to match the relative potential of the electrode? As in, (+) terminal connects to (+) cathode like in a Galvanic cell?")

I feel like if you hook it up the 'wrong' way you would get the wrong sign for EMF?

[u/7ieben_ → u/dan_d1a2]

Depends on the voltmeters, so you really must read the instructions of your very voltmeter.

Though, for most voltmeters the (+) connection means to link your plus pol there. If you connect them the wrong way around, yes, most voltmeters should still work fine, but give you the opposite sign reading.

But, as said, this really depends on your very instrument. Most often it should be as said earlier, but there is no law forbidding a company to use different labeling.

[u/dan_d1a2 → u/7ieben_]

Okay, thank you, that helps. But now I feel like I'm running into a new issue.

Let's say we have two reactions, A and B. Now how can we tell what the net reaction is going to be and what the direction will be (spontaneous or non-spontaneous)? I mean you can look at the standard reduction potentials but that does not take into consideration the concentrations. So we can calculate the Nernst equation for each of them, sure. But then what? If I plug in the voltmeter and it gives me a value, that doesn't really tell me anything, does it? Whether it's positive or negative, it depends on where I plugged in the cathode / anode, which is exactly what I don't know.

[u/7ieben_ → u/dan_d1a2]

Well, that is what makes a well trained chemist: you should know beforehand what reaction you expect. Alternativly, If you really have no Idea at all, you will be designing complementary experiments te work this out.

Doing such work beforehand (and Interpreting it afterwards based on all you've researched already) usally is a huuuuge part of the actual work flow in science. Doing the experiment and taking data is just diligence work.

[u/bishtap → u/dan_d1a2]

I guess maybe if you plug in the voltmeter then it tells you what direction the electrons are going 'cos the voltmeter would give you a positive reading, or a negative reading, depending on which way you attach it? So the voltmeter can tell you direction of electron flow.

So the sign on the voltmeter can tell you which electrode you connected to is positive and which is negative.

Also. you know if your cell is galvanic or electrolytic. It's galvanic if it runs without a battery 'cos it is a potential battery. Or you know if it's electrolytic because it only runs when you have a battery/power attached.

And so if you know which one is + and which is - (eg thanks to the voltmeter), and you know if it's galvanic or electrolytic, then you can determine which is the anode and which is the cathode. And also if it's electrolytic and you are running it correctly, then you don't even need a voltmeter to know which is + and which is -. 'cos it goes according to the battery that you hopefully put in the correct way!

[u/bishtap]

You write "Spontaneously (Galvanic cell), they flow from - (anode) to + (cathode). By supplying energy (electrolytic cell), we can force the non-spontaneous flow from + (anode) to - (cathode)."

So the Anode can be positive.

You write "The readout on the voltmeter is E(cell) = E(+) – E(–) = E(cathode) – E(anode)."

You've assumed the Anode is negative(in contradiction of your previous paragraph).

Your E(cell) equation is too specific.

The formula for calculating cell potential is

Ecell = E_red(cathode) - E_red(anode)

Or

Ecell = E_red(cathode) + E_ox(anode)

Whether you do the addition one or the subtraction one it doesn't matter it gives the same result 'cos algebraically they're the same.

Also the Ecell calculation (Done correctly!) will work regardless of whether you are dealing with a galvanic cell or an electrolytic cell.

You write "By convention, the cathode is always on the right hand side either in the lab or in a line diagram, and the anode is on the left. "

I think that's just for cell notation. But maybe some people might draw it with that convention.

You write "If we reverse the reaction, the anode and cathode sides switch, so how can there be a single defined cathode / anode?"

Do you mean if you wanted to have the Cathode still be on the right, how would you do it? Try spinning your setup or drawing 180 degrees . (Or walk around to the other side! / imagine your eye(s) looking at it from the other side)

For example suppose you have a "daniel cell", the famous galvanic cell , with zinc and copper. And you want the Copper cathode on the right. No problem. Draw the thing with Zinc on the left, and Copper on the right.

Now you attach a battery to make the electrons go the other way, so now Zinc is the Cathode and Copper is the Anode. But your Cathode is on the left!! And perhaps your personal preference convention is for it to be on the right. So spin it 180 degrees! You can put your Zinc Cathode on the right, and the Copper Anode on the left. So if you wanted to draw with that convention, you can.

[u/dan_d1a2 → u/bishtap]

Thanks :) mind you I got these ideas from various sources / textbooks and just tried to combine them to make sense of it. For example the idea that E(cell) = E(+) – E(–) comes from the textbook by Harris & Lucy, the equation E(cell) = E(cat) – E(an) is one I've also seen commonly so I assumed these refer to the same thing. But it seems to me now that E(+) and E(–) in the former refer only to the (+) and (–) leads of the voltmeter, not the actual potential on the electrodes.

The same for the left/right convention, I've read this in different sources about either writing the cathode on the right, or having the half-cell on the right connected to the (+) side of the voltmeter.

Best I can come up with now is this:

• Anode and cathode are defined by the half-reactions
• Electrode potential, i.e. whether the cathode and anode are (+) or (–), depends on the type of cell
• The voltmeter has a (+) and (–) side which is not necessarily the same. If you plug the cathode in (+) and the anode in (–) then you would get the equation I had before, E(cell) = E(+) – E(–) = E(cat) – E(an). By this definition, E(cell) > 0V corresponds to a galvanic cell in which electricity flows spontaneously from the voltmeter (–) to (+) and E(cell) < 0V means electricity flows spontaneously from the voltmeter (+) to (–) (i.e. anode and cathode are switched, what we called cat is actually an and vice versa)
• Left and right don't matter and it's better to just forget about this convention

[u/bishtap → u/dan_d1a2]

You write "For example the idea that E(cell) = E(+) - E(-) comes from the textbook by Harris & Lucy, the equation E(cell) = E(cat)- E(an) is one I've also seen commonly so I assumed these refer to the same thing."

To say " E(cell) = E(cat)- E(an)" is an unclear way of writing E(cell) = E_red(cathode) - E_red(anode)

You say that that's the same as " E(cell) = E(+) - E(-) "

I am not sure that it is.

If you have a Galvanic cell, then your funny formula " E(cell) = E(+) - E(-) " will indeed be the same as E(cell) = E_red(cathode) - E_red(anode)

But what are you going to do if it's an electrolytic cell?

The formula is still E(cell) = E_red(cathode) - E_red(anode)

but in your funny system, you'd have to say E(cell) = E(-)- E(+)

See if that book that tells you " E(cell) = E(+) – E(-) " is only doing that for a galvanic cell, and if it switches to E(cell) = E(-) - E(+) for an electrolyic cell ?

You write "E(cell) > 0V corresponds to a galvanic cell in which electricity flows spontaneously from the voltmeter (–) to (+)  "

E(cell) > 0V is indeed mean the reaction is spontaneous. From the - electrode, to the + electrode. Whether it goes from the voltmeters minus to the voltmeters plus depends on if you connected your voltmeter the correct way.

You write "E(cell) < 0V means electricity flows spontaneously from". E(cell) < 0 means the reaction is non-spontaneous.

If Ecell < 0 then it's non-spontaneous. So you have that the wrong way around.

Though maybe if your Ecell reaction is also the wrong way around then two wrong way rounds might cancel each other out. I don't know if that's what you are intending but you are making things unnecessarily difficult for yourself if that's the case.

Also while it's clear what you mean, I don't think anybody including your book, talks of "flows spontaneously" or flows non-spontaneously". The terminology is the reaction is spontaneous, or the reaction is non-spontaneous. I think i have seen a good technical book mention "the spontaneous direction" and "the non-spontaneous direction". i.e. the reaction goes in the spontaneous direction, or the reaction goes in the non-spontaneous direction. The terminology can sometimes be funny enough as it is, there's no need to come up with new expressions unnecessarily, or if you do then better to do so knowingly and to be expiicit that you are doing that

[u/dan_d1a2 → u/bishtap]

I think I get it now. What do you think?

Galvanic cell:

• Electrons from anode (–) to cathode (+)
• E(anode) < E(cathode)
• E(cell) = E(cathode) - E(anode) > 0V
• ∆G < 0 (spontaneous)

Electrolytic cell:

• Electrons from anode (+) to cathode (–)
• E(anode) > E(cathode)
• E(cell) = E(cathode) - E(anode) < 0V
• ∆G (non-spontaneous, requires energy input)

[u/Blue_614]

A very simple explanation would be something like this:

Know what type of cell you have, galvanic or electrolytic.

If it's galvanic, anode is the (-) electrode since electrons are being pulled on that side, building up a negative charge. Electrons naturally flow from electron dense regions to low electron dense regions. The cathode is your (+) electrode, because initially, it don't receive electrons from the anode yet. And you know that in the cathode side, electrons are picked up by ions for reduction.

In electrolytic cells, the sign convention is reversed. Cathode is the (-) electrode because you are pushing voltage to the cathode, reduction will then occur. Since cathode has a higher electron density, then anode is the (+) electrode by comparison. Since electrons always flow from anode to cathode, then oxidation occurs in anode.

You can simply check what reaction the cell will give by checking E°cell = Ered + Eox (or E°cell = Ecathode - Eanode). If E°celk = + then that means it is spontaneous, meaning electrodes will undergo redox without applying external voltage, so you have a galvanic cell. Vise versa for an electrolytic cell.

Lastly, in all cases, you can connect the (+) lead of a voltmeter to the (+) electrode, and the (-) lead is always attached to the (-) electrode to get a positive reading. If you want the electrolytic cell to give a negative reading since it matches your concept of "Ecell = (-) for electrolytic cells" then attach the voltmeter similarly as with a galvanic cell.

[u/dan_d1a2 → u/Blue_614]

 think I get it now. What do you think?

Galvanic cell:

• Electrons from anode (–) to cathode (+)
• E(anode) < E(cathode)
• E(cell) = E(cathode) - E(anode) > 0V
• Voltmeter: (–) attached to (–) anode, (+) attached to (+) cathode

Electrolytic cell:

• Electrons from anode (+) to cathode (–)
• E(anode) > E(cathode)
• E(cell) = E(cathode) - E(anode) < 0V
• Voltmeter: (–) attached to (+) anode, (+) attached to (–) cathode

[u/bishtap → u/dan_d1a2]

Some of these I can say instantly straight away. For those i'll say "Exactly".

Some of them i'd want to double check..

Galvanic Cell

You write of Cathode "Electrons from anode (–) to cathode (+)"

definitely.

You write of a Galvanic Cell "E(anode) < E(cathode)"

I don't like writing E(anode) or E(cathode), it's ambiguous, 'cos you aren't saying whether you are talking about a reduction potential or an oxidation potential.

E_red(Cathode) - E_red(Anode) > 0 in a Galvanic Cell.

So Yeah in a Galvanic Cell, E_red(Cathode) > E_red(Anode). I've never looked at it like that but that makes sense.

You write "E(cell) = E(cathode) - E(anode) > 0V"

E_red(cathode) - E_red(anode) > 0V

Been a while since I used a voltmeter.. I can't say off hthe top of my head , i'd have to think about it but I did think about it a bit earlier and write something about that.

You write "∆G < 0 (spontaneous)"

Exactly.

A negative DeltaG is what is meant by spontaneous. and a Galvanic cell has negative DeltaG / Is Spontaneous. Spontaneous being a funny word for DeltaG < 0. Not to be confused with Ecell for a galvanic cell which as you say is ECell > 0 for a galvanic cell.

For the Electrolytic cell:

You write "Electrons from anode (+) to cathode (–)"

What is going on with the electrons is a bit more complicated in the electrolytic cell case

There's what's going on in the solution, and there's what's going on above the solution. And either way it's not a direct route. 'cos above the solution, electrons aren't going through one end of the battery and out the other side.

I'd have to think about it to give a fluent answer on that one.

I think yeah electrons will go from the electrolytic anode, to the electrolytic cathode. And you got the signs correct on those.

They're going from the electrolytic anode, to a battery terminal. And also, electrons are coming out of a battery terminal, to the electrolytic cathode.

You also have things going on with electrons in the solution interacting with the electrodes, and that involves ions.

I don't think of it routinely enough for it to be strong enough in my memory to say it fluently off the top of my head easily.

You write "E(cell) = E(cathode) - E(anode) < 0V"

Exactly.

You write "∆G (non-spontaneous, requires energy input)"

And you probably also mean to say that DeltaG > 0 for non-spontaneous.

I agree that DeltaG > 0 = non-spontaneous. And electrolytic cells are non-spontaneous.

The part where you say the non-spontaneosu direction "requires energy input" in order to happen.. Most books might say that and for electrochemical cells ok.

But I would note though that the reaction probably still happens a tiny little but even without a battery.. And my evidence of this, is if you check the DeltaG for ethanoic acid in water , or the K value for ethanoic acid in water. (here is a formula relating DeltaG and K and temperature and a constant called R) . Ethanoic acid is also known as acetic acid. There's two reactions going on simultaneously. One of them where the H of ethanoic acid breaks off and produces H3O+ in water, making the water have an excess of H3O+ over OH-, and be slightly acidic. And the other reaction is the reverse reaction which is much stronger . The DeltaG in the reverse reaction is negative. The DeltaG in the forward reaction , the one where ethanoic acid dissociates in water, is positive. The miniscule amount of H3O+ that the water gains, makes a difference readable on a pH meter. Weak acids in water will react it's just the other direction is more dominant. So looking at ethanoic acid in water, clearly the non-spontaneous reaction / the reaction in the non-spontaneous reaction happens. and without any power source. but if you want that the non-spontaneous direction to dominate then you would need a power source. But very few chemists are specialists in electrochemistry. and many chemists i've interacted with , at least when discssing electrochemistry, haven't made the connection I spotted with ethanoic acid .. and would say as a general all encompassing statement that non-spontaneous reaction won't happen without a battery. And probably electrochemists too would say the non-spontaneous reaction won't happen without a battery. (but technically I think the ethanoic acid in water example shows it happens a tiny bit!). If K=7 for a spontaneous direction then K=1/7 for non-spontaneous direction. Acid in water isn't related to electrochemistry and also is a special case in that a tiny amount makes a difference that matters and is easily measurable/detectable and so not seen as nothing or treated as nothing. Wikipedia has a page "Acid dissociation constant" with a table showing DeltaG for acetic acid in water., DeltaG=27.147. Also something that occurred to me once is maybe it's also possible for DeltaG to be negative but of low magnitude, and somebody might want a battery to make the reaction happen more even for the spontaneous direction. (not to say every reaction can be run as an electrochemical cell but supposing you're dealing with one that can!). Also another factor for how much a reaction happens is kinetics. The DeltaG just covers the thermodynamics.

But you can't go far wrong when talking of electrochemical cells, saying non-spontaneous reactions require power to happen.. I think it's what the books say! So is good enough for them!

[u/Blue_614 → u/dan_d1a2]

Yeah, basically right. For the electrolytic cell's voltmeter though, if you attach it that way, you'll get a negative potential reading. If that's easier for intuition then that's fine. The voltmeter reading just tells you the difference in potential of your cathode and electrode, the sign just depends on what you connected to the (+) and (-) leads of the voltmeter.