Internal Resistance of the Battery - Ohmic Resistance

[Dario56]

When current starts flowing in a circuit, voltage of the battery drops from EMF value. There are three sources of this however here I am only interested in ohmic resistance of the battery.

EMF of the battery is created because of electrochemical processes on both electrodes which create potential difference on electrode/electrolyte interface due to charge separation.

When talking about ohmic resistance of the battery I am not sure how does ohmic resistance of battery components affect charge separation on the interface. What does ohmic resistance of electrolyte or electrodes have to do with equilibrium of electrochemical reactions on the electrodes?

 

[hutchphd]

Here is a clue: the resistance of a particular type battery is roughly inversely proportional to the active area of the plate surfaces.

 

[Dario56]

Here is a clue: the resistance of a particular type battery is roughly inversely proportional to the active area of the plate surfaces.

While I do agree with this, it didn't really answer my question.

 

[Dale]

What does ohmic resistance of electrolyte or electrodes have to do with equilibrium of electrochemical reactions on the electrodes?

Nothing. The Ohmic resistance does not change the electrochemical reactions on the electrodes, it changes the electric field between the electrodes. By Ohm’s law, if there is a current through a resistor there is also a voltage across the resistor and the two are related by $V=IR$.

Suppose that the circuit is grounded at the conductor attached to the negative terminal of the battery and that the half cell at the negative side is 0.5 V and the half cell at the positive side is 1 V.

Now, in an open circuit condition there is no current so there is no Ohmic voltage drop. So the negative terminal is at 0 V, the electrolyte is at 0.5 V, and the positive terminal is at 1.5 V.

Suppose the circuit is closed and there is enough current to produce a 0.1 V Ohmic voltage drop (and neglect the other electrochemical effects at the electrodes). Now, the negative terminal is at 0 V and the electrolyte immediately adjacent is at 0.5 V. However, because of Ohm’s law the electrolyte is not all at 0.5 V, but instead the electrolyte adjacent to the positive terminal is at 0.4 V, and thus the positive terminal is at 1.4 V.

So Ohmic resistance gives a voltage drop without changing the electrochemistry at the electrodes, but instead by changing the E field between the electrodes.

 

[BvU]

Hi,

You should know better than to create confusion by posting the same question three times !
Now you have two answers in two different threads, intractable ...

$\$

 

[Dario56]

Hi,

You should know better than to create confusion by posting the same question three times !

$\$

Hahaha, yes. I've posted it on three different forums.

 

[berkeman]

Hahaha, yes. I've posted it on three different forums.

Not funny. You'll be receiving an infraction for multiple posting as soon as we can sort this out. Sheesh.

 

[Dario56]

Not funny. You'll be receiving an infraction for multiple posting as soon as we can sort this out. Sheesh.

Yes, well sorry. I've posted on multiple forums because usually I get very few answers if any.

 

[phinds]

Yes, well sorry. I've posted on multiple forums because usually I get very few answers if any.

Please read the forum rules. We actually take them seriously here and multiple posts are NOT funny nor are they acceptable, regardless of your rationale. It wastes the time of the moderators who have to clean up your mess.

 

[Dario56]

Please read the forum rules. We actually take them seriously here and multiple posts are NOT funny nor are they acceptable, regardless of your rationale. It wastes the time of the moderators who have to clean up your mess.

Yes, sorry. I am actually new here, so I am not well aware of the rules. But, yeah you know why I posted on more forums now.

 

[berkeman]

So to get this thread back on-topic, did the reply by @Dale above answer your question?

 

[Dario56]

Nothing. The Ohmic resistance does not change the electrochemical reactions on the electrodes, it changes the electric field between the electrodes. By Ohm’s law, if there is a current through a resistor there is also a voltage across the resistor and the two are related by $V=IR$.

Suppose that the circuit is grounded at the conductor attached to the negative terminal of the battery and that the half cell at the negative side is 0.5 V and the half cell at the positive side is 1 V.

Now, in an open circuit condition there is no current so there is no Ohmic voltage drop. So the negative terminal is at 0 V, the electrolyte is at 0.5 V, and the positive terminal is at 1.5 V.

Suppose the circuit is closed and there is enough current to produce a 0.1 V Ohmic voltage drop (and neglect the other electrochemical effects at the electrodes). Now, the negative terminal is at 0 V and the electrolyte immediately adjacent is at 0.5 V. However, because of Ohm’s law the electrolyte is not all at 0.5 V, but instead the electrolyte adjacent to the positive terminal is at 0.4 V, and thus the positive terminal is at 1.4 V.

So Ohmic resistance gives a voltage drop without changing the electrochemistry at the electrodes, but instead by changing the E field between the electrodes.

Didn't you say firstly that negative terminal is on 0,5 V and in the next paragraph that it is on 0 V? Is that a mistake?

 

[Dario56]

So to get this thread back on-topic, did the reply by @Dale above answer your question?

Still discussing.

 

[Dale]

Didn't you say firstly that negative terminal is on 0,5 V and in the next paragraph that it is on 0 V? Is that a mistake?

No. I always said that the negative terminal (meaning the metal) is at 0 V and the electrolyte immediately adjacent to the negative terminal is at 0.5 V.

 

[Dario56]

No. I always said that the negative terminal (meaning the metal) is at 0 V and the electrolyte immediately adjacent to the negative terminal is at 0.5 V.

Sorry, I don't get what you want to say.
Quote from your answer: Suppose that the circuit is grounded at the conductor attached to the negative terminal of the battery and that the half cell at the negative side is 0.5 V and the half cell at the positive side is 1 V.

Now, in an open circuit condition there is no current so there is no Ohmic voltage drop. So the negative terminal is at 0 V, the electrolyte is at 0.5 V, and the positive terminal is at 1.5 V.

You said that negative half cell is at 0,5 V and in the next paragraph that the negative terminal is at 0 V. I thought negative half cell and terminal is the same thing. Same story for positive terminal.

 

[Dale]

You said that negative half cell is at 0,5 V

No, I said the half cell at the negative terminal IS 0.5 V. Meaning that the half cell produces a voltage difference of 0.5 V across the interface (e.g. across the bilayer). So if one side is AT 1000 V then the other side will be AT 1000.5 V.

 

[Dario56]

No, I said the half cell at the negative terminal IS 0.5 V. Meaning that the half cell produces a voltage difference of 0.5 V across the interface (e.g. across the bilayer). So if one side is AT 1000 V then the other side will be AT 1000.5 V.

Oh, okay. That makes sense. Thank you.

 

[Dario56]

No, I said the half cell at the negative terminal IS 0.5 V. Meaning that the half cell produces a voltage difference of 0.5 V across the interface (e.g. across the bilayer). So if one side is AT 1000 V then the other side will be AT 1000.5 V.

To conclude, ohmic resistance of electrolyte (which is the biggest source of ohmic resistance in the battery as far as I know) causes that potential difference between terminals on which we connect conductors drops due to reason you explained clearly through an example. Ohmic resistance doesn't affect potential difference developed in electrochemical double layer (it is still 0.5 V and 1 V difference on electrodes).

What about resistance of electrodes themselves? As electrodes are made of materials which also have resistance there should be potential gradient in the electrode itself. So, potential difference between terminals may depend on where we connect condutor to the electrode as potential isn't the same through electrode.

 

[Dale]

What about resistance of electrodes themselves?

That resistance is typically negligible compared to the electrolyte resistance. But in any case even though it is physically located in a different material it is electrically in series with the electrolyte resistance. So it simply gets lumped in there too.

 

[dlgoff]

When current starts flowing in a circuit, voltage of the battery drops from EMF value. There are three sources of this however here I am only interested in ohmic resistance of the battery.

EMF of the battery is created because of electrochemical processes on both electrodes which create potential difference on electrode/electrolyte interface due to charge separation.

When talking about ohmic resistance of the battery I am not sure how does ohmic resistance of battery components affect charge separation on the interface. What does ohmic resistance of electrolyte or electrodes have to do with equilibrium of electrochemical reactions on the electrodes?

This site, Battery University, should be useful for you.

If you're talking about lead-acid batteries, take a look at this page from their site.

 

[hutchphd]

That resistance is typically negligible compared to the electrolyte resistance. But in any case even though it is physically located in a different material it is electrically in series with the electrolyte resistance. So it simply gets lumped in there too.

The other piece of this is that for linear systems, Thevenin's theorem tells us we can simplify the battery into an ideal voltage source and a resistor.
So it really doesn't matter that it is not that simple and lots of complicated things are going on internally in the battery. We choose this representation because it works well and has some more or less tenuous relation to the internal physics. Worrying about "where is the resistor?" is not a useful enterprise, unless you are designing batteries. Then the answer is not trivial and you can usefully spend a career on it.

.

 

[cmb]

When current starts flowing in a circuit, voltage of the battery drops from EMF value. There are three sources of this however here I am only interested in ohmic resistance of the battery.

EMF of the battery is created because of electrochemical processes on both electrodes which create potential difference on electrode/electrolyte interface due to charge separation.

When talking about ohmic resistance of the battery I am not sure how does ohmic resistance of battery components affect charge separation on the interface. What does ohmic resistance of electrolyte or electrodes have to do with equilibrium of electrochemical reactions on the electrodes?

I'm unclear what your question actually is.

A voltaic cell is such that there are positive charge carriers able to cross some medium that is impermeable (or as impermeable as possible) to electrons.

In a way, a cell that is part of an electrical circuit is a 'positive charge carrier only' and prevents electrons passing across that part of the circuit, while the electrical circuit outside the cell prevents positive charge carriers from running around it.

Thus, the resistance in the external circuit to the circuit current is the sum total of all the limitations and restrictions that prevent electron motion in it.

Likewise in the cell, the resistance is the sum total of all the limitations and restrictions to the motion of positive charge carriers within it.

This would mostly be the electrolyte that carries the positive charge carriers and the interfaces between the electrodes and the electrolyte. In the case of lithium ion batteries particularly, there is also a restriction of the lithium ions intercalating in and out of the [generally graphite] electrodes.