Electromotive force (or E cell?) during charging and discharging

[etotheipi]

The EMF of the cell is the potential of the cathode subtract that of the anode, consistent with $\Delta G = -nF\varepsilon$.

If the cell is running as a galvanic cell/discharging, the EMF should be positive. If it is charging, the cathode (the one gaining electrons) is now negative and at lower potential, so the EMF should be negative.

So if we draw a circuit diagram consisting of cells in different orientations, some should have positive EMFs and some should have negative EMFs. However, on circuit diagrams EMFs appear to always be labelled on cells as a positive value. Why is this?

There do exist some complications - if we don’t know the direction of current, then we also don’t know the sign of the EMF...

 

[mjc123]

If the cell is running as a galvanic cell/discharging, the EMF should be positive. If it is charging, the cathode (the one gaining electrons) is now negative and at lower potential, so the EMF should be negative.

When the cell is charging, the "cathode" is acting as an anode (oxidation takes place there) and vice versa. A positive potential is imposed on the cathode, and a negative one on the anode, to induce the charging reaction to occur.

 

[etotheipi]

When the cell is charging, the "cathode" is acting as an anode (oxidation takes place there) and vice versa. A positive potential is imposed on the cathode, and a negative one on the anode, to induce the charging reaction to occur.

I think I understand this part and the merits of referring to the two terminals of a cell as “poles” for this reason.

But if we set up a circuit with two cells connected in opposite directions, one should have a positive EMF and the other a negative one (I.e. we’re fixing one end of the reversed cell to do oxidation, etc.). However, both cells would just be labelled with a positive number of volts.

However, surely one should be negative?

 

[mjc123]

What do you mean? Can you draw a circuit diagram to illustrate?

 

[etotheipi]

For instance, usually we’d just call these cells of EMF 5V and 3V.

However, the EMF of the lower cell should actually be -3V, since the cathode is on the right hand side (smaller bar) in this configuration. But I’ve never seen anyone label EMF’s like this as negatives, even though it seems more correct to do so.

 

[mjc123]

No, but I think common sense would tell you that one EMF should be subtracted from the other, as they are in opposite senses. The voltage across the resistor is 2V (rather than 8V, if the 3V cell was the other way round). Subtracting 3V is the same as adding -3V.

 

[zoki85]

View attachment 258464

For instance, usually we’d just call these cells of EMF 5V and 3V.

However, the EMF of the lower cell should actually be -3V, since the cathode is on the right hand side (smaller bar) in this configuration. But I’ve never seen anyone label EMF’s like this as negatives, even though it seems more correct to do so.

-3V with respect to what?
As you said, the orientation of EMFs in the closed circuit is important, and it is obvious that applied EMF between terminals of "R" is 2 V.

 

[etotheipi]

-3V with respect to what?
As you said, the orientation of EMFs in the closed circuit is important, and it is obvious that applied EMF between terminals of "R" is 2 V.

As in the EMF is defined electrochemically as the potential of the cathode subtract that of the anode.

In this respect, the EMF is signed.

All of the usual circuit analysis rules still apply, e.g. KVL by adding up changes in potential, though I’m more concerned about the slightly more semantic question of labelling EMFs.

A positive EMF can only result of positive current flows out of the positive pole. If the cell is being charged, the EMF is negative, and the reaction is not spontaneous.

One tends to label EMFs on circuit diagrams always as positive numbers, however this can’t be right.

 

[zoki85]

One tends to label EMFs on circuit diagrams always as positive numbers, however this can’t be right.

Why not? If one uses conventional symbols for DC sources in circuit diagrams, or arrows next to positive voltage numbers, everything is ok

 

[etotheipi]

Why not? If one uses conventional symbols for DC sources in circuit diagrams, or arrows next to positive voltage numbers, everything is ok

But if the cell is charging, it would incorrect to label the EMF +3V, since it is -3V.

If we instead label the potential difference along with a reference direction, you’re correct: there is no ambiguity.

However from my understanding, the labels next to cells represent the EMF, and there don’t ever seem to be reference directions drawn in.

 

[zoki85]

But if the cell is charging, it would incorrect to label the EMF +3V, since it is -3V.

What if two 3V identical cells are put in parallel? They don't charge and don't discharge (ie. current is 0), but the difference of potentials between terminals is still 3V (not 0V)

 

[etotheipi]

One of the problems is that Wikipedia says that

$$\varepsilon = -\int_{A}^{B} \vec{E} \cdot d\vec{l}$$ The path is taken from the negative terminal to the positive terminal to yield a positive emf, indicating work done on the electrons moving in the circuit.

However the next section refers to the relation $\Delta G = -nF\varepsilon$, which of course requires $\epsilon$ to be negative if the cell is charging.

According to chemistry, the integral is taken from the terminal at which oxidation occurs to that at which reduction occurs, which could just as easily be from the positive terminal to the negative terminal during charging!

So I don't know how to reconcile the chemical definition of EMF with the physics one!

 

[zoki85]

Not quite sure what is bothering you. A dry cell is regularly characterized by its' capacity (Ah) and voltage at zero current (EMF). Both, E-field inside of the battery and "EMF" exists regardless of current flow ( EMF in reality is not constant as current changes but that's another story). As regards reversed roles of anode and cathode, maybe this link about rechargeable batteries can help:
https://www.chemicool.com/definition/cathode-anode.html
How exactly chemical energy gets transformed in electrical energy is irrelevant for external circuit analysis.

 

[etotheipi]

My problem is that within electrochemistry there exists a very specific definition of EMF, namely

$\varepsilon = E_{red} - E_{ox}$

where $E_{red}$ and $E_{ox}$ are the potentials of the cathode and anode respectively. If a cell is operating galvanically, $\varepsilon$ comes out as positive (since the electrode undergoing reduction is also the one that happens to be at higher potential). If the cell is being driven backward by a larger external cell, then $\varepsilon$ is negative.

The sign of the EMF is important because it directly relates to the change in Gibbs free energy, $\Delta G = -nF\varepsilon$.

If we consider any particular cell in a circuit diagram, the larger bar and smaller bar represent the pole at higher and lower potential respectively, no matter the direction of current. However, if positive current flows into the positive terminal, the EMF of the cell as computed with the electrochemical relation will be negative, and $\Delta G > 0$ making the reaction non-spontaneous - as expected.

However, the bit I quoted from Wikipedia makes a point to specify that the integral is from lower to higher potential. This agrees with the chemical definition in the case of discharging, but appears wrong for the charging case!

Furthermore, when we label a cell with a certain number of volts, I'm not sure whether this refers to the magnitude of the EMF or just the EMF. I've argued that it can't be the EMF since I've never seen the label take negative values.

 

[zoki85]

Something seems wrong with your reasoning. EMF of a normal battery will not change direction when the current is forced to flow in the opposite direction through it (it can change magnitude though). It will still try to separate charges as before. Forget potentials of terminals.That's misleading.

 

[etotheipi]

Something seems wrong with your reasoning. EMF of a normal battery will not change direction when the current is forced to flow in the opposite direction through it (it can change magnitude though). It will still try to separate charges as before. Forget potentials of terminals.That's misleading.

This appears to be fairly well founded electrochemistry though, as noted in here and here. Bard and Faulkner, the electrochemistry textbook I'm referring to, also gives $E_{cell} = E_{red} - E_{ox}$.

As far as I'm aware, the EMF must change sign if the direction of current is reversed since the electrodes performing reduction and oxidation are also reversed.

 

[zoki85]

This appears to be fairly well founded electrochemistry though, as noted in here and here. Bard and Faulkner, the electrochemistry textbook I'm referring to, also gives $E_{cell} = E_{red} - E_{ox}$.

Read on the Wiki page chapter Sign Conventions.

 

[etotheipi]

Read on the Wiki page chapter Sign Conventions.

I'm aware of the passive and active sign conventions, however my question is of a different nature, namely the conventions regarding the actual EMF.

Bard and Faulkner write this,

The cell reaction emf, $E_{rxn}$ , is then defined as the electrostatic potential of the electrode written on the right in the cell schematic with respect to that on the left.

By adopting this convention, we have managed to rationalize an (observable) electrostatic quantity (the cell potential difference), which is not sensitive to the direction of the cell's operation, with a (defined) thermodynamic quantity (the Gibbs free energy), which is sensitive to that direction. One can avoid completely the common confusion about sign conventions of cell potentials if one understands this formal relationship between electrostatic measurements and thermodynamic concepts (3,4).

Like I made reference to in an earlier post, the relative potentials of the two electrodes do not change if we reverse the current. If we traverse the cell in the same direction, we'll measure the same change in potential.

However, the EMF will reverse sign, since now we've fixed the electrode that was the cathode to now the anode, and vice versa. As Bard and Faulkner make reference to, the equation $\Delta G = -nF\mathcal{E}$ makes specific use of this fact.

 

[zoki85]

Let me put it in this way. With series connection of two battery cells of opposite orientation, and corresponding EMFs 5V and 3V , assuming R>>, the voltage delivered to the external circuit is E=+5V+(-3)V or V=+5V-(+3)V. The result is the same

 

[etotheipi]

Let me put it in this way. With series connection of two battery cells of opposite orientation, and corresponding EMFs 5V and 3V , assuming R>>, the voltage delivered to the external circuit is E=+5V+(-3)V or V=+5V-(+3)V. The result is the same

You're right, at the end of the day common sense prevails and - given we know which is the positive pole and which is the negative pole - we can apply Kirchhoff's voltage laws and do everything we need.

However, I was just looking to gain a deeper understanding of the physics to chemistry relationship. Evidently in an electrochemical sense, the sign of the EMF is very meaningful and it is stringently defined. It is cathode minus anode, or reduction minus oxidation.

In Physics, I haven't seen the same distinction made, as it appears one is more inclined to just do the integral in a direction of choice, and give EMFs as positive values no matter whether the cell is charging or discharging.

I am getting confused because there shouldn't be any disagreement between the two disciplines, however evidently the terms are used differently!