0
$\begingroup$

A defining equation of electrochemical cells is that the EMF, $\mathcal{E} = E_{red/cat} - E_{an/ox}$, equals the electric potential of the cathode minus that of the anode.

This means that if a cell is discharging, $\mathcal{E}$ is positive and if it is charging, $\mathcal{E}$ should be negative. This is consistent with the relation $\Delta G = -nF\mathcal{E}$, since charging should be non-spontaneous and as such requires a negative EMF.

The EMF from A to B may be expressed as an integral

$$\mathcal{E} = -\int_{A}^{B} \vec{E} \cdot d\vec{l}$$

Electrochemically, A should always be at the anode and B always at the cathode (in a redox sense).

However, Wikipedia states that

The path is taken from the negative terminal to the positive terminal to yield a positive emf

Indeed, I've only seen positive values of volts labelled next to cells no matter their orientation. That is, in circuit theory, the EMF appears to be given as e.g. $+3V$ no matter whether the cell is charging or discharging.

This doesn't seem correct, since couldn't and shouldn't the EMF be negative during charging/when the cell is connected in the opposite direction relative to the positive current? I don't know how to reconcile the electrochemical definition of EMF with that given in Physics.

Why is it that in Physics, EMFs are only ever expressed as positive numbers - for instance, in labels on cells in circuit diagrams - even when the actual EMF of a cell is negative?

$\endgroup$
3
  • $\begingroup$ The work described here is the external work required to move positive charge from the negative terminal to the positive terminal which yields an increase in potential (positive emf) and potential energy of the positive charge. So I'm not sure why it talks about electrons. $\endgroup$
    – Bob D
    Commented Mar 10, 2020 at 21:06
  • $\begingroup$ @BobD Good catch, I've now abridged the quote so as to avoid any confusion! $\endgroup$
    – 13509
    Commented Mar 10, 2020 at 21:22
  • $\begingroup$ Charging is a lot like having 2 unequal voltages in parallel see some good resources here: youtu.be/XoLzn49IZjM See the about section for theory $\endgroup$
    – ChemEng
    Commented Mar 10, 2020 at 21:39

1 Answer 1

0
+50
$\begingroup$

Yes, when the cell is charging, it will have a negative potential difference (Potential of Cathode with respect to Anode). This can be easily verified by my measuring the Potential Difference across a charging cell with a standard Multimeter.

For the second part, Sometimes negative EMFs are also given in circuit diagrams. It does happen. Its just that it is not common.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.