Clarification about the sign of EMF in batteries and electrochemistry

[etotheipi]

In electrochemistry, we define $E_{cell} = E_{cat} - E_{an}$, the difference between the electrode potentials of the cathode and the anode. This has the effect that if the reaction is spontaneous, we obtain a positive $E_{cell}$ and if it is not - i.e. we need an external driving voltage source in order to fix one as the cathode and the other as the anode - $E_{cell}$ comes out as negative. This is fine, since if we are driving a cell "backward" then the half cell at lower potential is in this case fixed as the cathode, and the change in potential going from the anode to the cathode (as is the definition of $E_{cell}$) is negative.

In a Physical sense, the EMF from point A to point B is given by
$$\varepsilon = -\int_{A}^{B} {\vec{E}} \cdot d\vec{l}$$
With this said, I'm slightly confused as to how $\varepsilon$ relates to $E_{cell}$. If it is always the case that $\varepsilon = E_{cell}$, then the implication is that the integral is always from the anode to the cathode (again this would be fine, but I haven't been able to find a reference that says this!).

The problem is that most of the time (in Physics anyway) $\varepsilon$ is given as a positive value even when the cell is being driven backward/charged - which doesn't seem consistent with the electrochemical definition of $\varepsilon$.

So I was wondering if somebody could clarify whether $\varepsilon$ does indeed always equal $E_{cell}$, or whether it represents the magnitude of $E_{cell}$. Thanks!

 

[mitochan]

During discharge
$$V_{cell}=\phi_{cat}-\phi_{an}=\int_a^c \mathbf{E}\cdot\mathbf{dl}>0$$
For circuits of no EM induction
$$\oint \mathbf{E}\cdot\mathbf{dl}=0$$
$$V_{cell}=\int_a^c \mathbf{E}\cdot\mathbf{dl}=-\int_c^a\mathbf{E}\cdot\mathbf{dl}$$
If you think of more complicated case of EM induction, EMF and battery $\phi_{cell}$ work additionary on the circuit.

 

[etotheipi]

During discharge
$$V_{cell}=\phi_{cat}-\phi_{an}=\int_a^c \mathbf{E}\cdot\mathbf{dl}>0$$

Thanks for your reply; though if the cathode is at higher potential during discharge, then shouldn’t the electric field vector point from the cathode to the anode? In which case $\vec{E}$ and $d\vec{l}$ are in opposite directions so their dot product is negative, and we would instead require a negative sign out the front?

 

[mitochan]

Ah, you are right.
$$\mathbf{E}=-\nabla \phi$$
minus sign.