2
$\begingroup$

When galvanic cell operates or when current starts flowing in a circuit, voltage drops from electromotive force of the cell (EMF) because of internal resistance of the battery.

There are three sources of internal resistance or sources of polarization:

  1. Ohmic polarization
  2. Activation polarization
  3. Concentration or Diffusion polarization

Working voltage of the cell is given by following equation: $$ E = EMF - I(R_o+R_a+R_c) $$

where R represents polarizations mentioned and I is current in the cell.

EMF of the cell is created because of charge separation in electrochemical double layer on anode and cathode due to electrochemical processes of certain species or redox systems present. More charge is separated, bigger the absolute value of electric potential developed on electrode. EMF of the cell can be calculated from Nernst equation if equilibrium is established on both interfaces/electrodes.

My question is how do these three types of polarizations mentioned affect amount of charge separated on interfaces when current starts flowing in a circuit physically speaking? What processes happen on electrodes so that voltage of the cell drops from EMF value?

For diffusion polarization that is quite clear since if activity of some species like ions is different close to electrode compared to bulk of electrolyte if diffusion is slow process, from Nernst equation electrode potential is different compared expected value calculated from activity of species in the bulk.

If we look at ohmic polarization it seems that electrode potential developed is affected by ohmic resistance of conductors in the cell. Since voltage of the cell drops, this would mean that electric potential of cathode decreases while of anode increases compared to Nernst equation values. Why is it that potential of cathode drops while of anode increases?

Activation polarization represents resistance due to kinetics of electrochemical reactions. It also decreases voltage of the cell. Slower the reaction, bigger the voltage drop of the cell compared to EMF. What does activation resistance have to do with charge separation in double layer and why does it decrease cathode potential and increase anode potential?

$\endgroup$
7
  • 2
    $\begingroup$ Lots of questions, but in regard to internal cell resistance, you can model the system as an ideal voltaic cell in series with the internal resistance. The open circuit voltage is just the ideal cell voltage, as per Nernst. But if significant current is drawn, then you have a de facto voltage divider: the cell resistance and the external load resistance. So your measurement gives the voltage across the external load. No need to worry about the individual electrodes: only the difference matters. $\endgroup$
    – Ed V
    Jun 22, 2021 at 18:33
  • 1
    $\begingroup$ After Ed V, you should simply determine the tension measured on the cell before ($\ce{E_0}$) and after connecting the cell to an external resistance $\ce{R_1}$. The cell produces now a tension $\ce{E_1 < E_0}$ and a current $\ce{I_1}$ . The inner resistance $\ce{R_0}$ is given by : $\ce{\frac{E_0 - E_1}{I_1} - R_1}$ $\endgroup$
    – Maurice
    Jun 22, 2021 at 21:05
  • $\begingroup$ @Ed V, yes, however why is it that voltage drops from EMF, how does activation resistance for example affect voltage of the cell and why would it drop? $\endgroup$ Jun 25, 2021 at 8:51
  • $\begingroup$ It has to. Every voltage source, such as a Daniell cell, has an internal impedance that is, in general, complex. Even ignoring capacitance and just looking at the internal resistance, that resistance comes from the solutions in the cell reservoirs and the salt bridge. There must be a complete circuit, including the resistance of the solutions and that is not accounted for in the Nernst equation. Think about two cells with different salt bridges: one short, with large cross section, and the other very long and narrow. Then the latter cell has high cell resistance and cannot supply much current. $\endgroup$
    – Ed V
    Jun 25, 2021 at 10:05
  • $\begingroup$ Yes, that holds. What I am interested is what physically happens so that voltage of the battery drops since EMF is created due to charge separation in double layer on both electrodes. Why is amount of charge separated in double layer affected by activation and ohmic resistance? What processes and how do they affect this? $\endgroup$ Jun 25, 2021 at 10:48

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.