I did a lab in galvanic cells where I changed the surface area. As a result, the current increased, but I don't really understand why. I guessed it was because more are means that more ions from the anode are oxidized and more ions from the cathode reduced, more electrons are transferred and taking into account that current is the flow of electrons per unit time, the current increases.

Is that even correct?

Thank you all

  • 3
    $\begingroup$ You nailed it... $\endgroup$ – MaxW Mar 4 at 19:23
  • $\begingroup$ what I guessed is actually right? $\endgroup$ – user74908 Mar 4 at 19:29

Yeah, you're on the right track. In any battery reaction there is some rate-limiting step that determines the maximum power the battery can supply. Most of the possible rate-limiting steps involve the surface of the electrode.

As an example, imagine that the rate-limiting step is the diffusion of new ions from the electrolyte bulk to the cathode (or diffusion of ions away from the surface at the anode. The physics of diffusion in this case is governed by Nernst-Planck equation, where the quantity of interest is the ion flux (number of ions of unit area). However, the total current is determined by the total number of ions. To get from flux to total number of ions you have to integrate over the area of the electrode: if you double the area of the electrode you'll double the total number of ions (ie. current) if all else stays equal.

That's just one example that depends on the specifics of your cell, but many of the surface-related limited steps in a battery follow this type of behavior.

  • $\begingroup$ One of my electrochemistry professors used to give a nice example. Why can't you start a car with a 9 V Everyday Alkaline Batteries? It is all about the rate of current which can drawn from a battery. $\endgroup$ – M. Farooq Apr 6 at 20:49

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