# How does a change in the surface area of the electrodes affect the voltage produced by a galvanic cell?

I did a lab where I increased the surface area of the electrodes in a galvanic cell, that resulted in an increase in voltage. However, I have been reading and that is not supposed to happen. I am a bit confused, if you increase the surface area of the electrodes, the rate of reaction as well as the current increases, therefore, if Ohm's law ($$V = IR$$) is taken into account, the voltage should also increase. But according to some sources, the voltage should remain unchanged.

Could anyone please tell me if an increase in surface area will or will not increase the voltage produced by the cell, and why?

• @NightWriter Please don't use \mathrm{...} where it shouldn't be used (I noticed you really like to put it everywhere in your edits). Please (re)visit this page, this page, this one and Math.SE Meta post on how to format your future posts better with MathJax and Markdown. – andselisk Feb 21 at 22:05
• Ok, thanks @andselisk! Yeah, I have no idea when it should be used, I just go by appearances, like most people... – Buck Thorn Feb 22 at 7:32

The standard electrode potential, $$E^\circ$$, in volts, does not depend on the surface area of the electrodes. However the standard electrode potential is measured with an infinitesimal current flow. In the simplest model you can imagine a cell, galvanic or electrolytic, as having an internal resistance. So as the current of the cell increases the voltage decreases. Larger electrodes would effectively reduce the internal resistance of the cell.
• No, in the simple model $\mathrm{V}_{obs} = i(R_{battery} + R_{load})$ – MaxW Feb 21 at 18:23