Do I use the total surface area of
both electrodes or one of them to calculate the current density
Dear Chandan Kumar Tiwari
, I know that you know that this is possible in four-electrode configuration. You also need a two chamber cell with a membrane.I noticed that currently the majority of questions are elementary. Commonly, the answer is in text-books.
My question is about how to calculate the current density. for example, If I used a two-electrode cell containing one Ni foam electrode of 0.5 cm2 as the working electrode and the other Ni foam electrode of 0.5 cm2 (the same surface area) as both counter and reference. I used this system to measure the linear sweep voltammetry (LSV) for overall water splitting (current (mA) versus potential) I want to convert the plot to be (Current density (mA/cm2) versus potential). Do I divide the current (mA) by 0.5 cm2 (surface area of one electrode) or by (2*0.5) cm2 (total surface areas of the two electrodes)