In a two electrode system, is the applied current density referring to just the working electrode or to the auxiliary electrode as well?
It is normal in many laboratory electrochemistry experiments for the working electrode to have a much smaller surface area than the counter (auxiliary) electrode. This when the current density at the working electrode is 0.5 mA per square cm, the current density on the counter electrode is much smaller.

It is more common that someone will express the potential and the current density on the working electrode.

If for example we have a reference electrode which is at 0 volts (standard hydrogen electrode) and the potential of the working electrode is + 400 mV. We could have a flow of 1 mA out of the working electrode. As no current flows through the reference electrode all of this current flows through the counter electrode.

The electrical potential between the counter and the reference electrodes is known as the compliance voltage. This is the voltage required to create a flow of current which is just right to make the right amount of current flow through the working electrode to keep its potential at + 400 mV.

This is done using a potentistat, a range of different designs of such devices exist. I think that the most simple design and one of the best is the general design used by PAR (Princetown Applied Research) in the classic 173. It has been joked that at one point almost all electrochemists have used a 173, I rather like the 173 it is a nice machine. But we are not here to discuss classic electrochem machines.

Back to the design of a machine like a 173, If we set the drive at -400 mV then the top op amp will impose a current on the cell via the secondary electrode which will bring the potential of the reference electrode to -400 mV.



The bottom op amp which is a I to V converter will hold the working electrode at zero volts (ground or GND) as far as the electronics is concerned. Thus for the chemist if they hold the view that the refernece electrode is the zero volts point then the working electrode will be at +400 mV.

If 1 mA flows into the working electrode and the value of resistor R is 1000 ohms then the output of the I to V converter (bottom op amp) will have to be -1 volt for the sum of the currents coming to the inverting (-) input of the op amp to be zero. We assume that the input resistance of an op amp is infinity. Thus if we attach a x-y plotter to the drive terminals (x to E or potential axis) and the (y axis to the current indication terminal) we will get the plotter to plot -400 mV, -1 volt.

A word of warning the Europeans and the Americans often follow different conventions on what sign to give a reducing current (cathodic) and what sign to give a reducing potential.

If I recall rightly the European convention is for bottom left to be reducing current and reducing potential.