Hello everyone. First of all, I'm not an electrical engineer, but rather a chemical one, hence the possibility of the stupidity of my question.
There is an electrochemical reactor in which goes anodization process in a potentiostatic regime at 60 volts. Initial current is 200 mA than falls to 30 mA in fractions of a second and then it can drop approximately to 5 mA in a few hours. The task is to measure this current. I learned that sensors on Hall effect are used for these tasks, but I can't find an appropriate one with sufficient resolution for my case.
Can you suggest something to tackle this in a not very complicated way or all solutions are too complicated for a non-electrical engineer?
Small currents of this magnitude are most easily sensed with a shunt-resistor. You measure the voltage across the shunt.
A 2W or 3W 22 ohm resistor can be used (on the low voltage side of the cell), which will produce 4.4V for a current of 0.2A.
Connect high-side of shunt via a 10k resistor to an Arduino analog pin to protect it from over-voltages.
Fix the Arduino ground at the low side of the shunt, and use analogRead, scaling for the value of the shunt resistor.
You will lose a few volts from the cell of course, but this will decrease as the current falls so the constant voltage
condition isn't badly compromised.
Hall sensors come into there own for when you need to measure high currents with isolation to only low precision. They are often used for electric motor control, for instance.
Edit: Mark said exactly the same thing with different words...
Use a resistor. Calculate a resistor value that will give 5V for the maximum current. Select a standard resistor value just less than this. Wire that resistor between the most negative point in the circuit and the device under test.
That most-negative point is now "ground" for the Arduino. Connect a wire between the other end of the resistor and an Arduino analog input. analogRead() and Ohm's law are your friends.
There are a couple of reasons why you might require hall effect instead of a resistor. Primarily it means that the two circuits do not have to share a ground. The second reason is you may not want to add an extra 22 ohm resistance in the circuit.
Thank you all for the quick reply! This is turned out to be up a lot easier than I initially expected 
I would pick a resistor that gives about 1V at maximum current, and use the internal reference for better stability of the readings.