How come no voltage is meassured across my electrolysis tank?
You have already identified impedance to be the key word.

In this case, the (output) impedance of the battery charger is relatively high, while the resistance of your 'circuit' (the water tank) is very low.

Basically, you have a voltage divider made from the impedance of the charger and the resistance of the tank in series. What you are measuring is the voltage across the small resistance of the tank, because you probably cannot measure the voltage across the internal resistance of the charger.







simulate this circuit &ndash; Schematic created using CircuitLab

In more detail, the charger is able to provide, say, 14V maximum or 15A maximum. When the current draw tends towards zero, the voltage will max out at about 14V; however, when the current drawn approaches 15A, it will be limited there and the voltage will drop in response to Ohm's law (you cannot push 15A through a 0.1 Ohm resistance and at the same time have 14V drop across that resistance.)

The charger may either employ some 'intelligent' charging controlling both voltage and current, or it may just limit the voltage to 14V and provide as much current as it can, which is implicitly limited by the impedance of some parts of the charger like e.g. its mains transformer.

I assume the "analog voltmeter" you mention is the one integral to the charger. This voltmeter is made to display voltage when charging batteries, so will probably be most 'accurate' (or responsive) for voltages between 10 and 14V; nobody usually needs the very-low-end scale below, say 8V, and nobody would expect the meter to be precise anyway; it's just some indicator to coarsely estimate the state-of-charge of the battery. These often don't even have a linear scale so that you may have a very hard time distinguishing 1V vs. 2V; otoh, 11V vs. 12V will probably be much more visible.