[xchcui]

Hi.
There are tens of articles in the web that explains how to check the resistance of the coolant to galvanic process(anti-corrosion additives depletion)with voltmeter.
For example:

"You can also use an ordinary digital volt meter for the same purpose. With the engine off, touch the voltmeter positive test lead to the radiator or engine (making sure you get good metal-to-metal contact). Then open the radiator cap and insert the negative test lead into the coolant. A reading of up to 0.2 volts is considered acceptable and indicates the presence of reserve alkalinity in the coolant. If the coolant reads 0.3 to 0.6 volts, it is borderline and should be recycled or replaced. A reading of 0.7 volts or more would tell you the coolant is overdue for a change.

When i did that,i read 0.7V,but during the next 30 seconds the voltage drop slowly to 0.1V.
No article talked a bout the dropping of the voltage,so i don't understood if i should to relate to the 0.7V or to the 0.1V as the final right result.
I understand that the voltage potential between the aluminium(radiator,water pumps)and the nickel(the voltmeter probe)is about 0.6V.
The aluminium is the negative electrode,the nickel probe is the positive metal and the coolant is the electrolyte.
So it looks like that when i make this test,i read for the one second the voltage potential between those metal,while in the following seconds,it seems like a galvanic action happens,electrons flow through the wires and ions in the coolant,but the process stops after 20-30 seconds(0.1V).
1)what is the right result/what should i refer to,the 0.7V or the 0.1V?
2)Does the decreasing in voltage after 30 seconds means that the coolant does it job(prevent galvanic process)?
My logic says that if the coolant was not do its job,the voltage should remain 0.7V.shouldn't it?

Thanks in advance.

[Borek]

Not your fault, but the idea of measuring the resistance of a liquid by just dipping in the electrode is completely off. Even completely ignoring the chemistry, just sticking with basic physics, resistance is a function of the contact surface and the distance between the electrodes. It doesn't matter much in electronics, when the resistance at the contact surface is negligible compared to the resistance of the elements, but in electrochemistry geometry and the size of the electrode are one of the most important parameters of a measuring system.

Compare http://www.corrosion-doctors.org/Electrochem/conductivity.htm

[xchcui]

So,what is the value 0.7V that i got?
and why the voltage droped slowly to 0.1V during the 30 seconds and stopped?
maybe it is not accurate,but this results should indicate somthing,shouldn't it?

[Borek]

maybe it is not accurate,but this results should indicate somthing,shouldn't it?

I am not convinced it means anything. Just because the device used for measuring shows some number doesn't mean this number is a source of a valid information.

Perhaps there is some logic behind, but if so - I fail to see it. To me whole procedure looks like just some random urban legend.

[Intanjir]

We are measuring voltage here not resistance. (At least not directly)

A voltmeter works by having a great big resistance and a very sensitive current detector. Whatever voltage source you had was seemingly only able to apply the extra 0.6 volts across this resistance for 30 seconds before it ran out of charge. I can charge a tiny capacitor to 0.7 volts and expect the voltmeter to give me a reading for longer than that. This does not seem like a significant amount of charge. Still I'd be curious to know what the origin of it was.

Anyways, take some aluminium foil and stick it in a glass with some water and a bit of NaCl and measure the voltage as you did for the radiator. It won't decay down and it will be several times higher than 0.1.

[Intanjir]

Xchchui, have you considered that two electrodes in electrolyte have capacitance? What would you see if the RC constant is not too small and if the aluminium electrode is sufficiently passivated.

[xchcui]

Now that you mentioned a capacitor,it seems that my results similar to the results when i check voltage of a capacitor(or checking voltage of a low-capacity battery connected to high load consumer).
If the RC constant is not too small,i guess the voltage will drop slower and if the aluminium electrode will be passivated,it  means that the voltage potential between the two metals are different now,since the the aluminium electrode is  covered by a film material.
I know that in the coolant there are corrosion inhibitors that passivate the metals in the system by creating a film to prevent galvanic corrosion.
Is there any connection between my results to the passivation?
Since the voltage,drop so fast in my coolant test,it seems that there was very small charge that was involved in the process,but i can't figure out if it is related to the fact that the metals in the coolant system has a passivted film?
On the one hand if the aluminium passivated with a protecting film against galvanic corrosion,it should not release ions to the coolant,as this passivation should prevent from the aluminium to release ions to the coolants,but on the other hand when i did the test there is a flow of current through the voltmeter  ,as result,flow of ions in the coolant.
The charge seems to be very small,Does it mean that the corrosion inhibitors do there job?
If they does,why there is still a flowing of current through the voltmeter(even if small)when i made the coolant test?
By the way the voltmeter nickel probe is not passivated(i think it takes time),so i don't know,also,if it contribute somthing to this issue.

[Intanjir]

Even if just the aluminium electrode is passivated then the galvanic corrosion will be hindered since the metal has no access to hydroxide.
It is then a bit counter intuitive to increase the alkalinity in order to prevent galvanic corrosion. However a slightly alkaline solution probably results in a thicker passivation layer. Alkaline enough to encourage aluminium's oxidation without being so alkaline that you greatly increase its solubility.

Now just because the aluminium has no access to hydroxide doesn't immediately prevent it from donating electrons to the more electrophilic nickel. Simply by being in electrical contact with nickel, aluminium will become slightly positively charged and the nickel will be slightly negatively charged.

Of course this doesn't by itself go on without end, it stops when a certain potential difference accumulates across them.
https://en.wikipedia.org/wiki/Galvani_potential
(BTW this potential isn't the same as the electrode potentials that are normally measured for batteries. In fact every source I read claims you can't measure this, though a few bother to include the condition that you can't so long as electrons aren't net accumulating in any parts of your circuit.)

So you might ask yourself does the movement of electrons cause the oxidation and reduction of the electrodes or is the other way around? Does the oxidation/reduction cause the current?

Now my understanding is more than a little sketchy on this stuff, but assume that there is some source of potential difference, perhaps not the Galvani potential.

So assuming this can you see how it would interact with a capacitance in series with it? Keep in mind that neither electrode is charged initially.

And yes the nickel not being passivated is significant. I claim that if it was passivated your voltmeter would almost immediately read 0.1V if submerged or 0V otherwise.

[Intanjir]

Sorry Fermi Level not Galvanic Potential.
https://en.wikipedia.org/wiki/Fermi_level

[xchcui]

I see that i measure,actually,the fermi level with the voltmeter.
But if i connect the voltmeter and electrons flow from the aluminium(that become slightly positively charged)to the nickel electrode(that become slightly negatively charged),it looks like the voltmeter should show increasing in the potential difference value,So why shouldn't the  voltmeter show zero voltage at the beginning and increase in voltage  during the following seconds until a certain potential difference accumulated across them,oppose to my actually reading?(0.7v at first and decreasing to 0.1v)?

[Intanjir]

A voltmeter measures Fermi level differences. This is a bit different from the Electrostatic potential difference.
We have to make a distinction because electrons will flow for more reasons than just charge imbalances. After all, electric forces are only one of the kinds of force that an electron feels.
So Fermi level is defined as the overall potential for electrons that governs their movement amongst materials.
A voltmeter is actually an ammeter in disguise. It actually measures current and so the potential differences it derives from that are the Fermi level ones.
So in this scenario the Nickel and the Aluminium are both becoming more charged, so there is an increasing Electrostatic Potential difference between them, but the voltmeter doesn't measure this difference because it doesn't work by counting up all the charges.
The voltmeter will read 0 when current stops flowing through it. This will happen once the Electrostatic Potential between the two electrodes is big enough to counteract Nickel's intrinsic greater affinity for electrons.

[xchcui]

So,what should happens after  the current stops to flow towards the nickel,while there is equilibrium of the fermi level between the metals while the aluminium is slightly positive charged?
could it mean that,if the aluminium wasn't passivated and since it positive charged,some positive ions would release from the aluminium into the electrolyte?
And would make the reaction proceed?
I mean positive aluminium ions will released to the solution while more electrons will flow to the nickel,while the voltage reading will be steady?
But since the aluminium is passivated and it has no access to hydroxide,this action doesn't happens?
And,maybe,when the corrosion inhibitors in the coolant deplete and some parts of the aluminium loose the passivate film,maybe from that parts the aluminium release positive ions,while as results the voltage reading increase and become steady.
And as long as the aluminium loses more film,the reaction increased and the measured voltage increased while more steady current flow to the nickel.
Is it sound right?

BTW i didn't understood the capacitance issue and i didn't understand what did you mean about capacitance in series with it?
Did you mean that the coolant has capacitance or the two electrodes behave like capacitor?
I couldn't figure it out.

So assuming this can you see how it would interact with a capacitance in series with it? Keep in mind that neither electrode is charged initially.

Xchchui, have you considered that two electrodes in electrolyte have capacitance?

[Intanjir]

Yes, defeating the passivation would be able to account for the 0.1V steady voltage reading. Basically it allows for some of the built up excess positive charge to leave the aluminium electrode. This then allows nickel to steal more electrons from the aluminium so that it stays just as positively charged.

I am suggesting that the tiny point where the one nickel electrode meets the aluminium is the voltage source. As long as the other electrode is held in the air the circuit is open. When it is placed into the electrolyte the circuit is completed. It then behaves for a time in much the same way that it would if it was instead completed by a reasonably big capacitor placed in series with it. The only difference seemingly being that instead of going to 0V it ended at 0.1V, ie a very leaky capacitor.

So why would unpassivated nickel connected to passivated aluminium by an electrolyte have a high capacitance? You might look up the various types of capacitor...

BTW a real capacitor won't actually work the same way unless its electrodes were also made of nickel and aluminium, since otherwise we would just be introducing more voltage sources each time dissimilar metals touched.

[xchcui]

Yes, defeating the passivation would be able to account for the 0.1V steady voltage reading.

I meant ,if the aluminium will lose the passivations gradually(during weeks,months),will the voltage reading will be more than steady 0.1V?
Will the voltage increase more and more as the passivation film deplete from the aluminium,while more aluminium will be exposed to the coolant?
and each time that  more parts of aluminium will be exposed to the coolant,the voltage reading will be higher:0.2V,0.4V,0.7V...?
Or the voltage reading will be steady 0.1V and not more,even if all the aluminium will be exposed to the coolant with no passivate film?

...I am suggesting that the tiny point where the one nickel electrode meets the aluminium is the voltage source...

I don't understood that.
The nickel electrode is dip into the coolant,it is not meet the aluminium.The coolant is between them.
Maybe i didn't understood your intention?

So why would unpassivated nickel connected to passivated aluminium by an electrolyte have a high capacitance?

It guess that the passivate film on the aluminium act like a dielectric material,while the coolant act as one electrode and the aluminium is the other electrode like the electrolytic capacitors.
And since the electrodes(coolant/aluminium)so close each other(thin passivation layer)it can have high capacitance.
Now,it looks like when i put one probe of the voltmeter into the coolant,it is like i put the probe at one capacitor lead(which connected to an electrolyte)while when the other probe of the voltmeter on the aluminium,it is like i connected that probe to the other lead of the capacitor which has passivated metal.
This action is measuring the capacitor voltage,so,it looks like i measure the voltage between the coolant and the aluminium.

How does the depletion of the passivate film from the aluminium influence the reading voltage at the voltmeter?

[Intanjir]

Why would the Aluminium lose the passivation? Is your car made of Magnesium or Calcium metal? Is your coolant HCl? Under reasonable circumstances the passivation layer should only get better over time.

A thinner passivation layer will give you a larger voltage. The passivation layer is providing a large resistance to the movement of electrons/ions. This internal resistance will result in a smaller measured voltage.

The nickel electrode is dip into the coolant,it is not meet the aluminium.The coolant is between them.

Your voltmeter has two nickel electrodes, and I know that you know you are connecting the other one to the aluminium as you say it in the next paragraph. ;P

It guess that the passivate film on the aluminium act like a dielectric material,while the coolant act as one electrode and the aluminium is the other electrode like the electrolytic capacitors.

Yep! It is a classic aluminium electrolytic capacitor.

This action is measuring the capacitor voltage,so,it looks like i measure the voltage between the coolant and the aluminium.

Nope! There was no voltage between the coolant and the aluminium before you connected in the nickel, and yet the voltmeter read 0.7V initially.
You cannot measure the voltage between the coolant and the aluminium unless your wet electrode is made aluminium. You can do this btw by just inserting one end of a piece of aluminium foil into the electrolyte and then connecting the dry end to your nickel electrode.
You are measuring the voltage of a circuit and there is more than one voltage source. A leaky capacitor and a 'battery'.