[xchcui]

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.

I thought that the passivation layer degrades with time and released from the metal,and the functions of the corrosion inhibitors that flow in the coolant is to renew the places that the layer film was released from the metal.
This was my starting point to understand the connection between the voltage reading in the voltmeter and the condition of the coolant,while i based on the connection between the depletion of the passivation from the metals and the different reading voltage with regard to the amount of the passivation degradation.
So,if the aluminium doesn't lose the passivation with time,what is the reason that when the coolant is aged,the voltage reading should increase when i am making that test?

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.

In my test the voltage start with 0.7V and decrease to 0.1V during 30 seconds(as i mention before).
But it go down to steady 0.1V.
Does the steady 0.1V refer to,what you called,"leaky capacitor"?
Does the "leaky capacitor"refer to the parts on the aluminium that doesn't have passivation layer and the aluminium ions released from these parts to the coolant while electrons flow to the nickel probe through the wire all the time?(since it doesn't go down to 0V but stays 0.1V).
What should happen in the coolant in order that the voltage reading will be higher?(+0.3v)

[Intanjir]

The passivation layer is thin and a little porous. As a result electrons/ions can travel through the thin layer of aluminium oxide. This is the leakiness. It is the same way an aluminium electrolytic capacitor leaks.

You do not expect there to be any aluminium directly exposed to H2O for any length of time unless conditions are exceedingly acidic or basic. It will immediately passivate and the passivation layer will be reasonably impermeable. Hydroxide/Oxygen will still percolate through, but not quickly and each time it happens it just adds more relatively insoluble Aluminium Oxide/Hydroxide to the passivation layer. Areas which are thinner get thicker faster since more Aluminium Oxidizes underneath these places. As a result the layer tends to get thicker and more even over time.

Any region with exposed metal will have an extremely high reaction rate and will produce a solid passivation layer immediately.

The oxide/hydroxide is not totally insoluble. However the dissolution rate is quite slow. So some of the passivation layer can slowly dissolve. This rate will remain pretty much constant. An equilibrium thickness will then be approached. That thickness and perhaps the porosity of it as well will depend on pH.

Suppose I have a nice thick layer at a slightly alkaline pH thanks to using inhibited coolant. What would eventually happen to that layer if the pH was neutralized?

I think it is unlikely that chunks of passivation break off all at once and expose bare metal. The oxide/hydroxide should tend to dissolve on the outside where there is solvent. The scenario for the metal is more interesting. We already know from a recent thread that aluminium foil seems to dissolve from the inside out in acid. But while it is conceivable that Al3+ ions are be able to percolate through the lattice, whatever kind of Al(OH)x species at neutral/alkaline pH will not.
Recent Thread discussing acidic dissolution of Aluminium:
http://www.chemicalforums.com/index.php?topic=80732.0
Aluminium Speciation Diagram:
http://www.intechopen.com/source/html/40512/media/image2.png

Of course a positive charge on the Aluminium metal will help Hydroxide along somewhat as it attempts to percolate through. So it seems reasonable that we can measure how good of a barrier we have to hydroxide percolating in somehow with a voltmeter. It is interesting that we can do it just making use of the dissimilar metals.

[xchcui]

Suppose I have a nice thick layer at a slightly alkaline pH thanks to using inhibited coolant. What would eventually happen to that layer if the pH was neutralized?

If the pH was neutralized and became acidic i guess that the dissolution rate of the passivation layer will increase,the layer will be thinner and let more aluminium ions to released into the coolant.
Is this the moment that i should  read higher voltage than 0.1v in my voltmeter?as it becomes more leaky(porosity)?

I understand that the 0.1V shows a constant flow of electrons through the wire and ions into the coolant in small and slow rates.(leaky capacitor).

Is the initial 0.7V measurement(while decreasing voltage to 0.1V) happened only as result of the use of dissmilar  metals(nickel and aluminium)as they  got into fermi-level equilibrium,while the 0.1V is the main value that i need to consider?
And even if the voltmeter will stop after 30 seconds on 0.2V,0.4V,0,6V etc.Will the initial voltage will be always 0.7V?
If yes,i could understand that the first seconds that the aluminium donates electrons to the nickel until equilibrium has  nothing to dowith the condition of the coolant or the passivation layer.And only when the voltage gets steady(0.1V,0.2 etc.)it show the conditions of the passivation layer by telling us how much its leakage.
Am i right?

[Intanjir]

Yup!

The initial voltage should still be about 0.7 even as the layer gets thinner. The steady state voltage is what is important.

Now also yes the passivation layer should also get thinner(in the long run), but it is probably because the layer grows slower not because it dissolves faster. When things were a bit alkaline we had extra OHs percolating through and reacting with the aluminium, thus growing the layer faster. However on the other hand at the higher pH aluminium is somewhat more soluble, so the layer was probably also dissolving faster. So lowering the pH makes it dissolve slower, and in fact probably makes us exceed the solubility initially, so it starts precipitating. If some it precipitates on the layer then it will actually grow the layer a bit at first. But this won't last for long.

This is a closed system and the total quantity of aluminium oxide/hydroxide is always increasing. So at some point you have to ask if it isn't redepositing on the passivation layer where else is it accumulating? It is barely soluble, almost none of it is in solution. The aluminium surface dominates the surface area of the enclosure so accumulation on other surfaces is probably irrelevant. It could also accumulate as a useless floc. This would be a bit surprising. Normally when given enough time things try and minimize surface area and tiny little particles of floc have a great big surface area.

I think the answer might be that the passivation layer is increasing in thickness in some places. Another way to state that surface area is minimized is to state that curvature is equalized. A substance would rather precipitate in a indentation since it could be supported on more sides and would rather dissolve from a b_ump since it is less supported there. However, while the fins of a radiator are quite serpentine with lots of differences in curvature, the parts actually containing the coolant are not. So this explanation seems unlikely.

So I am having a hard time seeing how the passivation layer can ever actually get thinner in a spot given that there is no obvious place for the aluminium to go.

[xchcui]

So,we can say that checking the coolant condition with voltmeter is possible,but it is indirectly.
While the steady voltage value that we get is not the voltage between the coolant and the metals,but it shows,indirectly, the leakage through the passivation layer.
When the passivation layer goes thinner the voltage measument will increase,what is indicate,indirectly,about the decreasing of the PH and corrosive inhibitor depletion.

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'.

If the voltmeter probes were made from aluminium,i assume that the results will be as if i insert a piece of aluminium foil like you mention.
If it is right,my question is:why wouldn't i get the same 0.1V when i use similar metals as with the dissimilar metals.I understand that with similar metals(aluminium probes)i wont get 0.7V with decreasing to 0.1V,(since there will be no need to get into equilibrium),but shouldn't i get the same results,final steady 0.1V,using similar metals ?

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.

When the aluminium electrode become slightly positive,Do positive aluminium ions fall into the coolant or the negative oxide/hydroxide ions in the coolant connected with these positive charge on the electrode?
In the post i saw that you said"...positive charge leave the aluminium electrode..",but i would like to clarify this,since it looks like the aluminium positive charge in the electrode connected with the hydroxide/oxide and form a layer than drop into the coolant,doesn't it?or maybe them both happens?or in a different conditions?

[Intanjir]

shouldn't i get the same results,final steady 0.1V,using similar metals ?

There is no voltage source when using similar metals.

When the aluminium electrode become slightly positive,Do positive aluminium ions fall into the coolant or the negative oxide/hydroxide ions in the coolant connected with these positive charge on the electrode?

It should depend on the sort of ion of aluminium exists at the prevailing conditions.
http://www.intechopen.com/source/html/40512/media/image2.png
Neutral or positive species of aluminium could flow away from the positive electrode, while a negatively charged species would only be attracted.
At a pH of 8 almost all dissolved aluminium is in the form of Al(OH)₄^-.

Perhaps this is the more relevant fact for why they want the coolant to be alkaline, so long that they are principally worried about galvanic corrosion. That is, perhaps it has little or even nothing to do with the thickness of the passivation layer and more to do with whether a galvanic process can encourage aluminium to leave from a spot.
At a pH of 6 an AlOH₂⁺ could leave from the surface of the passivation layer, and nothing would have to cross through the layer's thickness.
H⁺+AlOH₃ AlOH₂⁺+H₂O

In this case the measurement at different pH of different voltages would still correspond to a different leakage rate of the capacitor when galvanically challenged by nickel but this would no longer be seen as due to a change in thickness of the passivation.

[xchcui]

From your last post,can i understand that the passivation layer contains only AlOH3?
I understand why in acidic solution the AlOH3 breaks into water and Al(OH)2+ and dissolve in the coolant,but how does the high  volume  of Al(OH)4- in high PH coolant prevent from the aluminium to corrode?
Is it because the Al(OH)4- is negative charged?
BTW when we pour alkaline coolant into the aluminium at the first time,it looks like that in order to have all that Al(OH)4- in the solution and Al(OH)3 as a passivation layer,the aluminium,initialy,lost alot of material to the coolant until it get into a kind of equilibrium,doesn't it?

[Intanjir]

The passivation layer is a network of aluminium oxide and hydroxide. There isn't really discrete bits of AlOH₃ or Al₂O₃. It is all one big network of aluminums connected through oxygens or if there is no aluminium to connect to then terminating with a hydroxide. I couldn't find a good diagram for alumina so here is the corresponding one for silica:
http://pubs.rsc.org/services/images/RSCpubs.ePlatform.Service.FreeContent.ImageService.svc/ImageService/ChapterImage/bk9781849738293/BK9781849738293-00001/bk9781849738293-00001-s1.gif


Aluminium is poorly soluble. There is very little dissolved in neutral water.
Similarly the passivation layer should also be quite thin, not much aluminium metal is lost.

A negative aluminium ion wouldn't migrate to the negative nickel electrode and would preferentially precipitate at the positive aluminium electrode.

[xchcui]

shouldn't i get the same results,final steady 0.1V,using similar metals ?

There is no voltage source when using similar metals

I am confused.Here you said that there is no voltage source using aluminium probes,while before you said there will be even more than one voltage source

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'.

BTW,if i am not wrong,the presence of an electrolyte and an electrical conducting path between the metals is essential for galvanic corrosion to occur.
In the coolant system the iron engine and the aluminium radiator are in the same coolant(electrolyte),the engine is ground but the aluminium is isolated from the ground,so there isn't electrical conducting path between the radiator to other parts.So why does the galvanic reaction,still,may occur?(i believe,it might be the first question to be ask)

[xchcui]

Thank you very much for your very informative explanation.The main question has been solved and even more and we could close this thread.
But you said two opposite things(or maybe you was talking about different cases,i don't know)that it is very important that you will clarify,since it is influence alot at understanding the all issue.
when i said that,dipping  one nickel voltmeter probe to the coolant and one probe on the aluminium probably,measure the voltage between the coolant and the aluminium radiator,You said that it is possible only if your wet electrode is made of aluminium...
So,i asked:shouldn't i get the same final steady 0.1V,whether i use aluminium voltmeter probes or nickels probes?
You said:There is no voltage source when using similar metals.
And you said that,one moment after you explained that i will get more than one voltage source if i use aluminium probe/foil.

[Intanjir]

When you use the nickel probe you create a voltage galvanically. This voltage then charges up the capacitor, which in turn then also has a voltage. So when analyzing that circuit you need to account for both of these voltage sources. So you have 0.7V from nickel, and -0.6V from the capacitor once charged up. This leaves you with 0.1V to drive a leakage current. So two components of the circuit produce a voltage, however one of them causes the other.

When you use the aluminium probe no voltages are created because the metals of the electrodes in the electrolyte are similar. This lets you get an honest measurement of the voltage across the capacitor, which should be 0 since nothing is actively charging it and it is leaky.

You can't measure the capacitor's voltage with the nickel electrode because the dissimilar metal will act as a battery. If you want to measure the voltage between the electrolyte and the radiator while using the nickel electrode then you will need to use a second voltmeter made of aluminium at the same time. In this case it should read around -0.6V.

They design the car to avoid galvanic coupling between components of the car. So any dissimilar metals that might induce corrode should be separated by a high resistance. If everything goes according to design then galvanic corrosion shouldn't be a problem. The real world is messy though. Electrical barriers can be removed or accidentally bridged, bits of rusty iron can get introduced, etc.

[xchcui]

When you use the nickel probe you create a voltage galvanically. This voltage then charges up the capacitor, which in turn then also has a voltage. So when analyzing that circuit you need to account for both of these voltage sources. So you have 0.7V from nickel, and -0.6V from the capacitor once charged up. This leaves you with 0.1V to drive a leakage current. So two components of the circuit produce a voltage, however one of them causes the other.

When you use the aluminium probe no voltages are created because the metals of the electrodes in the electrolyte are similar. This lets you get an honest measurement of the voltage across the capacitor, which should be 0 since nothing is actively charging it and it is leaky.

You can't measure the capacitor's voltage with the nickel electrode because the dissimilar metal will act as a battery. If you want to measure the voltage between the electrolyte and the radiator while using the nickel electrode then you will need to use a second voltmeter made of aluminium at the same time. In this case it should read around -0.6V.

They design the car to avoid galvanic coupling between components of the car. So any dissimilar metals that might induce corrode should be separated by a high resistance. If everything goes according to design then galvanic corrosion shouldn't be a problem. The real world is messy though. Electrical barriers can be removed or accidentally bridged, bits of rusty iron can get introduced, etc.

This is a great explanation that solved the details that i didn't understand and also it is summarizing all the issue that we were talking about.
Though,in order to understand your last explanation with high level of understanding,it was necessary to understand,first, all the details that were explained along the all thread.
Thanks,again,for your help and for your patience.

[Intanjir]