[xchcui]

Hi Tittywahah.
I don't want to push aside the issue that i try to understand and spend time on the electro plating or other aspects of that issue.
The electro cleaning process is the main issue that i am trying to understand.
And I would like to focus on the questions that i directed to eazye1334:

"""...1)When i make the part anodic for a brief period and oxygen is liberated at the part surface.Doesn't the negative oxygen ions(O--)oxidize the part again and make the part surface to be copper oxide again(CuO)?
I know that the O-- ions give electron to the anode and become a gas,but what happens next,that cause the Cu++ ion to drop into the solution,rather than be connected with the oxygen ions in order to form CuO again?
2)When i make the part cathodic and the H+ ions take electrons from the copper oxide surface part and turns to bubbles gas,How does this action cause the oxygen from the part surface to be removed?
Also,when the oxygen is removed from the part surface in that process,the part is also being plated with Cu++ ions and with other loose ions in the solution,Isn't it disturb to the cleaning process?..."""

The answer for those questions will help me alot to understand the issue.

[Tittywahah]

)When i make the part anodic for a brief period and oxygen is liberated at the part surface.Doesn't the negative oxygen ions(O--)oxidize the part again and make the part surface to be copper oxide again(CuO)?
===I have to assume by 'PART' you mean the dirty copper piece, if so then No, that piece of copper gets quite clean from my observations and the rubbish simply drops out of solution.====
As I understand the chemistry (and I have not looked this up so I may well be misguided, but I am learning as well), physical copper does not really drop and travel anywhere, you have electrons moving away from the anode and attaching themselves to copper cathode.  Elemental copper has no charge, neither the cathode or the anode, both pieces are ZERO charge.  So when an electrical current is passed electrons from the copper anode head towards the copper cathode. As the copper anode starts to lose electrons it is being Oxidized and each moment of oxidation the copper becomes negatively charged and an oxide is formed and this drops out straight away into the solution.

The rest of your questioning I find difficult to follow, call it a lack of experience, but this is as far as I can take you with my understanding.  If I am mistaken then I will be glad to be corrected.  Here is another link for you to watch _ I don't have time at the moment to watch it, but hope there is something here.
Kind regards  http://www.youtube.com/watch?v=MTqxhUGf6s4

[eazye1334]

"""...1)When i make the part anodic for a brief period and oxygen is liberated at the part surface.Doesn't the negative oxygen ions(O--)oxidize the part again and make the part surface to be copper oxide again(CuO)?
I know that the O-- ions give electron to the anode and become a gas,but what happens next,that cause the Cu++ ion to drop into the solution,rather than be connected with the oxygen ions in order to form CuO again?

Ok, so backing up, normally you would be using an alkaline cleaner for an electrocleaner. The normal process involves electrocleaning the part and then following up with an acid dip. Yes, through anodic cleaning, you will produce somewhat of an oxide film at the surface of the part, but a dip into acid will remove the oxide film because it has very low adhesion. The de-plating action of the anodic cleaning will help remove all other impurities on the surface, leaving you will a clean part surface and a very thin oxide layer, which as I said can be removed with an acid dip. This oxide layer can also be left on as a passivation layer that helps with corrosion resistance. However, it should not affect the appearance of the part.

2)When i make the part cathodic and the H+ ions take electrons from the copper oxide surface part and turns to bubbles gas,How does this action cause the oxygen from the part surface to be removed?
Also,when the oxygen is removed from the part surface in that process,the part is also being plated with Cu++ ions and with other loose ions in the solution,Isn't it disturb to the cleaning process?..."""

Water gets broken down at the part surface, which is where the hydrogen comes from. The hydrogen at the part surface is reducing the oxides and forming OH- ions. I would guess that if you check the pH of your cleaning solution after cathodic cleaning that the pH would be higher than when you started. I say guess because I've never used an acid for electrocleaning, it's just a thought.

The part will be plated with some ions, whether copper or anything else you have in the solution, but keep in mind that these we have very, very low adhesion. The scrubbing done by the hydrogen gas helps clean the surface and may help remove some of those plated ions. Keep in mind that the hydrogen evolution during cathodic cleaning is twice as high as the oxygen evolution during an anodic cleaning, so it is a pretty violent scrubbing action. But this is exactly why you do the anodic cleaning as the final step, because that will remove what was loosely plated. You should always end with anodic cleaning.

[Tittywahah]

Obviously I missed the boat then, misunderstood everything?

[eazye1334]

Obviously I missed the boat then, misunderstood everything?

Honestly I couldn't tell you. I'm a chemical engineer, not a chemist, so I can't confirm or deny exactly what you said. I just know about this because I do it for a living.

[xchcui]

Thanks for your explanations.

[Ok, so backing up, normally you would be using an alkaline cleaner for an electrocleaner. The normal process involves electrocleaning the part and then following up with an acid dip. Yes, through anodic cleaning, you will produce somewhat of an oxide film at the surface of the part, but a dip into acid will remove the oxide film because it has very low adhesion. The de-plating action of the anodic cleaning

As i said i uses vinegar+pinch of salt(NaCl)as elctrolyte.This is cheap method for me and safer.
So after i will do the second step by connecting my object(part)to the anode and it will be covered with very thin oxide layer,Will a fresh vinegar+salt solution be enough acidic in order to remove this layer from the part by dip it in?
Can you give me the chemical reaction at the anode that shows how the oxide layer is created?
I saw in the link(that Tittywahah attached)this reaction at the anode:
4OH- ions becomes O2+2H2O+4e-.
So i get water and oxygen gas bubbles,but how does the oxygen react with the copper and become that thin oxide layer if i get only plain oxygen gas?

[Tittywahah]

Ok, I will give it my best:  If wrong I hope someone will correct me:
Cu in its elemental state has no charge.  In this reaction Oxygen has -2 charge, these are the oxidation states:
Now here is the reaction I believe:  Cu + O2 = CuO   what is really happening is this:
Cu⁰ + O₂^4- = Cu²⁺O^2-
There are TWO oxygen atoms (making ONE oxygen molecule) and only ONE oxygen atom is bonding with the copper, where does the other one go?

[Borek]

Cu in its elemental state has no charge.

OK

In this reaction Oxygen has -2 charge

Not exactly, initially it is in the elemental state and has no charge as well (oxidation number of zero).

Now here is the reaction I believe:  Cu + O2 = CuO   what is really happening is this:
Cu⁰ + O₂^4- = Cu²⁺O^2-
There are TWO oxygen atoms (making ONE oxygen molecule) and only ONE oxygen atom is bonding with the copper, where does the other one go?

This is wrong. Correct reaction equation is

2Cu + O₂ 2CuO

(which already explains what happens to the other oxygen).

This is a redox reaction, in which copper gets oxidized

Cu  Cu²⁺ + 2e^-

and oxygen gets reduced:

O₂ + 4e^- 2O^2-

[xchcui]

ok.i see that the oxygen react with the copper and become CuO,but where is the reaction that creates the oxygen bubbles?

And refer to the second part of my question in my last post relating to easye1334:

... The de-plating action of the anodic cleaning will help remove all other impurities on the surface, leaving you will a clean part surface and a very thin oxide layer, which as I said can be removed with an acid dip.

Does a fresh vinegar 5%+salt(NaCl)will do this job?

[billnotgatez]

Just curious

What kind of power source are you using?
What voltage are you using?

[eazye1334]

Does a fresh vinegar 5%+salt(NaCl)will do this job?

Yes, it should be enough and you don't need the salt. Just the vinegar will do, it's actually what we use before passivating the part through other chemical means.

[xchcui]

Thanks,eazye1334.
 
Due to the fact that your explanations were the most understandable to me(without deprive any of the nice members that help me with my questions),can you refer to and explained to me what is going on with the oxygen that created in the anode?
When you mention that the anode get thin layer of oxide on it, Tittywahah and Borek explained that the reaction is 2Cu+O2 creates 2CuO.
But what with the O2 that get out from the solution as bubbles?Not all the O2 molecule is bonded to the anode,isn't it?


 

[eazye1334]

Thanks,eazye1334.
 
Due to the fact that your explanations were the most understandable to me(without deprive any of the nice members that help me with my questions),can you refer to and explained to me what is going on with the oxygen that created in the anode?
When you mention that the anode get thin layer of oxide on it, Tittywahah and Borek explained that the reaction is 2Cu+O2 creates 2CuO.
But what with the O2 that get out from the solution as bubbles?Not all the O2 molecule is bonded to the anode,isn't it?

That just escapes as a gas. The oxide layer you're forming in the anodic step is not a very good, as in it is very thin and poorly adhered to the part. That's why a weak acid will still remove it.

[xchcui]

So,we can conclude and say that the hydroxide negative ion(4HO-)give its 4 electrons to the copper-anode while half of the O2 molecules that have been creating(besides the water)stick to the copper(oxide layer)and the other half,turning to a bubbles gas that escape out.Am i right?

Just curious

What kind of power source are you using?
What voltage are you using?

This is my first time so i start with small objects,while i use standard power supply 12vdc for electro-cleaning.should i use for the electro-cleaning lower voltage?like 6v maybe?

[billnotgatez]

@xchcui

standard power supply 12vdc for electro-cleaning

Could you post make and model etc