# Calculating theoretical thickness while anodising aluminium oxide

• lforster02
If you are using the molar mass of Al, then you need to use the number of electrons consumed per Al produced, i.e. 2.In summary, the conversation is discussing the theoretical thickness of an anodised coating. The anode reaction involves 2 electrons being transferred, but the reaction with aluminium consumes 3O, resulting in 6 electrons being transferred. The formula for calculating the thickness depends on whether the molar mass of Al2O3 or Al is used, with 6 or 2 electrons being used, respectively.
lforster02
I am trying to calculate the theoretical thickness of an anodised coating. I know all values apart from I am unsure whether to use 2e or 6e for the amount of electrons.

Anode reaction as follows: 2OH = H2O + O + 2e

the aluminium reactions preferentially with the one O atom, using equation: 3O + 2Al (anode) = Al2O3

There is 2 electrons transferred in the anode reaction, although 3O consumed to produce the coating Al2O3, which means 6e is transferred. Does anyone have any idea whether to use 6 or 2 in the formula: thickness = (1/density) x ((molecular mass x current x time) / (area x number of e x faradays constant))

Attached are the relevant reactions

Last edited by a moderator:
If you are using the molar mass of Al2O3, then you need to use the number of electrons consumed per Al2O3 produced, i.e. 6.

• Biology and Chemistry Homework Help
Replies
4
Views
11K
• Materials and Chemical Engineering
Replies
5
Views
5K
• Chemistry
Replies
1
Views
6K
• General Engineering
Replies
2
Views
6K
• Biology and Chemistry Homework Help
Replies
1
Views
1K
• Biology and Chemistry Homework Help
Replies
16
Views
21K
Replies
1
Views
2K
• Chemistry
Replies
4
Views
7K
• Chemistry
Replies
4
Views
6K
• Biology and Chemistry Homework Help
Replies
4
Views
32K