Calculating theoretical thickness while anodising aluminium oxide

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To calculate the theoretical thickness of an anodised coating, the critical factor is determining the number of electrons involved in the reactions. The anode reaction transfers 2 electrons, but since 3 oxygen atoms are consumed to produce Al2O3, a total of 6 electrons are effectively transferred. Therefore, when using the formula for thickness, the number of electrons should be set to 6 to accurately reflect the production of Al2O3. This approach aligns with the molar mass of Al2O3, ensuring the calculation is correct. Understanding the electron transfer is essential for precise anodising thickness calculations.
lforster02
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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
unnamed.jpg
 
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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.
 
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