Calculating theoretical thickness while anodising aluminium oxide

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SUMMARY

The discussion focuses on calculating the theoretical thickness of anodised aluminium oxide (Al2O3) by determining the correct number of electrons to use in the formula. The anode reaction involves the transfer of 2 electrons, but since 3 oxygen atoms are consumed to produce one Al2O3 molecule, a total of 6 electrons are effectively transferred. Therefore, the formula for thickness should incorporate 6 electrons when using the molar mass of Al2O3. This clarification is crucial for accurate calculations in anodising processes.

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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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