Calculating theoretical thickness while anodising aluminium oxide

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.
  • #1
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
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  • #2
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.
 

Related to Calculating theoretical thickness while anodising aluminium oxide

What is the theoretical thickness of anodized aluminum oxide?

The theoretical thickness of anodized aluminum oxide is typically calculated based on the anodizing process parameters, such as current density, electrolyte composition, and time. A common rule of thumb is that for every 1 ampere per square foot (ASF) of current density applied for one hour, approximately 0.001 inches (1 mil) of oxide thickness is formed. However, this can vary depending on specific conditions.

How do you calculate the thickness of anodized aluminum oxide?

To calculate the thickness of anodized aluminum oxide, you can use the formula: Thickness (in microns) = Current Density (in ASF) × Time (in hours) × Efficiency Factor. The efficiency factor is typically around 0.3 to 0.5 for sulfuric acid anodizing. For example, if you anodize at 12 ASF for 2 hours with an efficiency factor of 0.4, the thickness would be 12 × 2 × 0.4 = 9.6 microns.

What factors affect the thickness of anodized aluminum oxide?

The thickness of anodized aluminum oxide is influenced by several factors, including the current density, anodizing time, electrolyte temperature, electrolyte composition, and the type of aluminum alloy being anodized. Higher current densities and longer anodizing times generally result in thicker oxide layers, while higher temperatures and certain alloy compositions can affect the growth rate and quality of the oxide layer.

Why is it important to control the thickness of anodized aluminum oxide?

Controlling the thickness of anodized aluminum oxide is important because it directly impacts the material's properties, such as corrosion resistance, wear resistance, and electrical insulation. Different applications may require specific thicknesses to achieve desired performance characteristics. For example, thin coatings may be used for decorative purposes, while thicker coatings are often required for industrial applications where durability is critical.

Can the thickness of anodized aluminum oxide be measured non-destructively?

Yes, the thickness of anodized aluminum oxide can be measured non-destructively using techniques such as eddy current testing, optical microscopy, and X-ray fluorescence (XRF). These methods allow for accurate measurement of the oxide layer without damaging the sample, making them suitable for quality control and inspection purposes in industrial settings.

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