Respuesta :
The charge passed in anodizing is 1.3×10^6 C
Surface area and thickness is 2.5 m² and 23X10⁻⁶ m respectively. Calculate the volume of Al2O3 coating as show below:
Volume =Surface area x thickness
=2.5 m²x (23×10⁻⁶ m)
=5.8×10^-5 m³
Density of Al, is 3.97 g/cm³
Calculate the mass of aluminum as shown below:
Density= Mass/Volume
3.97 = Mass×1/5.8×10^-5 ×10^6
Mass = 2.3×10^2g
The anodization of Aluminum is as shown below:
4AI (s)+ 302 (g) → 2Al2/O3 (s)
Now, oxidation number of Al in Al2O3 is +3 whi that in Al is zero. So, one Al3+ gains three electron to form one atom of Al.
Al (s)→ Al (s)+3e
Charge = 2.3×10^2× 6 × 9.65 × 10^6 /(101.95 × 1 × 1)
=1.3×10^6 C
Hence the charge passed in anodizing is 1.3×10^6 C
Learn more about charge here:
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