A piece of Al with a surface area of 2.5 m2 is anodized to produce a film of Al₂O₃ that is 23 μm (23X10⁻⁶ m) thick.(b) If it takes 18 min to produce this film, what current must flow through the cell?

Respuesta :

The current must flow through the cell is 1.2×[tex]10^{3} }[/tex].

A stream of charged particles, such as electrons or ions, traveling through an electrical conductor or a vacuum is known as an electric current. The net rate of electric charge flowing through a surface or into a control volume is how it is calculated.

The surface area and thickness are 2.5 m² and 23X10⁻⁶ m respectively.

Volume =Surface area x thickness

Volume =2.5 m²x (23×10⁻⁶ m)

Volume =5.8×10^-5 m³

The 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, the oxidation number of Al in Al2O3 is +3 that in Al is zero. So, one Al3+ gains three electrons 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)

Charge =1.3×10^6 C

Hence the charge passed in anodizing is 1.3×10^6 C

Current =  charge × time

Current = (1.3×10^6 C ×1 min) /(18 min ×60 s) = 1.2×10^3 A

Hence, the current that flows is 1.2×10^3

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