Question

A new aluminium (AI) container develop pits corrosion right through its wall with an average current density of 1.3x104 A/cm2. If the average pits is 0.70mm in diameter, the aluminium wall is 0.90 mm thick and assume the pits has a cylindrical shape and the corrosion currents acts uniformly over the surface area of the pits. (Given atomic mass for Al is 26.98 g/mol and density is 2.7g/cm3). [Satu bekas baru aluminium (AI) terhasil kakisan lubang melalui dinding dengan satu ketumpatan arus purata 1.3 x 10' A/cm2. Jika purata diameter lubang adalah 0.70 mm, tebal dinding aluminium adalah 0.90 mm dan ganggapkan lubang mempunyai bentuk silinder dan arus kakisan bertindak seragam di kawasan permukaan lubang. (Diberi jisim atom untuk Al adalah 26.98 g/mol dan ketumpatan adalah 2.7g/cm3).J (i)Calculate the surface area for the cylindrical pits. [Kirakan luas permukaan bagi lubang silinder.] (2 Marks / Markah) (ii) Evaluate the mass of pits [Nilaikan jisim lubang./ (2 Marks / Markah) (iii) Predict the time required for pits to corrode using Faraday's equation. [Ramalkan masa yang diperlukan bagi lubang untuk menghakis menggunakan persamaan Faraday 's.] (2 Marks / Markah)

Answer 1

Average current density = 1.3 times 10^4 A/cm^2 Diameter of average = 0.70 mm Rightarrow r = 0.35 mm thickness is h = 0.9 mm atomic mass of AL = 26.98 g/m Density = 2.7 g/cm^3 rightarrow (a) surface area of cylindrical pits = 2 pi r h + 2 pi r = 2 pi (0.35) (0.9) + 2 pi (0.35)^2 = 1.9792 + 0.7697 = 2.7488 mm^2 b) volume of cylindrical pits = pi r^2 h = pi times (0.35)^2 (0.9) = 0.3464 mm^3 Density = 2.7 g/m^3 = 2.7 g/10^3 mm^3 = 2.7 times 10^- 3 g/mm^3 Therefore mass of pits = volume of pits times Density = 0.3464 times 2.7 times 10^- 3 g = 0.93528 times 10^- 3 g