Question

How long does it take to deposit a coating of gold 1.00 μm thick on a disk-shaped medallion 4.00 cm in diameter and 2.00 mm thick at constant current of 85 A? The density of gold is 19.3 g/cm3 and the gold is from an Au(III) solution.

Answer 1

First.. think about the reaction...

Au(+3) + 3e's ---> Au(s)

meaning 1 mole Au = 3 moles e's

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next think about depositing gold on the edges. you will increase the diameter of the disk from 4.00cm to 4.00cm+2x1.00μm = 4.00cm + 0.0002cm, which with the sig figs given = 4.00cm. So we can ignore the increase diameter

And if you think about the volume of gold deposited ONE face of the disk

A = pi x r²

V = pi x (d/2)² x (height)

since there are 2 faces, moles of gold deposited on faces =

2 x 3.1416 x (2.00cm)² x (1.00μm) x (1cm / 10000 μm) x (19.3g / cm³) x (1 mole Au / 197.0g Au) = 2.462x10^-4 moles Au

And if you think about the moles of gold deposited on the edge,

c = 2 x pi x r = pi x d

A = pi x d x thickness of disk

V = pi x d x thickness of disk x thickness of gold

moles of gold deposited on edge of disk =

3.1416 x (4.00cm) x (2.00mm) x (1.00μm) x (1cm / 10mm) x (1cm / 10000 μm) x (19.3g / cm³) x (1 mole Au / 197.0g Au) = 2.462 x10^-5 moles Au

and total moles Au deposited = 2.462 x10^-4 moles Au + 2.462 x10^-5 moles Au = 2.708x10^-4 moles Au

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then...

since 1 amp = 1C/s, 85A = 85C/s

and since 1e = 1.602x10^-19 C

and since 1 mole electrons = 6.022x10^23 electrons...

via dimensional analysis...

(2.708x10^-4 moles Au) x (3 moles e / 1 mole Au) x (6.022x10^23 e / mole e) x (1.602x10^-19 C / 1e) x (1sec / 85C) = 0.922 seconds

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