Question

How long does it take to deposit a coating of gold 1.00 um thick on a disk-shaped medallion 4.00cm in diameter and 2.00mm thick at constant current of 86A? The density of gold 19.3 g/cm³ and the gold is from an Au(III) solution. (answer in seconds please!)

Answer 1

The reaction would be:

which means 1 mole gold requires 3 moles of electrons.

While depositing gold on the edges, the diameter of a disk will be increased from 4.00 cm to 4.00 cm + 2*1.00 um = 4.00 cm + 0.0002 cm, after using significant figures, it would be 4 cm approximately. So, increase in diameter can be neglected.

The volume of gold deposited on one face of the disk:

A =

V =

since there are 2 faces, moles of gold deposited on faces =
2 x 3.1416 x (2.00cm)² x (1.00?m) x (1cm / 10000 ?m) x (19.3g / cm³) x (1 mole Au / 197.0g Au) = 2.462x10^-4 moles Au

Now, moles of gold deposited on the edge:

c = ( r is the radius of disk and d is the diameter of a disk)

Area =

Volume =

Moles of gold deposited on edge of disk = 3.1416 x (4.00cm) x (2.00mm) x (1.00?m) x (1cm / 10mm) x (1cm / 10000 ?m) x (19.3g / cm³) x (1 mole Au / 197.0g Au) = 2.462 x10^-5 moles Au

Total moles of gold deposited = 2.462 x10^-4 moles Au + 2.462 x10^-5 moles Au = 2.708x10^-4 moles Au

1 Ampere = 1 Coulomb/second

Thus, 86 A = 86 C/s

1 electron has 1.6 * 10^-19 C

1 mole electrons = 6.022 * 10^23 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