What current is required to electroplate 6.07

Question

ug of gold in 30 min from a gold (III) chloride aqueous solution? answer in units of A

Answer 1

1A (ampere) = 1 coulomb/sec

Atomic weight: Au=197

6.07ugAu/30min x 1gAu/1x10^6ugAu x 1molAu/197gAu x 3eqwtAu/1molAu x 96,500coulomb/1eqwtAu x 1min/60sec = 2.97 x 10^-4 coulomb/sec (A)

Answer 2

6.07 ug = 6.07 x 10^-6 g

Moles Au = 6.07 x 10^-6 g / 197 g/mol = 3.1 x 10^-8

Au3+ + 3e- >> Au

For every 3 faradays consumed 1 mole of Au is produced

3.1 x 10^-8 x 3 = 9.2 x 10^-8 Faraday

96500 Coulombs/ Faraday / 9.2 x 10^-8 Faraday = 0.00892 Coulombs

30 min = 1800 s

0.00892 / 1800 =4.96 x 10^-6 Coulombs/s => 4.96 x 10^-6 A