What current is required to electroplate 6.07 ug of gold in 30 min from

Question

a gold (III) chloride aqueous solution? in units of A

Answer 1

6.07 µg of gold is 6.07/197 µmoles. This represents 6.07*6.02/197 * 10^17 = 1.85*10^16 atoms of gold, with each gold ion carrying the charge of 3 electrons. The total charge carried is then 3*1.6*10^-19*1.85*10^16 C = 8.9*10^-3 C. Amperes is coul/sec, so 30 min = 30*60 sec so the current is 8.9*10^-3/1800 A = 4.95*10^-6 A or 4.95 µA

You could also use Faraday's law which is m = Q/F * M/n, where M = molar mass (197 for gold), n = valence no. F = Faraday's constant = 96485 C/mole. Given m = 6.07*10^-6 g, solve for Q, and find current in A from Q/t, t = 1800 sec.