Question

A jeweler needs to electroplate gold onto a bracelet using an ionic solution. He knows that the charge carriers in the ionic solution are gold ions of charge e, and that each gold ion has a mass of 22g. The gold ions move through the solution, and are deposited on the bracelet. He has calculated that he must deposit 0.20 g of 901d to reach the necessary thickness. If each gold ion has a mass of g, what current should he use to electroplate the bracelet in three hours

Answer 1

Step-by-step explanation:

First, we will calculate the total charge as follows.

`q_(total) = \frac{\text{weight of gold to be deposited}}{\text{atomic mass of gold}} * N_(A) * \text{charge on electron}`

=
`(0.2 g)/(196.97 g/mol) * 6.02 * 10^(23) * 1.6 * 10^(-19) C`

= 97.80 C

Now, we will calculate the current required by the jeweler to plate the bracelet as follows.

i =
`(q_(total))/(t_(total))`

=
`(97.80 C)/(3 * 60 * 60 sec)`

=
`9.05 * 10^(-3)` A

or, = 9.05 mA

thus, we can conclude that 9.05 mA current should he use to electroplate the bracelet in three hours.