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

Explanation:

First, we will calculate the total charge as follows.

      $q_{total} = \frac{\text{weight of gold to be deposited}}{\text{atomic mass of gold}} \times N_{A} \times \text{charge on electron}$

                    = $\frac{0.2 g}{196.97 g/mol} \times 6.02 \times 10^{23} \times 1.6 \times 10^{-19} C$

                    = 97.80 C

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

                   i = $\frac{q_{total}}{t_{total}}$

                     = $\frac{97.80 C}{3 \times 60 \times 60 sec}$

                     = $9.05 \times 10^{-3}$ A

or,                  = 9.05 mA

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