Question

A sculptor has asked you to help electroplate gold onto a brass statue. You know that the charge carriers in the ionic solution are singly charged gold ions, and you've calculated that you must deposit $0.50\mathrm{g}$ of gold to reach the necessary thickness. How much current do you need, in $\mathrm{mA}$, to plate the statue in $3.0$hours?

Short answer

The amount of current need to plate the statue in$3.0hrs$is$23mA.$

Step-by-step solution

Given Information

Amount of gold$=0.50g$

Time$=3.0hrs$

Necessary current to plate the statue in$3.0hrs$$=?$

Explanation 

Due to the missing electron on slightly charged ions, they carry a positive charge $e$.

The total charge passing through the solution onto the statue is, therefore:

$Q=It$

Where $t=3.0\mathrm{h}$

Total number of ions that pass is,
localid="1649078297650" $N=\frac{Q}{e}=\frac{It}{e}\to \left(1\right)$

Total number of ions necessary for $m=0.50\mathrm{g}$is

$N=\frac{m}{\mu },$

Here $\mu$is the mass of one ion of gold.

Now, $\mu$can be calculated by dividing the molar mass of gold with Avogadro's constant

$\mu =\frac{M}{N_{\mathrm{A}}},$

Which yields,

$N=\frac{m}{M}{N}_{\mathrm{A}}$

Equating (equation 1) with (equation 2) we find

$\frac{m}{M}{N}_{\mathrm{A}}=\frac{It}{e}$

This finally gives the expression for the current

$I=\frac{me{N}_{\mathrm{A}}}{Mt}.$

Molar mass of gold is $M=197\mathrm{g}/\mathrm{mol}$,

$I=23mA.$

Final Answer

Hence, the amount of current need to plate the statue in$3.0hrs$is$23mA.$