A circular medallion is to be electroplated with gold by using an electrolyte of #sf(Au_((aq))^(3+))#. Given the data below, how long will it take to plate a layer of gold #sf(1mum)# thick?

Firstly I will find the volume of the plated medallion. Then subtract the volume of the unplated disc.

This will give me the total volume of gold. Given the density I can then get the mass of gold.

Since we know it is #Au^(3+)# I can find the quantity of charge needed to discharge this mass of gold.

Since current is the rate of flow of charge I can get the time taken to discharge this amount.

The first thing to do is put the lengths into a consistent unit as three different ones are used in the question. Since density is in #"g/cm"^(3)# I will put everything into #"cm"#.

Imagine a cross section like this:

The blue area is the gold. The orange area is the medallion.

The volume of the plated disc #V# is given by:

#V=piR^(2)H#

The thickness of the layer is #1mum=10^(-6)m=10^(-4)cm#

So #R=(4/2 + 10^(-4))=2.0001cm#

The medallion is #2mm# thick. #1mm=0.1cm:.h=0.2cm#

#:.H=(0.2+2xx0.0001)=0.2002cm#

So the volume of the plated disc becomes:

#V=pi(2.0001)^(2)xx0.2002" "cm^(3)#

#V=pi(0.800880082)" "cm^(3)" "color(red)((1))#

The volume of the medallion #v# is given by:

#v=pir^(2)h#

#v=pi[2^(2)xx0.2]=pi(0.8)" "cm^3" "color(red)((2))#

To get the volume of gold in blue we subtract #color(red)((2))# from #color(red)((1))rArr#

#V_(Au)=pi[0.800880082-0.8] " "cm^(3)#

#V_(Au)=pi0.000880082" "cm^(3)#

#:.V_(Au)=0.00276486" "cm^(3)#

The density #rho# is given by:

#rho=m_(Au)/V_(Au)#

#:.m_(Au)=rhoxxV_(Au)=19.3xx0.00276486=0.0533618"g"#

The electrode equation is:

#Au^(3+)+3erarrAu#

This tells us that 1 mole of #Au# will require 3 moles of electrons to discharge.

The charge on 3 moles of electrons = #3xxF# where #F# is the charge on a mole of electrons and is the Faraday Constant which is #9.6485xx10^(4)"C/mol".#

The #A_r# of #Au# is #196.96657#

#:.196.99657"g"# require #3xx9.6485xx10^(4)=28.945xx10^(4)"C"#

#:.1"g"# requires #(28.945xx10^(4))/(196.99657)"C"#

#:.0.0533618"g"# require #(28.945xx10^(4))/(196.99657)xx0.0533618"C"#

#=78.405"C"#

Electric current #I# is the rate of flow of charge:

#I=Q/t#

#:.t=Q/I=78.405/89 "s"#

#t=0.881"s"#

In reality a current of #89"A"# is way too high. As well as being highly dangerous it would not give a good deposit. For a good finish you need a low current density (I/A).

Otherwise you get a spongy deposit which falls away easily.