For a science experiment you need to electroplate

[rayhan619]

Homework Statement

For a science experiment you need to electroplate a 130 nm thick zinc coating onto both sides of a very thin, 3.0 cm*3.0 cm copper sheet. You know that the charge carriers in the ionic solution are divalent (charge 2e) zinc ions. The density of zinc is 7140 kg/m^3
If the electroplating apparatus operates at 1.3 mA, how long will it take the zinc to reach the desired thickness?

Homework Equations

not sure

The Attempt at a Solution

I have no idea how to solve this problem

 

[LowlyPion]

Figure that you are going to need to supply 2 electrons for each zinc atom you are to plate onto the copper.

Figure the volume of the plating as total surface area times the depth.

They give you the density so figure how many atoms need to be plated. That times 2 electrons then has to pass as the current doesn't it?
And they give you the rate that electrons can be supplied, so ...

 

[rayhan619]

so we know area and thickness of the plate and the density of the zinc, and the current.
so we can get the volume of the plate and using density we can figure out the mass.
volume = 3*10^-2* 3*10^-2* 130*10^-9 m = 1.17*10^-10 m^3
and the mass = 1.17*10^-12 m^3 * 7140 kg/m^3 = 8.35*10^-7 kg
what is the next step? how do i figure out number of atoms?

 

[LowlyPion]

so we know area and thickness of the plate and the density of the zinc, and the current.
so we can get the volume of the plate and using density we can figure out the mass.
volume = 3*10^-2* 3*10^-2* 130*10^-9 m = 1.17*10^-10 m^3
and the mass = 1.17*10^-12 m^3 * 7140 kg/m^3 = 8.35*10^-7 kg
what is the next step? how do i figure out number of atoms?

Looking up zinc I see the standard atomic weight is 65.38 g/mol.
http://en.wikipedia.org/wiki/Zinc

How many atoms in a mole?

And 2 electrons needed per atom and ...

 

[rayhan619]

so in 1 mole the weight of zinc is 65.38 g
so in 8.35*10^-7 kg 1.3*10^-8 mole
right?

 

[LowlyPion]

so in 1 mole the weight of zinc is 65.38 g
so in 8.35*10^-7 kg 1.3*10^-8 mole
right?

And now how many atoms per mole?

Does Avogadro ring a bell?

 

[rayhan619]

i don't know. its complicated for me.
in 1 mole we have 6.022*10^23 atoms
so in 1.3*10^-8 mole we have 1.3*10^-8 * 6.022*10^23 = 7.83*10^15 atoms
so number of electrons = 7.83*10^15 *2 = 1.57*10^16

then how do i figure out the time?

 

[LowlyPion]

i don't know. its complicated for me.
in 1 mole we have 6.022*10^23 atoms
so in 1.3*10^-8 mole we have 1.3*10^-8 * 6.022*10^23 = 7.83*10^15 atoms
so number of electrons = 7.83*10^15 *2 = 1.57*10^16

then how do i figure out the time?

Well what is the current? 1.3 ma right? So how many electrons per second is that?

1 ampere is 1 Coulomb / sec

and 1 Coulomb is equal to 6.241 *10¹⁸ electrons.

 

[rayhan619]

1 Coulomb is equal to 6.241*10^18 electrons.
so 1.57*10^16 electron = 2.5*10^-3 C
I = Q/t
so t = Q/I
= 2.5*10^-3/1.3*10^-3
= 1.92 s

is this right?

 

[LowlyPion]

so we know area and thickness of the plate and the density of the zinc, and the current.
so we can get the volume of the plate and using density we can figure out the mass.
volume = 3*10^-2* 3*10^-2* 130*10^-9 m = 1.17*10^-10 m^3
and the mass = 1.17*10^-12 m^3 * 7140 kg/m^3 = 8.35*10^-7 kg
what is the next step? how do i figure out number of atoms?

Oh, and you want me to check numbers too?

Well, now that I look ... there are 2 sides to be plated. So that one is not quite right.

 

[LowlyPion]

1 Coulomb is equal to 6.241*10^18 electrons.
so 1.57*10^16 electron = 2.5*10^-3 C
I = Q/t
so t = Q/I
= 2.5*10^-3/1.3*10^-3
= 1.92 s

is this right?

Looks like it propagates the factor of 2 all the way. I get 3.785s.

 

[rayhan619]

ok so for both side the total time would be 1.92*2 = 3.84 s
I tried plugging that in but that didnt work.
any idea what I am doing wrong here?

 

[LowlyPion]

ok so for both side the total time would be 1.92*2 = 3.84 s
I tried plugging that in but that didnt work.
any idea what I am doing wrong here?

Not sure why. Not without redoing the whole daisy chain.