Question

How many faradays of electrons and coulombs of charge are needed to deposit 20 g of Mg from molten magnesium chloride (MgCl2)?

It would be much appreciated if you could please show the work that you had to arrive at that answer... It would help me understand how you got there. Thank you! :)

Answer 1

We need to find out how many electrons there are - in moles. We know that Magnesium forms ions with charge 2+, therefore one Magnesium cation needs two electrons in order to turn into an electroneutral atom. To find out the number of electrons necessary to produce 20 g of Magnesium, we divide mass by molecular mass and get moles and multiply by 2:
\[n=\frac{ m }{ M }=\frac{ 20 }{ 24 } = 0.833 mol x 2 = 1,67 mol\]
In one mole there are \[6.02*10^{23}\] particles, therefore the number of electrons we have is
\[N = N _{A}*n=6.02*10^{23}*1,67 = 10^{24}\]

Since the charge of one electron is \[1.6*10^{-19} C\]
The total charge we have to apply in order to obtain 20 g of Mg is
\[Q = 10^{24}*1.6*10^{-19}=1.6*10^{5} C = 160 kC\]

I'm a bit confused about finding the Farads as it is defined as the charge applied at a given voltage. Hope that helps.

Answer 2

Thank you SO MUCH! That did help immensely. I understand it now. Thanks again for such a detailed response!

Answer 3

Faradays's constant is F = 96500 C/mol

or : 1 faraday = 96500 coulombs

Answer 4

Thank you :)

Answer 5

I can't give you both best response but you guys were both very helpful!