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Topic: Electrolysis Plating  (Read 2441 times)

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Offline Timeless Thinker

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Electrolysis Plating
« on: April 28, 2020, 07:17:42 PM »
Aluminum is a soft, silvery-white metal used in a variety of items, from window frames and kitchen utensils to aircraft.  Aluminum must be processed via Al2O3(s)  from Bauxite. Our bauxite source claims their product contains about 65 % Al2O3(s) by mass.  The equation for electrochemical production of aluminum is

2 Al2O3(aq) + 3 C(s) ⟶ 4 Al(l) + 3 CO2(g)

(a) How much bauxite is required to produce 2.0 million metric tons of aluminum metal per year by electrolysis? (a metric ton is 1000 kg)

(b) Estimate the energy amount needed to produce this amount of aluminum in 1 year if the operating voltage for a typical aluminum generating cell is 6.0 V.

What formulas do I use for these questions?  Where does voltage come into play in part b?

Offline chenbeier

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Re: Electrolysis Plating
« Reply #1 on: April 29, 2020, 01:23:34 AM »
How is  electrical energy described. Which unit does it has?
You need current calculated by law of Faraday from a. and the voltage.

Offline Timeless Thinker

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Re: Electrolysis Plating
« Reply #2 on: April 29, 2020, 01:54:12 AM »
But How do I find amps for part a? There seem to be two variables there.
The equation I have so far is:
Amps x Seconds/year x 1C/Amp·Second x 1/96485C x 4 mol Al/12 mol e- x 203.92 g of 2Al2O3

But should I use 2(Al2O3) as molar mass or should I just use 101.96?

Offline chenbeier

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Re: Electrolysis Plating
« Reply #3 on: April 29, 2020, 02:47:23 AM »
What is the formula for faraday law?

Which variables do you have?

Offline Borek

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Re: Electrolysis Plating
« Reply #4 on: April 29, 2020, 03:12:11 AM »
But How do I find amps for part a? There seem to be two variables there.

No, this is just a simple stoichiometry, one that even doesn't require balanced reaction equation. All you have to do is to follow aluminum.

If you have 1 kg of bauxite, what mass of Al2O3 does it contain? And how much Al?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline Enthalpy

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Re: Electrolysis Plating
« Reply #5 on: April 29, 2020, 06:06:01 AM »
The equation for electrochemical production of aluminum is
2 Al2O3(aq) + 3 C(s) ⟶ 4 Al(l) + 3 CO2(g)
WHAT??

Offline chenbeier

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Re: Electrolysis Plating
« Reply #6 on: April 29, 2020, 06:27:33 AM »
This can be used, because the Carbon rods will be oxidiesed as anodes by the oxide ions

The cathion reaction is the reduction of aluminium ions. The anode reaction is the oxidation of carbon.

Combination of both gives a normal redox reaction. Even current is used for an electrolysis.


Offline Timeless Thinker

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Re: Electrolysis Plating
« Reply #7 on: April 29, 2020, 10:17:27 AM »
1kg of bauxite has .650 kg of Al2O3 and that has .344 kg of Al.
Do I just set up a proportion of .344/1kg = 2 million tonnes/ x tonnes?
So then the answer is 5.812 million metric tonnes?
 

Offline chenbeier

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Re: Electrolysis Plating
« Reply #8 on: April 29, 2020, 11:32:09 AM »
1kg of bauxite has .650 kg of Al2O3 and that has .344 kg of Al.
Do I just set up a proportion of .344/1kg = 2 million tonnes/ x tonnes?
So then the answer is 5.812 million metric tonnes?

Yes

Offline Timeless Thinker

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Re: Electrolysis Plating
« Reply #9 on: April 29, 2020, 11:57:11 AM »
And part b, where does the voltage come into to equation? How do I use that to solve for amps?

Offline chenbeier

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Re: Electrolysis Plating
« Reply #10 on: April 29, 2020, 12:01:09 PM »
Do you know Faraday Law ?
How do you calculate Power and Energy . 
The inventer of the steam engines name multiplied by time.

Offline Timeless Thinker

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Re: Electrolysis Plating
« Reply #11 on: April 29, 2020, 08:50:03 PM »
Yes: Joules = Voltage x Current x Time
But I'm not sure if my equation to solve for Amps is correct. This is what I have so far:

grams Al x (1 mol Al/101.96g) x (A·S)/C x 96486C/1 mol e- x 12 mol e-/4 mol Al = seconds/yr

Should I use 1 mol Al/101.96g? In the reaction, there is a coefficient before the 2Al2O3. Also, is my mols of e-/mols of Al correct?

Offline chenbeier

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Re: Electrolysis Plating
« Reply #12 on: April 30, 2020, 12:54:20 AM »
The Faraday law is given m = I* t * M/(z*96485 As)
m = mass [g], I = current [A], t = time [s ], z = amount of electron for Al it is 3, M = molar mass [g/mol]

Solve for I.

The reaction is Al3+ + 3 e- => Al

So no need for coefficients.



Offline Timeless Thinker

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Re: Electrolysis Plating
« Reply #13 on: April 30, 2020, 01:22:24 AM »
That makes much more sense, thank you. Should my final answer then be in Amps?

Offline chenbeier

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Re: Electrolysis Plating
« Reply #14 on: April 30, 2020, 01:56:06 AM »
Your final answer should be in something like watt hour or joule.
After getting the amps you have to calculate the energy.

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