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Electrolysis

Electrolysis, the purpose of which is to produce aluminium from alumina (aluminium oxide), is carried out with a potential difference of 4.5V between the two electrodes and the current passing through the cell is 180kA with the current efficiency of 95%.
Calculate the amount of energy required for production of 100kg of aluminium.

I first wrote the reaction equation: 2 Al2O3 (s) = 2 Al (s) + 3 O2 (g) with transfer of 12e-. Then I calculated the moles of aluminium that need to be produced, then moles of electrons that would take, then charge passed, then multiplied by Q to get E. My final value was

E = (100*103*96485*4.5)/26.9815 * 12/2 = 9.6551*109 J, around 9.66 GJ

But the answer given was 5.07 GJ. Where have I gone wrong?
(edited 10 years ago)
Reply 1
Avatar for Borek
Borek
4
Original post by Big-Daddy
Where have I gone wrong?


In two places. Check your reaction, don't ignore information included in the problem.
Reply 2
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Big-Daddy
OP
13
I see ... that was dumb.

So now 2 Al2O3 (s) = 4 Al (s) + 3 O2 (g) with transfer of 12e-, E = (100*103*F*V)/26.9815 * 12/4 = 4.82756 * 109 J, i.e. around 4.83 GJ. Is the difference just rounding? It feels like they've multiplied 4.83*1/0.95 = 5.08 GJ, i.e. the final issue is probably with the current efficiency, but what's going wrong? I think E = QV and Q is a direct function of how much Al (s) is produced so the fact that current is actually 180*103*0.95 A shouldn't matter.

In other words, for some mass of Al (s) to be produced we need some number of electrons to pass. This number of electrons (in moles) is translated directly to charge by multiplying by Faraday's constant, thus we find Q. And E=QV. Why do we need the current or "current efficiency" at all?

(You do need it in another part of the problem, which I got right - and now I can see why I got it right, but this one wrong, as they asked to find the rate of production of CO2 (s) when the anode is graphite, and because I got the right coefficient for O2 (g) I was fine here.)
Reply 3
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Borek
4
What is the definition of the current efficiency?
Reply 4
Avatar for Big-Daddy
Big-Daddy
OP
13
Original post by Borek
What is the definition of the current efficiency?


Ah I see. It's more than just "current says its 180 kA but is actually lower by some factor due to power losses". It is an explicit statement of what fraction of the electrons that pass are being used for "efficient current" i.e. are being used to drive specifically the reaction being considered.

One question about current efficiency - is it purely a function of the circuit being used? Or does it depend on the exact system, including composition, which redox reaction is occurring, etc.? If I want to take that same apparatus as in the question and move it to a new electrolytic cell, with new starting concentrations and maybe even completely different reagents, will current efficiency be the same as before or do I need to re-measure it for the system now at hand?
Reply 5
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Borek
4
It will depend on the system, as it depends on side reactions.

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