Masses or Volumes from Electrolysis.
The electrolysis of molten aluminium
oxide - Al2O3
produces aluminium at the cathode and oxygen at the anode.
See the extraction of aluminium for more detail.
A current of 10
amps is allowed to flow
through molten aluminium oxide for 5 hours.
What mass of aluminium is deposited at the cathode?
1) Write the half equation for the electrolysis.
Al3+ + 3e- Al
3 moles of electrons (3 Faradays)
are required to deposit 1 mole of aluminium.
2) Find how
many Faradays have passed through
the aluminium oxide in 5 hours.
Q = I x t
hours contains (5 x
60 x 60)
= 18,000 seconds.
10 x 18,000
= 180,000 Coulombs.
Faraday = 96,500 coulombs.
180,000 coulombs = 180,000 ÷ 96,500 Faradays
= 1·865 Faradays.
3) From the
proportion in 1
3 Faradays are required to deposit 1 mole of aluminium.
1·865 Faradays will deposit (1·865 ÷ 3) moles of aluminium
= 0·622 moles of aluminium.
mass = moles x RAM
RAM of aluminium = 27
= 0·622 x 27
= 16·79 g of aluminium.
Compare this figure with 193 g of lead, 11·5 times the mass
from the same quantity of electricity (see the previous page).
The high charge on the aluminium ion (Al3+) requiring
3 Faradays per mole makes aluminium expensive to extract.
In addition, aluminium has a light nucleus (RAM = 27).
One mole of aluminium has a mass of only 27 g.
One mole of lead has a mass of 207 g.
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