1. Engineering Unit – Chemistry Section
Electrolysis
Valentim M. B. Nunes
Engineering Unit – Chemistry Section
March 2018

2. Electrolysis
The electrolysis is a process in which electrical energy is used to force a non spontaneous chemical reaction.

3. Electrolysis of molten sodium chloride
The electrolysis of molten NaCl is done in large scale in the Downs Cell
810 °C
Anode: 2 Cl-(l)  Cl2(g) + 2 e- production of chlorine
Cathode: 2 Na+(l) + 2 e-  Na(l) production of sodium

4. Electrolysis of water
In pure water the ionic concentration is very low ([H3O+]  110-7 M), so there are no sufficient ions for electrolysis.
Adding an acid, for instance H2SO4 or HCl 0.1 M:
anode: 2 H2O(l)  O2(g) + 4H+(aq) + 4 e-
cathode: H+ + e-  1/2 H2(g)
Global: 2 H2O(l)  2 H2(g) + O2(g)

5. Purification of cooper
The electrolysis has many other applications like the extraction and purification of metals.
Impure cooper contains impurities like Zn, Fe, etc…

6. Production of aluminum
Anode: 2 O2-  O2 + 4 e-
Cathode: Al e-  Al(l)
Global: 2 Al2O3  4 Al + 3 O2
The Hall-Héroult made aluminum metal extremely economical and with numerous applications: construction, soft drinks cans, aircraft...

7. Metallic Electrodeposition
The electrodeposition of a metal over another has the objective to achieve certain properties such as appearance (ex: brightness), hardness, corrosion resistance, etc.
Au+(aq) + e-  Au(s)

8. Quantitative aspects: Faraday law
Faraday discovered that the mass of product formed or consumed in an electrode is directly proportional to the amount of electricity transferred and the molar mass of substance.
Current (A)  time (s) Charge in Coulomb (C)
Number of Faraday´s
Nº of moles of substance oxidized or reduced
Mass of substance oxidized or reduced

9. Faraday law
Example: Calculate the mass of aluminum that is deposited in an electrolysis, in a bath containing Al3+, with a current with 40 A, during 30 minutes.
Total charge = 40 C/s  30 min  60 s/min = C
nº of Faradays = C / C.mol-1  0.75 mol de e-
Since Al e-  Al
nº of moles of Al = 0.75/3 = 0.25 mol
mass of Al = 0.25 mol  27 g.mol-1  6.75 g