ELECTROCHEMISTRY AS
 Redox reactions
 Oxidation : loses electrons/oxidation
number increases /loses
hydrogen/accepts oxygen
 Reduction : accepts electrons/oxidation
number decreases /accepts
hydrogen/loses oxygen
 Calculation of oxidation numbers
 APPLICATIONS OF ELECTROLYSIS
 1.Electrolysis is the chemical
decomposition of a substance
(electrolyte) by an electric current
 2.Electrodes :
 Anode ( + )
 Cathode ( -)
 3.Electrolyte : ionic compound ( molten
or aqueous solution )
 Consists of cations ( positive ions )
 And negative ions ( anions )
 4.During electrolysis :
 a.Cations → cathode ( reduction )
 b.Anions → anode ( oxidation )
 5.Example :
 Electrolyte : molten PbBr2
 Cathode (reduction) :
Pb2+ (l) + 2e- → Pb (s)
 Anode ( oxidation ) :

2 Br- (l) - 2e- → Br2 (g)
 More than one cation and / or anion
  selective discharge

EXTRACTION OF
ALUMINIUM
 a.Extracted by electrolysis of molten
salts ( not aqueous solutions )
 b.Importance : Al widely used due to its
properties
 Light , strong , good electrical conductor
,does not corrode
 c.Extracted from ore : bauxite
 d.Electrolyte : pure alumina ( Al2O3 ) and
cryolite ( Na3AlF6 )
 Function of cryolite : to lower m.p of
Al2O3 ( from 2050o C to about 950o C )
 e.Electrodes : graphite anode and
cathode
 f.Equations :
 i)Cathode : Al3+ (l) + 3e- → Al (s)
 ii)Anode : 2 O2- (l) → O2 (g) + 4e Iii)Overall equation :
2 Al2O3 (l) → 4 Al(s) + 3 O2 (g)
 g.Al which is more dense sinks to the
bottom and is syphoned off

 h.At the high operating temp ,carbon anode






blocks replaced often due to oxidation to CO2
by O2 evolved
C (s) + O2 (g) → CO2 (g) ,
H highly exothermic ,heat evolved helps to
partly maintain electrolyte in molten state
i.Pollutants :
i) CO(g) : from incomplete combustion of anode
ii) fluorine : from cryolite : 2F- → F2 + 2eFluorine is corrosive and toxic
PURIFICATION OF COPPER
 a.Impure Cu obtained by roasting its ore
in air , then purified by electrolysis.
 b.Electrodes :
 Anode ( impure copper )
 Cathode ( pure copper )
 c.Electrolyte : aqueous Cu2+ (eg
aqueous CuSO4 )
 d.Equations :
 i)Anode : Cu (s) → Cu2+ (aq) + 2e-
 ii)Cathode : Cu2+ (aq) + 2e- → Cu (s)
 Copper transferred from anode to
cathode
 Observations :
 Anode dissolves
 Pure copper deposited at cathode
 e.Impurities : example
 i) metals more reactive/electropositive than Cu :




Fe and Zn
ii) metals less reactive/electropositive than Cu :
Ag and Au
f.Fe and Zn also ionises , enters solution as
Fe2+ (aq) and Zn2+ (aq)
However at cathode : only Cu2+ discharged.
Fe2+ and Zn2+ remains in solution
 g.Ag and Au remain undissolved
 and fall to the cell bottom as anode
sludge ( from which they can be
recovered )
ELECTROLYSIS OF BRINE
(concentrated NaCl)
 a.Using the diaphragm cell
 Consists of 2 chambers ( anode and
cathode chamber )
 b.Electrolyte : purified brine
 Purification removes Mg2+ and Ca2+
which may form insoluble hydroxides
that then clogs the diaphragm
 c.Electrodes :
 i)Anode : titanium or inert electrode
(graphite)
 it resists corrosion by the very reactive
chlorine formed
 ii)Cathode : steel or graphite
Diagram
Steel cathode (1)
Titanium anode (1)
Container + compartment +
electrodes + diaphragm (1)
 d. Equations :
 i) Anode :
2Cl- -> Cl2(g) + 2e-
 ii) Cathode : 2H+ (aq) + 2e- → H2 (g)
 Another possible way of writing equations :
 From H2O : H+ (discharged) and OH- (unchanged)
2H2O + 2e- → H2 + 2OHFrom NaCl : Cl- (discharged) and Na+(unchanged)
Anode :
2Na+ + 2Cl- - 2e- → Cl2 + 2Na+
Overall equation :
2H2O + 2NaCl → H2 + 2NaOH + Cl2
 Cathode :




 Molar ratio of products :

H2 : NaOH : Cl2 = 1 : 2 : 1
 e. Na+ goes through diaphragm to
cathode chamber
 f. NaOH forms through the following
reaction :
 Na+ + OH- → NaOH (aq)
 and flows out of cell
 NaOH used as detergent , soap , paper
industries
 g.Level of brine on left side (anode) is
deliberately higher than the right side
( cathode) ………. WHY?
 SO THAT THE BRINE WILL
SLOWLY FLOW THROUGH THE
ASBESTOS DIAPHRAGM
TOWARDS THE CATHODE,
CARRYING THE SODIUM IONS
WITH IT AND PREVENTING
THE REVERSE FLOW OF
SODIUM HYDROXIDE
TOWARDS THE ANODE
WHERE IT WILL REACT WITH
THE CHLORINE.
 h. Products obtained :
 i) chlorine
 Uses:
 purify water supply
 disinfectant , bleach
 used in plastics , polymers (eg PVC)
 ii) hydrogen
 Uses : manufacture of ammonia ,
margarine and HCl , as fuel
 iii) aqueous NaOH
 Uses : manufacture of soap, paper and
detergent
 Note :
 If question specifies manufacture of
chlorine from electrolysis of brine.
 Chlorine ( main product )
 Hydrogen and NaOH ( by products )
 i. Other products :
 i)H2 and Cl2 can be combined to make HCl:
H2 + Cl2 -> 2HCl
ii) Cl2 and NaOH(aq) :
 (1)Cl2 and cold NaOH(aq) ( 15o C) produces
sodium chlorate(I), NaClO
 NaClO used as : bleach and disinfectant
 Cl2 + 2NaOH -> NaCl + NaClO + H2O
 Or Cl2 + 2OH- -> Cl- + ClO- + H2O
0
-1
+1
Type of reaction : Chlorine undergoes
disproportionation
Oxidation no of Cl increases from 0 to +1(oxd)
and decreases from 0 to -1(red)
 (2)Cl2 and hot NaOH(aq) ( 70o C) produces
sodium chlorate(V), NaClO3
 NaClO3 used as : weedkiller
 3Cl2 + 6 NaOH -> 5NaCl + NaClO3 + 3H2O
3Cl2 + 6OH- -> 5Cl- + ClO3- + 3H2O
0
-1
+5
Type of reaction : Chlorine undergoes
disproportionation
Or