1. 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

2.  APPLICATIONS OF ELECTROLYSIS  1.Electrolysis is the chemical decomposition of a substance (electrolyte) by an electric current  2.Electrodes :  Anode ( + )  Cathode ( -)

3.  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 )

4.  5.Example :  Electrolyte : molten PbBr 2  Cathode (reduction) :  Pb 2+ (l) + 2e - → Pb (s)  Anode ( oxidation ) :  2 Br - (l) - 2e - → Br 2 (g)  More than one cation and / or anion   selective discharge

5. 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

6.  c.Extracted from ore : bauxite  d.Electrolyte : pure alumina ( Al 2 O 3 ) and cryolite ( Na 3 AlF 6 )  Function of cryolite : to lower m.p of Al 2 O 3 ( from 2050 o C to about 950 o C )  e.Electrodes : graphite anode and cathode

9.  f.Equations :  i)Cathode : Al 3+ (l) + 3e - → Al (s)  ii)Anode : 2 O 2- (l) → O 2 (g) + 4e -  Iii)Overall equation :  2 Al 2 O 3 (l) → 4 Al(s) + 3 O 2 (g)  g.Al which is more dense sinks to the bottom and is syphoned off

10.  h.At the high operating temp,carbon anode blocks replaced often due to oxidation to CO 2 by O 2 evolved  C (s) + O 2 (g) → CO 2 (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 - → F 2 + 2e -  Fluorine is corrosive and toxic

11. 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 Cu 2+ (eg aqueous CuSO 4 )

14.  d.Equations :  i)Anode : Cu (s) → Cu 2+ (aq) + 2e -  ii)Cathode : Cu 2+ (aq) + 2e - → Cu (s)  Copper transferred from anode to cathode  Observations :  Anode dissolves  Pure copper deposited at cathode

15.  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 Fe 2+ (aq) and Zn 2+ (aq)  However at cathode : only Cu 2+ discharged.  Fe 2+ and Zn 2+ remains in solution

16.  g.Ag and Au remain undissolved  and fall to the cell bottom as anode sludge ( from which they can be recovered )

17. ELECTROLYSIS OF BRINE (concentrated NaCl)  a.Using the diaphragm cell  Consists of 2 chambers ( anode and cathode chamber )  b.Electrolyte : purified brine  Purification removes Mg 2+ and Ca 2+ which may form insoluble hydroxides that then clogs the diaphragm

18.  c.Electrodes :  i)Anode : titanium or inert electrode (graphite)  it resists corrosion by the very reactive chlorine formed  ii)Cathode : steel or graphite

19. Diagram

21. Steel cathode (1) Titanium anode (1) Container + compartment + electrodes + diaphragm (1)

22.  d. Equations :  i) Anode : 2Cl - -> Cl 2 (g) + 2e -  ii) Cathode : 2H + (aq) + 2e - → H 2 (g)

23.  Another possible way of writing equations :  From H 2 O : H + (discharged) and OH - (unchanged)  Cathode : 2H 2 O + 2e - → H 2 + 2OH -  From NaCl : Cl - (discharged) and Na + (unchanged)  Anode : 2Na + + 2Cl - - 2e - → Cl 2 + 2Na +  Overall equation :  2H 2 O + 2NaCl → H 2 + 2NaOH + Cl 2  Molar ratio of products :  H 2 : NaOH : Cl 2 = 1 : 2 : 1

24.  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

25.  g.Level of brine on left side (anode) is deliberately higher than the right side ( cathode) ………. WHY?

26. S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.

27.  h. Products obtained :  i) chlorine  Uses:  purify water supply  disinfectant, bleach  used in plastics, polymers (eg PVC)

28.  ii) hydrogen  Uses : manufacture of ammonia, margarine and HCl, as fuel  iii) aqueous NaOH  Uses : manufacture of soap, paper and detergent

29.  Note :  If question specifies manufacture of chlorine from electrolysis of brine.  Chlorine ( main product )  Hydrogen and NaOH ( by products )

30.  i. Other products :  i)H 2 and Cl 2 can be combined to make HCl: H 2 + Cl 2 -> 2HCl ii) Cl 2 and NaOH(aq) :  (1)Cl 2 and cold NaOH(aq) ( 15 o C) produces sodium chlorate(I), NaClO  NaClO used as : bleach and disinfectant

31.  Cl 2 + 2NaOH -> NaCl + NaClO + H 2 O  Or Cl 2 + 2OH - -> Cl - + ClO - + H 2 O 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)

32.  (2)Cl 2 and hot NaOH(aq) ( 70 o C) produces sodium chlorate(V), NaClO 3  NaClO 3 used as : weedkiller  3Cl 2 + 6 NaOH -> 5NaCl + NaClO 3 + 3H 2 O Or 3Cl 2 + 6OH - -> 5Cl - + ClO 3 - + 3H 2 O 0 -1 +5 Type of reaction : Chlorine undergoes disproportionation