power point on batteries NCERT class 12 chemistry

1. Products of Electrolysis
What are the factors affecting products of electrolytes?

2. Products of Electrolysis- Factors
a) Nature of the electrode
b) Reduction potential of ions
Types of electrodes
Active Takes part in reaction (Cu, Ni, Ag)
Inert  do not take part in reaction(Pt, C)

3. Predict the products of electrolysis at cathode & anode.
a) Molten sodium chloride is electrolyzed using Pt electrode.
Cathode: Sodium metal
Anode: Chlorine gas

4. Predict the products
b) Molten Copper chloride is electrolyzed using Cu
electrode.
Cathode: Copper metal
Anode: Cu2+

5. Products of Electrolysis- Factors
b) Reduction potential of ions
A reaction occurs at
cathode higher 0E
anode  lower 0E

6. Predict the products of electrolysis of an aqueous
solution of sodium chloride using Pt electrode.
Cathode reaction
Hydrogen gas is liberated

7. Products of Electrolysis – continued …..
Predict the products at anode when NaCl (aq) is electrolyzed?
Anode
Liberation of chlorine gas takes place
Over potential
The minimum excess potential over the standard reduction potential
require to discharge an ion.

9. One or more galvanic cells connected in series
and used as a source of electricity
Compact
Light weight
Constant voltage

10. Battery
Primary
Not rechargeable
Secondary
Rechargeable

11. Dry cell(Leclanche cell)
Anode: Zinc container
Cathode: Graphite + MnO2
Electrolyte: NH4Cl + ZnCl2 paste
Cell potential = 1.5 V
Primary battery 1

12. Mercury Cell
Anode: Zn(Hg)
Cathode: HgO + C
Electrolyte: KOH + ZnO (Paste)
Primary battery 2

13. Mercury Cell
Cell potential = 1.35 V (constant voltage)

14. Anode: Pb
Cathode: PbO2 + Pb
Electrolyte: H2SO4(38%)
Lead storage Battery
Secondary Battery 1

15. Cell potential = 2 V (per cell)
Lead storage Battery
Battery in use
On Charging the Battery
Reverse above reaction

16. Anode: Cd
Cathode: Ni(OH)3
Electrolyte: KOH
Cell potential = 1.4 V
Nickel Cadmium Battery
Secondary Battery 2
Overall reaction

17. HOME WORKS
a) Write the uses of primary & secondary cells
b) Write the overall reaction taking place in those cells

18. Fuel cells
Galvanic cells which convert energy of combustion of fuel
directly into electrical energy are called fuel cells
e.g.
Hydrogen – Oxygen fuel cell
Methanol – Oxygen fuel cell

19. Hydrogen – Oxygen fuel cell
NaOH

20. Rusting

21. Rusting
A process in which iron reacts with water and oxygen from its
surrounding to form hydrated ferric oxide(rust).Fe2O3.xH2O
Rusting – a redox reaction

22. Do other metals react with water or air in its surrounding?
It is the gradual destruction of metals by electrochemical reaction
with their environment
Corrosion

23. Prevention of rusting
1) Painting
2) Galvanizing(Zn Coating)
3) Cathodic protection

24. Cathodic Protection

25. REVISION- ELECTROCHEMISTRY
Pattern
Max. Marks:8
Part A  1 Mark x 1 Question
Part B  2 Marks x 1 Question
Part D  5 Marks x 1 question

26. REVISION- ELECTROCHEMISTRY
Important topics
1. Daniel cell
2. SHE
3. Nernst equation problems
4. Conductivity and molar conductivity
5. Kohlrausch’s law
6. Faraday’s laws
7. Batteries ( Dry cell, Pb Acid Battery)
8. Fuel cell
9. Rusting of iron

27. 1) Explain the construction and working of SHE
2) Define Molar conductivity
3) State Kohlrausch’s law
4) State Faraday’s first law and second law
5) Explain the construction and working of H2-O2 Fuel cell.
6) Explain cathode and anode reactions in a Dry cell
7) Explain the over all reaction in a Lead storage battery when it is
in use.
Learn – Answer - Write

28. The Gibb’s energy for the decomposition of aluminium oxide at
500 oC is –966 kJmol-1. What is the minimum potential
difference needed for electrolytic decomposition of Al2O3.
⅔ Al2O3  4/3 Al + 3O2
Number of electrons involved = n
⅔ Al3+  4/3 Al0 hence (2/3)* 3 = 2 e and (4/3) * 0 = 0e
Number of electrons per aluminium atom is = 2
Hence for two aluminium atoms = 4 electrons
n = 4
∆G0 = -nFE0
Hence E0 = - ∆G0 /nF
On substituting the values E0 cell = - (-966kJ/4 * 96500)
Minimum potential difference = 2.5 V

29. Calculate the emf of the cell
Fe/Fe2+(0.6M)//Sn2+(0.2M)/Sn
E0
Fe = -0.44V and E0
Sn = + 0.14 V
Answer:
n = 2
E0cell = 0.58 V
Emf of cell = 0.566 V

30. 0.2964 g of copper was deposited on passage of a current of
0.5 A for 30 minutes through a solution of copper sulphate.
Calculate the atomic mass of copper.
Answer: Atomic mass of copper = 63.56