1. Chemistry

2. Session Electrochemistry - 2

3. Session Objectives Electrolysis Faradays Laws of electrolysis Electrode Potential Electromotive force Electrochemical cells

4. Electrolysis The process of decomposition of an electrolyte by the passage of electricity is called electrolysis. In electrolysis electrical energy is used to cause a chemical reaction.

5. Electrolysis At cathode At anode For example, Electrolysis of molten sodium chloride

6. Electrolysis of aqueous sodium chloride H + ions are discharged at cathode because discharge potential of H + ion is much lower than Na + ion At anode, Cl – is discharge as Cl 2 (gas) because discharge potential of Cl – is much lower than that of OH – ion.

7. Electrolysis of aqueous copper sulphate Electrolysis of aqueous copper sulphate solution using inert electrodes At cathode, Cu 2+ ions are discharged in preference to H + ions because discharge potential of Cu 2+ is much lower than H + ions. At anode, OH – ions are discharged in preference to SO 4 2– ions because discharge potential of OH – is much lower than SO 4 2– ions.

8. Faraday’s law If W grams of the substance is deposited by Q coulombs of electricity, then But Q = it, Hence W i t or W = Z it I = current in amperes t = time in seconds. Z = constant of proportionality (electrochemical equivalent.)

9. Faraday’s Law E=Equivalent mass of the substance 1 Faraday=96500 coulomb

10. Illustrative Example Find the total charge in coulombs on 1 g ion of N 3–. Electronic charge on one mole ion N 3– = 3 ×1.602 ×10 –19 × 6.023 × 10 23 coulombs = 2.89 ×10 5 coulombs Solution : No. of moles in 1g N 3- ion = 1/14 =0.0714286. Therefore, charge on 1g N 3– ion =0.0714286 x 2.89 ×10 5 coulombs =2.06 x 10 4 Coulombs

11. Illustrative Example On passing 0.1 Faraday of electricity through aluminium chloride what will be the amount of aluminium metal deposited on cathode (Al = 27) Solution:

12. Illustrative Example How many atoms of calcium will be deposited from a solution of CaCl 2 by a current of 25 milliamperes flowing for 60 s? Number of Faraday of electricity passed = 4.68 × 10 18 atoms of calcium. Solution: moles of Ca atoms atoms of Ca

13. Illustrative Example What current strength in amperes will be required to liberate 10 g of bromine from KBr solution in half an hour? Bromine is liberated by the reaction at anode as follows Solution: i = 6.701 ampere

14. Faraday’s second law of electrolysis When the same quantity of electricity is passed through different electrolytes the masses of different ions liberated at the electrodes are directly proportional to their chemical equivalence

15. Illustrative Example The electrolytic cells, one containing acidified ferrous chloride and another acidified ferric chloride are connected in series. What will be the ratio of iron deposited at cathodes in the two cells when electricity is passed through the cells ? Solution Ratio of iron deposited at cathode will be in their ratio of equivalents

16. Illustrative Example A 100 W, 110 V incandescent lamp is connected in series with an electrolyte cell containing cadmium sulphate solution. What mass of cadmium will be deposited by the current flowing for 10 hr? (Gram atomic mass of Cd = 112.4 g)

17. Solution Watt = Volt × Current 100 = 110 × Current But we know, Q = i × t Number of equivalent = = 0.339 Mass of cadmium deposited = 19.06 g

18. Electrode Potential The electrode potential depends upon: Nature of the metal Concentration of the metallic ions in solution Temperature of the solution. Zn (s) Zn ++ (aq) + 2e – Zn ++ (aq.) + 2e – Zn (s) Oxidation electrode potential Or Reduction electrode potential

19. Standard Hydrogen Electrode OR Pt, H 2 (g) (1 atm)/ H + (aq) (c = 1 M)

20. When elements are arranged in increasing order of standard electrode potential as compared to that of standard hydrogen electrode, It is called electrochemical series. Electrochemical series Higher the RP, more tendency to get reduced (strong oxidising agent) and vice versa. Note: Oxidation-Reduction-Potential of an element have same magnitude and different sign

21. (a)To compare the relative oxidizing and reducing powers. (b) To compare the relative activities of metals. (c)To calculate the standard EMF of any electrochemical cell. (d)To predict whether a redox reaction is spontaneous. Applications of electrochemical series:

22. Applications of Electrochemical Series Higher the standard reduction potential, the lesser will be its reducing strength. Li is strongest reducing agent in aqueous solution. The lesser the standard reduction potential of an element, greater will be its activity. A more active metal displaces a less active one from its salt solution. Those metals which have positive oxidation potential will displace hydrogen from acids. The metals above hydrogen are easily rusted and those situated below are not rusted.

23. E 0 values of Mg +2 /Mg is –2.37V, Zn +2 /Zn is –0.76V, Fe +2 /Fe is –0.44V.Using this information predict whether Zn will reduce iron or not? Illustrative Example Zinc has lower reduction potential than iron. Therefore, it can reduce iron. Solution:

24. Standard electrode potential When the ions are at unit activity and the temperature is 25°C (298 K), the potential difference is called the standard electrode potential (E°).

25. Electrochemical cell Daniel cell Chemical energyelectrical energy. Galvanic cell:-

26. Oxidation at anode Reduction at cathode Cell reaction Electromotive force(EMF) E 0 cell =E 0 cathode -E 0 anode Reactions involved

27. Electromotive force(EMF) of a cell As per IUPAC convention if we consider standard reduction potential of both the electrodes then Emf = E R.H.Electrode - E L.H.Electrode

28. Salt Bridge Salt bridge Agar—agar mixed with KCl, KNO 3, NH 4 NO 3) etc. Eliminates liquid - liquid junction potential Maintains the electrical neutrality of solutions. Completes the circuit.

29. Illustrative Example Calculate the emf of the cell = –0.25 V for and 1.5 V Au 3+ /Au. = 1.5 – (–0.25) = 1.75 V Solution:

30. Illustrative Example The standard oxidation potential E o for the half reaction are given below. Calculate E 0 for the following cell reaction Zn + Fe ++  Zn ++ + Fe Solution = –0.41 – (–0.76) = +0.35 V

31. Thank you