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Chemistry

Chemistry. Session. Electrochemistry - 2. Session Objectives. Electrolysis Faradays Laws of electrolysis Electrode Potential Electromotive force Electrochemical cells. Electrolysis.

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Chemistry

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  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. At cathode At anode Electrolysis 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 Cl2 (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, Cu2+ ions are discharged in preference to H+ ions because discharge potential of Cu2+ is much lower than H+ ions. At anode, OH– ions are discharged in preference to SO42– ions because discharge potential of OH– is much lower than SO42– ions.

  8. Hence W i t Faraday’s law If W grams of the substance is deposited by Q coulombs of electricity, then But Q = it, 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 coulombson 1 g ion of N3– . Solution : No. of moles in 1g N3- ion = 1/14 =0.0714286. Electronic charge on one mole ion N3– = 3 ×1.602 ×10–19 × 6.023 × 1023 coulombs = 2.89 ×105 coulombs Therefore, charge on 1g N3– ion =0.0714286 x 2.89 ×105 coulombs =2.06 x 104 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 CaCl2 by a current of 25 milliamperes flowing for 60 s? Solution: Number of Faraday of electricity passed moles of Ca atoms atoms of Ca = 4.68 × 1018 atoms of calcium.

  13. Illustrative Example What current strength in amperes will be required to liberate 10 g of bromine from KBr solution in half an hour? Solution: Bromine is liberated by the reaction at anode as follows 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. Number of equivalent = = 0.339 = 19.06 g Solution Watt = Volt × Current 100 = 110 × Current But we know, Q = i × t Mass of cadmium deposited

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

  19. Standard Hydrogen Electrode OR Pt, H2 (g) (1 atm)/ H+ (aq) (c = 1 M)

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

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

  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. Illustrative Example E0 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? Solution: Zinc has lower reduction potential than iron. Therefore, it can reduce iron.

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

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

  26. Reactions involved Oxidation at anode Reduction at cathode Cell reaction Electromotive force(EMF) E0cell=E0cathode-E0anode

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

  28. Salt Bridge • Salt bridge Agar—agar mixed with KCl, KNO3, NH4NO3) etc. • Eliminates liquid - liquid junction potential • Maintains the electrical neutrality of solutions. • Completes the circuit.

  29. Calculate the emf of the cell = –0.25 V for and 1.5 V Au3+/Au. Illustrative Example Solution: = 1.5 – (–0.25) = 1.75 V

  30. The standard oxidation potential Eo for the half reaction are given below. Calculate E0for the following cell reaction Zn + Fe++ Zn++ + Fe Illustrative Example Solution = –0.41 – (–0.76) = +0.35 V

  31. Thank you

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