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الموضوع: Quantitative Aspects of Electrolysis

  1. #1
    عضو ذهبي الصورة الرمزية ماجده
    تاريخ التسجيل
    Aug 2011
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    افتراضي Quantitative Aspects of Electrolysis

    Consider the reaction in the molten NaCl electrolysis cell.
    2Cl ® Cl2(g) + 2e. 2 moles of electrons per mole of Cl2(g)
    Na+ + e ® Na(l). 1 mole of electrons per mole of Na metal
    Recall that the charge on the electron is Q = nF where F is the Faraday constant (96485 C/mol), named after Michael Faraday.
    Thus, to produce one mole (23 g) of sodium, we need Q = 1×96485 C = 96500 C.
    To produce one mole (70.9 g) of chlorine gas, we need 2×96500 = 1,930,000 C.
    Current is defined as the amount of charge passing a point in a circuit in one second.
    I = Q/s The units are 1 Ampere = 1 Coulomb/second (1A = 1C/s).
    Now, if we have a current of 50.0 A passing through an NaCl(l) electrolysis cell in 1 hour, how much sodium and chlorine will we produce.?
    50.0 A × 3600 s = 180,000 C (NOTE: 1As = 1C)
    n = Q/F = 180,000/96,485 = 1.87 mol e
    It takes lots of current to produce very little sodium and chlorine.

    We can do other types of calculations similar to this.
    What mass of aluminum will be produced in 1.00 h by electrolysis of Molten AlCl3 using a current of 10.0 A?
    Q = I×t = 10.0 A × 3600 s = 3.60×104 C.
    It takes three moles of electrons for every mole of Aluminum. So...
    n = Q/F = 3.60×104 C / 96485C/mol e ×1/3 × 27.0 g/mol = 3.36 g Al.

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  2. #2
    عضو ذهبي الصورة الرمزية ماجده
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    Aug 2011
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    افتراضي رد: Quantitative Aspects of Electrolysis

    What volume of Cl2(g) at STP (0°C, 1 atm) will be produced by a current of 20.0 A in 2.00 h in the same cell as used in the previous example?
    Note that in this problem, we used
    R=0.08206 Latm/mol K.


    This was to facilitate the conversion to volume. You could have also used the more standard value of R=8.314 J/mol K but conversion to volume would have involved one more step of calculations.

    If we had tried to electrolyse sodium Chloride in water, we would need to consider not only the Na+ and Cl as possible reactants but also the water. We can never produce sodium in aqueous solution, because it will spontaneously react with the water to produce Na+(aq). Whether we tried to produce sodium metal either chemically or electrolytically, the chance of success would be equally bad. Thus, at the cathode of an aqueous NaCl cell, we would get the reduction of water happening
    2H2O + 2e ® H2 + 2OH E°red = –0.83 V
    rather than the reduction of the sodium
    Na+ + e ® Na E°red = –2.71 V
    Notice that the reduction potential for the water reduction is a lot more positive than that for the sodium. This is a way we can look up reduction potentials and tell what will happen in our solution before hand.
    Looking at the two possible reactions at the anode of the NaCl (aq) cell, we see that the Cl and the H2O are both candidates for oxidation. The two possible half-reactions are:
    2H2O ® O2 + 4H+ + 4e E°ox = –E°red = –1.23
    2Cl ® Cl2 + 2e E°ox = –E°red = –1.36
    Since the water has a more positive oxidation potential than the chlorine reaction, it should be oxidized more readily. There is a complication, however. The water oxidation requires a considerable over-voltage (extra voltage) to make the reaction rate appreciable. Since the two possible reactions are quite close in potentials the extra voltage can quickly be sufficient to cause the chlorine oxidation to occur. Since this latter reaction is quite rapid in comparison to the water ® oxygen one, it will dominate at only slightly elevated voltages.
    If we had used Na2SO4 rather than NaCl as the electrolyte, then the anode reaction would have been the water one since SO42– is very difficult to oxidize (very negative oxidation potential in comparison to water).

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  3. #3
    عضو ذهبي الصورة الرمزية م.عبد الرحمن
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    افتراضي رد: Quantitative Aspects of Electrolysis

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  4. #4
    عضو مميز الصورة الرمزية قمر بلحاج
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    افتراضي رد: Quantitative Aspects of Electrolysis

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