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Calculating the mole of electrons form the current

  1. Dec 10, 2004 #1
    In the chlor-alkali industry, Cl2(g) and NaOH(aq) are produced by electrolysis of an aqueous solution ofo sodium chloride. If a 1000L tank of NaCl(aq) at pH 7 is electrolysed for one hour with a current of 1.00x104A, the final pH will be:

    I am completely stuck on how to find the solution to the problem aside from calculating the mole of electrons form the current:

    I = C/t
    10000 = C/3600s
    C = 3600000C

    mol e- = 360000000/96490Cmol-1
    mol e- = 373.1

    After that I am stuck.
     
  2. jcsd
  3. Dec 10, 2004 #2
    Huh? You're stuck?
    Well, what you want to know is the amount OH- produced, right?

    How many electrons do you need to produce one OH- ion?
    For this you should look at the reactions occurring at the anode and at the cathode.
    At the anode, Cl- is converted to 1/2 Cl2(g) and an electron, at the cathode, water is reduced to hydrogen gas, leaving OH- behind. It is up to you to calculate how many. Hint: look only at the charged species in your cathode reaction and you'll see immediately how many OH- are produced for each electron? (oh, did I say too much? :wink: )

    From this you get the pH... if you assume the volume of the water to be unchanged (see below)!

    (Maybe we should check if there still are 1000L of water? We are after all consuming water and any decrease of the water volume would of course affect the pH - but given the question I guess we can assume the volume is unchanged. As an exercise, however, and to get an idea of the different proportions, I think you should calculate the amount of water consumed. Hint: how many oxygens are there in water? and in OH-? Given that water is reduced to OH- and H2, how many water molecules correspond to one OH- ion? What is the volume corresponding to this amount, given the density 1 g/ml and molar mass 18 g/mol?)
     
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