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Half equations for molten hydroxide electrolysis

  1. Dec 1, 2015 #1
    I found a thread on electrolyzing molten sodium hydroxide: http://www.thenakedscientists.com/forum/index.php?topic=51331.0

    Question: does it proceed via Reaction 1 or 2?

    1) NaOH -> Na + OH
    2) NaOH -> Na + O + H

    Does Na separation from NaOH by electrolysis of the molten NaOH proceed by Reaction 1 or 2?

    This process starts with NaOH material as a solid. The idea is that this is heated to become molten.
    Then electricity is applied to it to electrolyse out the Sodium, but what of the hydroxyl = OH?

    Does the OH come off as some sort of OH bound state, a Hydroxyl? some sort of "Brown's Gas" like gas? is it standalone OH gas? or as separate O and separate H? i.e. Can the O come off separately from the H or only bound?

    Does writing down the "half equations" break down this process and answer this question?
    Can someone provide the "half equations" analysis for Reactions 1 and 2 and under what conditions does it proceed via Reaction 1 and under what conditions does it proceed via Reaction 2?
  2. jcsd
  3. Dec 2, 2015 #2


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    Staff: Mentor

    The thread you have mentioned contains both basic reactions taking place listed, have you read it?
  4. Dec 7, 2015 #3
    Dear Borek
    my prior knowledge determines what I can and cannot absorb from it

    your prior knowledge determines what you can and cannot absorb from it

    whatever is there I understood only what I can and don't understand what I can't which is why I asked for help here

    if you can please point out where you think both reactions are presented in the initial thread, please point out which of the comments has it and which part of the comment it is, just to be clear since right now the equations provided in the thread do not appear to me to be like those below which were what I was hoping to find

    The half equations a friend told me by email:

    4NaOH = 4Na+ + 4OH- (1)

    cathode 4Na+2 + 4e = 4Na (2)

    anode 4OH - 4e = 2H20 + O2 (3)

    2Na + 2H2O = 2Na(OH) + H2 (4)

    Actually his Equation 3 above is similar to one of the equations in the thread (the second in the 10/05/2014 18:50:12 comment, but they differ in that there is free 2H and free O2 but here is H2O. So technically these are different equations. How do we know which one takes place, that is what fraction proceed via this equation 3 (making water) and what fraction proceed via the second equation there (making separate O2 & 2H)?

    Given these 4 reactions above does anyone know how to use them to determine what fraction proceeds via

    A) NaOH -> Na + OH
    B) NaOH -> Na + O + H

    So what fraction proceeds via A and what fraction proceeds via B?

    And if the reaction B could be forced to be 100% is there a way to separate the O & H so they do not instantly recombine?
    Last edited: Dec 7, 2015
  5. Dec 8, 2015 #4


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    Science Advisor

    In the link you were providing these two half-equations are given:
    ##\mathrm{e^– + K^+ \to K}## E0 = –2.93 V
    ##\mathrm{2 OH^– \to O_2 + 2H^+ + 4 e^-}## E0 = –0.40 V

    In the first equation, replace Na for K.
    The second equation I had to correct and, as we are working in liquid NaOH, the hydrogen ions will react immediately
    ##\mathrm{H^++OH^- \to H_2O}##.
    So no, there will be no free OH radicals and also no free O or H atoms.
    Oxygen and sodium are formed at different electrodes, so they won't recombine.
    Part of the water formed may react with the Na metal to give hydrogen
    ##\mathrm{2 Na+2H_2O \to 2Na^+ +2OH^- +H_2}##.
    So there will also be some hydrogen gas being produced, but also at the cathode, so that it won't react with the oxygen formed at the anode:
    In sum you get something like ##\mathrm{2*(2-x)NaOH \to 2*(1-x)Na +O_2 +2*(1-x)H_2O+xH_2}##, where x is somewhere between 0 and 1 (worst case), depending on how well the regions of the two electrodes are separated.
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