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Urgent help needed for question on electrolysis & comparing Eo values

  1. Nov 5, 2009 #1
    Here is the question:

    Predict the products formed at the anode and at the cathode when NaBr(aq) is electrolysed using inert electrodes

    Answer:

    Somehow i got 2 different answers by different answer sheets

    One of them stated that Br2 gas (anode) and H2 gas (cathode) would be liberated

    However, the other one writes H2(cathode) , H2O and O2 (anode) are formed

    There is no arguments for the products formed at the cathode as clearly it is H2 gas which is liberated there. However, the question lies in the products formed at the anode. Which one is correct?

    Eo values if needed:
    Br2 + 2e ::equil:: 2Br- Eo= +1.07V
    Na+ + e ::equil:: Na Eo= -2.71V
    O2 + 4H+ + 4e ::equil:: 2H2O Eo= +1.23V
    O2 + 2H2O + 4e ::equil:: 4OH- Eo= +0.40V

    2H2O + 2e ::equil:: H2 + 2OH- Eo= -0.83V

    The ones in red are the equations in question. Which one should i use to compare Eo values with Br at the anode (oxidation)? And why?

    The answer which states that bromine gas is liberated is most probably the right one (as it is provided by my school) but it used the O2 + 4H+ + 4e ::equil:: 2H2O equation to compare instead of O2 + 2H2O + 4e ::equil:: 4OH- why is this so?
     
  2. jcsd
  3. Nov 5, 2009 #2
    Where would you get your "4OH-" from in an aqueous solution of NaBr? The equilibrium concentration of it from the self-ionisation of water is going to be very insignificant and negligible (hence E value of the half-cell would be very positive actually, since equilibrium position lies very far to the right)
     
  4. Nov 5, 2009 #3
    Aren't OH- ions produced in the cathode?
     
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