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Homework Help: Electrolysis of Iron(II)Sulphate

  1. May 12, 2013 #1
    1. The problem statement, all variables and given/known data

    I am given a 1 M solution of Iron(II)sulphate(aq) and it is being electrolyzed using inert electrodes . What will form on the cathode ?

    2. Relevant equations

    Reduction potentials of : Fe =-0.44V
    of water to from hydrogen : -0.83V
    3. The attempt at a solution
    Since iron(II) ion has a more POSITIVE value it is easily reduced at the cathode to form Iron (solid).

    Unfortunately this answer is wrong .
    The correct answer is Hydrogen will be formed on the cathode ... Why ?
  2. jcsd
  3. May 12, 2013 #2
    Please help me !
  4. May 12, 2013 #3


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    Staff: Mentor

    Please elaborate.

    Generally speaking hydrogen gets reduced at 0 V by definition, doesn't it?
  5. May 12, 2013 #4
    It does indeed but that is something different. In this particular example we have :

    Fe(II) ions , water molecules, extremely low number of Hydrogen ions and hydroxide ions, Sulphate ions .

    Since the concentration of Hydrogen ions is almost negligible (as is the case in any aqueous solution) it should be OK to ignore them . So to form hydrogen gas at the cathode the reaction that occurs (in reality) is :

    2 water molecules + 2 electrons → one molecule of hydrogen gas + 2 hydroxide ions with a standard redox potential of -0.83V
  6. May 12, 2013 #5


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    Staff: Mentor

    No, that's not a correct approach. To be exact you should not just ignore H+ presence, but take into account everything, including partial pressure of hydrogen (good luck with, as it is initially zero), and plug it all into the Nernst equation. You should do the same with your other reaction, remembering that initial concentration of OH- is not zero either (actually it is identical to the initial concentration of H+), and that -0.83 V is given for a solution in a standard state (which means concentration of OH- equal to 1M).

    This is by no means a trivial prediction, but as long as you are not expected to do a very thorough analysis, just comparing -0.44 V with 0 V should do. Perhaps with some estimate of the H+ reduction potential based on just pH.
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