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What is the mass of H2 and O2 produced water electrolysis?

  1. May 13, 2015 #1
    1. The problem statement, all variables and given/known data
    2 moles of H2O undergoes electrolysis at 15 amps for 1 hour. What is the mass of H2 and O2 gas produced?

    2. Relevant equations
    4 H+(aq) + 4e−→ 2H2(g)

    2 H2O(l) → O2(g) + 4 H+(aq) + 4e−

    Faraday's Law of Electrolysis
    Q = n(e-) x F

    3. The attempt at a solution

    F = 96500C

    (15*3600s*4g)/(F*4)=0.55g H2

    (15*3600s*32g)/(F*4)=4.47g O2
     
  2. jcsd
  3. May 13, 2015 #2
    Isn't that the correct solution?
     
  4. May 13, 2015 #3
    I think so. I've seen a few other methods, but this one seems the simplest.
     
  5. May 13, 2015 #4

    James Pelezo

    User Avatar
    Gold Member

    You can check your calculation accuracy by converting the gram - values to moles. For water, moles H2(g) produced should be 2x moles of O2 produced. Your numbers do confirm this => 0.55g H2 = 0.28 mole H2 and 4.47g O2 = 0.14 mole O2 which is 2:1 ratio of H to O. Good job.
     
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