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Electrolysis of water and charges help

  1. May 4, 2005 #1
    Hi i am confused about on how to solve this question.

    The question asks: By the electrolysis of water, 11.2L oxygen at STP was prepared. a) What charge was required? b) if a current of 0.05A was used, how long did it take?

    I have done the question but i think i did it wrong.

    2H2O(l) ---> O2(g) + 4H +(aq) + 4e- (oxidation)
    4H +(aq) + 4e- ---> 2H2O(g) (reduction)

    a) 1mol/22.4L = 0.0446
    0.0446* 96500= 4303.9 4.3* 10^3

    b) 0.5A = 4.3* 10^3/t
    t = 8607.8s

    Is this completely wrong???? :confused:
  2. jcsd
  3. May 4, 2005 #2
    First your thread title is misleading since this is a question about electrochemistry. Second I'm not sure what is meant by "What charge was required?" Could you mean what potential (voltage) is required ? If so simply solve the redox reaction for the potential. Finally for b) simply convert the O2 to moles and convert the .05 A to moles of electrons/sec and due the math :)
  4. May 4, 2005 #3
    I think your full reaction should be [tex]2H_2O (l) \longrightarrow O_2 (g) + 2H_2 (g)[/tex]

    Your standard half reactions are [tex] O_2 (g) + 4 H^+ (aq) + 4 e^- \longrightarrow 2H_2O \ E^0 = +1.23 V [/tex] and
    [tex] 2H_2O (l) + 2 e^- \longrightarrow H_2 (g) + 2 OH^- (aq)\ E^0 = -0.83 V [/tex]

    I got these reduction potentions from http://www.jesuitnola.org/upload/clark/Refs/red_pot.htm Of course, during electrolysis the first reaction occurs in reverse. So what do you think the total potential would be?

    Edit: Actually you don't need to bother about the potentials.

    a) You have 11.2 L of O2 at 1 atm and 273 K. How many moles of O2 is this? Hint: PV = nRT. Now how many electorns are required to reduce 1 molecule of O2 ? 1 mol of O2? How many coulombs are there in x moles of electrons ?

    b) This one should be easy once you have (a) 1 ampere = 1 coulomb / second. You have coulombs and amps. Solve for seconds.
    Last edited: May 4, 2005
  5. May 5, 2005 #4


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    a)at STP 1mole/22.4L (I think), convert to moles

    -using the half reaction equation, find out the mole of electrons required for every 1 mole of oxygen. Convert to moles of electrons using factor labeling.

    -convert this mole value to avogadros number. From this convert to charge using the charge of one electron.

  6. Jul 8, 2006 #5
    Hi... I am working on this one too and would appreciate it if anyone could take the time to check my answers.

    By the electrolysis of water, 11.2L oxygen at STP was prepared.
    a) What charge was required?
    b) if a current of 0.05A was used, how long did it take?

    A) the text lists the following reaction
    H2O <-> 2H+ + .5O2 + 2e- -0.82
    since half a mol of O2 is produced the charge was the listed (-0.82)value in the equation as the equation is for half a mol of O2.

    B)Q=It or t = Q/I
    t= .5 mol / .5 amp = 1 second
    (not too sure on this one for the value of Q - the quantity of electric charge transferred through a cell)

    Thanks in advance :)
  7. May 30, 2007 #6
    Maybe I am missing something but I did this question very differently than Gymstar and Yellowduck:

    a) The molar volume of any gas at STP is 22.4 L
    Therefore 11.2 L of oxygen at STP = 1/2 mole
    1 molecule of oxygen requires 4 electrons (each O requires 2 electrons [tex]O^2^-[/tex]and oxygen is diatomic [tex]O_2[/tex])
    Therefore 1/2 mole of oxygen requires 4*1/2*1 mole of electrons
    1 mole of electrons = 9.64*10^4 C

    (4)(1/2)(9.64*10^4C) = 1.93*10^5 C

    b) t = Q/I

    t = (1.93*10^5 C) / 0.5 A = 386000 seconds or 107.2 hours
    Last edited: May 30, 2007
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