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    Any advice on how to remember what happens at the anode and cathode in a hydrogen oxygen fuel cell.
    The overall equation is easy .5O2+H2>H20 but even its the single equations that I can't remember
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    Can you create the redox half equation for e.g. MNO4- -> Mn2+ (in alkaline conditions)?

    If you can, all you have to realise is that one half equation is O2 -> OH- and the other is H2 to H2O.
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    And if I can't ? I can do it for normal conditions
    Mn04- 8H+ + 5e- -> Mn2+ 4H2O
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    I only teach it in acidic conditions and then teach how to convert to alkaline.

    To do that add as many OH- to BOTH sides as there are H+. In this case, you'd add 8OH- to both sides. The 8H+ and 8OH- will turn into 8H2O. The last step is to cancel out the waters, in this case, remove 4H2O from both sides.

    There is a trick to doing it faster, but I'll let you spot it. It should be obvious when you convert a few acidic half equations to alkaline.
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    (Original post by Pigster)
    I only teach it in acidic conditions and then teach how to convert to alkaline.

    To do that add as many OH- to BOTH sides as there are H+. In this case, you'd add 8OH- to both sides. The 8H+ and 8OH- will turn into 8H2O. The last step is to cancel out the waters, in this case, remove 4H2O from both sides.

    There is a trick to doing it faster, but I'll let you spot it. It should be obvious when you convert a few acidic half equations to alkaline.
    Yes is the trick that you the number of H+ (in the reactants) ions then replace it with H20 and then double the number of H20 (in products) and replace with OH-
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    (Original post by Pigster)
    Can you create the redox half equation for e.g. MNO4- -> Mn2+ (in alkaline conditions)?

    If you can, all you have to realise is that one half equation is O2 -> OH- and the other is H2 to H2O.
    However, being able to do the equations im still unable to see that
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    Starting with O2 -> OH- and H2 -> H2O...

    In acid conditions:
    O2 + 3H+ + 4e- -> OH- + H2O
    H2 -> 2H+ + 2e-

    In alkaline conditions:
    O2 + 2H2O + 4e- -> 4OH-
    H2 + 2OH- -> 2H2O + 2e-

    Multiplying the second one balances the e-. Combining and cancelling leaves you with 2H2 + O2 -> 2H2O.
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    I the tried.
    O2 -> OH- and H2 > H20 (this one is a bit odd since its not really what happens as the half eq.)????

    Acidic
    4e- + 02 +2H+ ->2 OH-
    And H2 -> 2H+ + 2e-

    Times second eq. By 2 and cancel
    2H2 + O2 -> (2H+ + 2OH-) 2H20

    Ok that's fine also doing this made me realise I don't know what the quick way to do it is?
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    Ok two questions what's the quick method. I thought I knew what it was but couldnt apply it here. And if H2 -> H20 why isn't the half equation done like that? (It doesn't have h20 in it )
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    Re-looking at my first acidic equation, it should have read:
    O2 + 4H+ + 4e- -> 2H2O.
    I was wrong because there was an OH- on the right, which you could not make in acidic conditions. I have a migraine this weekend, that's my excuse.

    You will need to read the question carefully, usually H2 cells are alkaline, so you'd have to convert.

    When I referred to a 'quick way' I meant converting acidic half equations to alkaline ones. You don't need to know it, it only is a little faster: the number of waters will half and move to the other side.
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    (Original post by Pigster)
    Re-looking at my first acidic equation, it should have read:
    O2 + 4H+ + 4e- -> 2H2O.
    I was wrong because there was an OH- on the right, which you could not make in acidic conditions. I have a migraine this weekend, that's my excuse.

    You will need to read the question carefully, usually H2 cells are alkaline, so you'd have to convert.

    When I referred to a 'quick way' I meant converting acidic half equations to alkaline ones. You don't need to know it, it only is a little faster: the number of waters will half and move to the other side.
    What I did above was for acidic and I forgot to write the conversion to alkaline. But even doing the conversion gives you the correct answer.
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    Please could you look at what I've done and correct and mistakes?
    Also, thank you since I didn't know how to convert to alkaline before.
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    Here (and glad) to help.

    The only issue is the O2 half equation in acid conditions is that OH- wouldn't form. I know my first post said that O2 -> OH-, but that is only true in alkaline conditions (OH- formed would react with acid). Strictly, you should use the equation I wrote a couple of hours ago (O2 -> H2O), but your scanned image gets you to the two correct half equations, so would get you the marks.
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    Ok so this should be completely correct (attachment). But why does reduction occur at the anode and oxidation at the cathode
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    Looks correct (to my sleepy head).

    ACID - anode current into device, e- flow is opposite to current direction.

    Electrode are being grabbed by O2 and removed from the wire connected to the voltmeter, e- flow into this electrode. Current therefore flows out of this electrode, which makes it the cathode. Your diagram is the wrong way around. I hope. I am half asleep.

    Everyone loves Wikipedia: http://en.wikipedia.org/wiki/Alkaline_fuel_cell

    If the cell is using electricity, e.g. electrolysis of Al, the positive electrode is the anode. If the cell is making electricity, e.g. a fuel cell, the positive electrode is the cathode. Handily.
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    (Original post by Pigster)
    Looks correct (to my sleepy head).

    ACID - anode current into device, e- flow is opposite to current direction.

    Electrode are being grabbed by O2 and removed from the wire connected to the voltmeter, e- flow into this electrode. Current therefore flows out of this electrode, which makes it the cathode. Your diagram is the wrong way around. I hope. I am half asleep.

    Everyone loves Wikipedia: http://en.wikipedia.org/wiki/Alkaline_fuel_cell

    If the cell is using electricity, e.g. electrolysis of Al, the positive electrode is the anode. If the cell is making electricity, e.g. a fuel cell, the positive electrode is the cathode. Handily.
    Which part of the diagram was incorrect please....
    I tried this question and stumbled greatly and the bit in pencil at the bottom is what I attempted to make sense of it but didn't help
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    Effectively, it's just the reverse reactions of those that take place in an alkaline hydrogen fuel cell.

    OH- -> O2 AND
    H2O -> H2

    Balanced the same way.
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    (Original post by Pigster)
    Effectively, it's just the reverse reactions of those that take place in an alkaline hydrogen fuel cell.

    OH- -> O2 AND
    H2O -> H2

    Balanced the same way.
    you mean H20 ->O2 ( as you said before where you corrected yourself )
    and H20 -> H2

    silly question how do I recognise is the reverse!??
    (thank you for replying quickly and consistently)
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    When a half cell generates electricity all the reactions go in the feasible directions.

    When electricity is pushed through, i.e. when recharging, all the reactions go in the opposite direction.
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    And it was hinted in the previous question I didn't recognise that. Thank you !!!
 
 
 
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