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# Hydrogen oxygen fuel cells watch

1. 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
2. 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.
3. And if I can't ? I can do it for normal conditions
Mn04- 8H+ + 5e- -> Mn2+ 4H2O
4. 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.
5. (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-
6. (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
7. 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.
8. 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?
9. 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 )
10. 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.
11. (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.
12. 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.
Attached Images

13. 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.
14. Ok so this should be completely correct (attachment). But why does reduction occur at the anode and oxidation at the cathode
Attached Images

15. 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.
16. (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
Attached Images

17. 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.
18. (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)
19. 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.
20. And it was hinted in the previous question I didn't recognise that. Thank you !!!

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