Merdan
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Ok so if I know one half of the reaction and the overall reaction is there a method of finding the second half. Here is an example:
1st half equation = O2 + 4H+ + 4e- = 2H2O
Overall reaction = CH3OH + 3/2O2 = 2H2O + CO2
Find the second half of the reaction its only one mark question so there is a quick method?
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Pigster
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1st half equation + 2nd half equation = overall reaction (after you've cancelled)

you'll need on the left: another 1/2O2 and a CH3OH
on the right: a CO2
And to make it balance - on the right: 4H+ and 4e- (which will cancel out)
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Borek
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(Original post by Merdan)
Ok so if I know one half of the reaction and the overall reaction is there a method of finding the second half.
You can try to subtract.

1st half equation = O2 + 4H+ + 4e- = 2H2O
Overall reaction = CH3OH + 3/2O2 = 2H2O + CO2
Apparently overall uses 3/2O2, so 1st half as used is

3/2O2 + 6H+ + 6e- = 3H2O

Subtract that from the overall, see what you are left with. In case you get negative coefficients, just move to the other side of the equation.
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Merdan
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(Original post by Pigster)
1st half equation + 2nd half equation = overall reaction (after you've cancelled)

you'll need on the left: another 1/2O2 and a CH3OH
on the right: a CO2
And to make it balance - on the right: 4H+ and 4e- (which will cancel out)
So I subtracted to get 1/2O2 + CH3OH = CO2 + 4H+ + 4e-
But that's not the right answer it should be CH3OH +H2O = CO2 + 6H+ + 6e-

I don't know how they got 6H+ and 6e- lol...
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Pigster
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I suspect there is more information in the question that you omitted.

The way to do it is to identify in the overall equation what the other redox is. In this case CH3OH -> CO2. Then just create the half equation as usual: add water to balance O; add H+ to balance the H; add e- to the more +ve side.
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Merdan
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(Original post by Pigster)
I suspect there is more information in the question that you omitted.

The way to do it is to identify in the overall equation what the other redox is. In this case CH3OH -> CO2. Then just create the half equation as usual: add water to balance O; add H+ to balance the H; add e- to the more +ve side.
Yes that's right.... it works thanks.
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Borek
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(Original post by Merdan)
I don't know how they got 6H+ and 6e- lol...
By multiplying 4 by 3/2. You have to do that to convert O2 to 3/2O2. You multiply everything in the equation, just like you do when adding half equations.
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zed963
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The best way is to do what merdan said.

Recognise what the reactant is converting to and then balance it from there.


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