Electrolysis

[Zippora]

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can some one please help me with this:

Write the half equations that would occur in the electrolysis of AlF3

I said it would be:

anode: H2O -> 1/2O2 + 2H+ + 2e-

cathode: H2O + e- -> 1/2H2 + OH-

because these water reactions have more positive voltages than the ones involving aluminium and flourine, but the answers said that it would be

anode: 2F- -> F2 + 2e-

cathode: Al3+ + 3e- -> Al(s)

can someone please help??????

Also, when writing a reaction does it matter whether you it like this:

2F- -> F2 + 2e-

or like this..

F- -> 1/2F2 + e-

?

Thank you for your help in advance

[Zippora]

can some one please answer, thanks

[Azreil]

I'm guessing (cause typing AlF3 into google didn't help much at all) that AlF3 can't occur in aqueous solution; ie it can only be molten. So there is no water involved hence the equations have to be of the elements themselved.

Write it the way you find easier/the way it needs to be balanced to write the overall equation.

Hope that helps.