[pear]

Hello!

I am having difficulty understanding the following problem.
    Write cathode & anode 1/2-reactions for the following Bronsted neutralization reaction:
    H₃O⁺(aq) + OH^-  2H₂O(l)


What I wrote before checking the solution manual:
    H₃O⁺    H₂O(l) + H⁺
    OH^-  + H⁺    H₂O(l)
but quickly recognized a lack of e^- exchange.


Well, the solution manual says:
    rewrite as:  H⁺(aq) + OH^-(aq)    H₂O(l)
    oxidation@anode:  O₂(g) + 2H₂O(l) + 4e^-    4OH^-(aq)
    reduction@cathode:  O₂(g) + 4H⁺ + 4e^-    2H₂O(l)

So.. How do I know to do the re-write?
Why does ...
    H⁺    H₂O(l)
not follow the procedure balance: A) elements other than H & O;  B) O w/ H₂O;  C) H w/ H⁺;  D) charge w/ e, 
which would give you (at the anode) ...
    OH^- + H⁺    H₂O  ?


Thanks in advance!

[Borek]

I don't like the question.

but quickly recognized a lack of e^- exchange.

Hardly suprprising - there is NO electron exchange, as this is not a redox reaction.

I guess they wan't to separate reactants yet somehow allow for the neutralization to occur. For that you need two separate reactions - one consuming H⁺, other one consuming OH^-. Net reaction will be neutralization, but for me that's putting things on the head.