Hello!
I am having difficulty understanding the following problem.
Write cathode & anode 1/2-reactions for the following Bronsted neutralization reaction:
H
3O
+(aq) + OH
- 2H
2O(l)
What I wrote before checking the solution manual:
H
3O
+ H
2O(l) + H
+ OH
- + H
+ H
2O(l)
but quickly recognized a lack of e
- exchange.
Well, the solution manual says:
rewrite as: H
+(aq) + OH
-(aq)
H
2O(l)
oxidation@anode: O
2(g) + 2H
2O(l) + 4e
- 4OH
-(aq)
reduction@cathode: O
2(g) + 4H
+ + 4e
- 2H
2O(l)
So.. How do I know to do the re-write?
Why does ...
H
+ H
2O(l)
not follow the procedure balance: A) elements other than H & O; B) O w/ H
2O; C) H w/ H
+; D) charge w/ e,
which would give you (at the anode) ...
OH
- + H
+ H
2O ?
Thanks in advance!