Reduction: 2 H(+) + 2 e(-) ==> H2
oxidation: 2 H2O ==> O2 + 4 H(+) + 4 e(-)
Balances as:
Reduction: 4 H(+) + 4 e(-) ==> 2 H2
oxidation: 2 H2O ==> O2 + 4 H(+) + 4 e(-)
Which gives :
2 H2O & 4 H(+) ==> 2 H2 & O2 + 4 H(+)
Reduced answer is:
2 H2O ==> 2H2 & O2