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### Question 14.15b

Posted: Tue Feb 07, 2017 10:07 pm
For 14.15 b, could I have done it by using the half reactions:
2H+ + 2e- -> H2
and
2H2O + 2e- -> H2+2OH- ?
I got the same E cell potential but my cell diagram doesn't include O2 (g) like it would have if I had used:
O2 + 4H+ +4e- -> 2H2O
O2 + 2H2O + 4e- -> 4OH-

Can anyone explain if it makes any difference and if so which one we should use?

### Re: Question 14.15b

Posted: Tue Feb 07, 2017 11:58 pm
It says that they are using the Bronsted Neutralization reaction and so that incorporates the half reactions

O2 + 4H+ +4e- $\rightarrow$ 2H2O
O2 + 2H2O + 4e- -> 4OH-

While the other half reactions might yield the same electric potential, it does not satisfy the neutralization reaction as it does not involve and acid and a base neutralizing one another.