6L7 (b) HW

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Hannah Lee 2F
Posts: 117
Joined: Thu Jul 11, 2019 12:15 am

6L7 (b) HW

Postby Hannah Lee 2F » Thu Feb 27, 2020 4:56 pm

How can neutralization of water be considered a redox reaction despite the fact that the oxidation #s haven't changed of either hydrogen or oxygen? And how could you easily identify which half-rxns in the table to use? When I first solved this, I mistakenly used the half-rxns involving H2 instead of O2.

(b) H+(aq) + OH-(aq) --> H2O(l), the Brønsted-Lowry neutralization reaction

Anisha Chandra 1K
Posts: 118
Joined: Thu Jul 11, 2019 12:17 am

Re: 6L7 (b) HW

Postby Anisha Chandra 1K » Thu Feb 27, 2020 9:15 pm

Yeah I was a bit confused about how the oxidation numbers don't change, but it can be broken down to reduction and oxidation half reactions:

reduction: 4e- + O2 + 2H2O --> 4OH-
oxidation: 2H2O --> 4H+ + O2 + 2e-

Daria Azizad 1K
Posts: 116
Joined: Thu Jul 25, 2019 12:15 am

Re: 6L7 (b) HW

Postby Daria Azizad 1K » Thu Feb 27, 2020 10:20 pm

Anisha Chandra 1K wrote:Yeah I was a bit confused about how the oxidation numbers don't change, but it can be broken down to reduction and oxidation half reactions:

reduction: 4e- + O2 + 2H2O --> 4OH-
oxidation: 2H2O --> 4H+ + O2 + 2e-

How do we know that O2 is involved? Like why is it not
Red: 2H+ + 2e- -> H2
Oxid: 2OH- + H2 -> 2H2O + 2e-

Anisha Chandra 1K
Posts: 118
Joined: Thu Jul 11, 2019 12:17 am

Re: 6L7 (b) HW

Postby Anisha Chandra 1K » Sun Mar 08, 2020 12:15 pm

I think O2 is an intermediate that we see cancels out of the final chemical equation.


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