H2O->H+ + OH-

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Yinhan_Liu_1D
Posts: 51
Joined: Sat Sep 24, 2016 3:00 am

H2O->H+ + OH-

Postby Yinhan_Liu_1D » Sat Feb 04, 2017 12:19 pm

Using O2 in the galvanic cell

Can anybody tell me what's been reduced and what's being oxidized in this reaction?

Since O2 does not appear in the overall reaction, how does the reduction and oxidation happen?

I am really confused about this one, so I would really appreciate your help.

Same with Ag+ + Br- -> AgBr , shouldn't the Ag and Br in AgBr already have the oxidation number as Ag+ and Br- ? How does this reflect a Redox RXN???


Thank you!

Navi Sidhu 1F
Posts: 25
Joined: Wed Sep 21, 2016 2:59 pm

Re: H2O->H+ + OH-

Postby Navi Sidhu 1F » Sat Feb 04, 2017 1:03 pm

This was a response by Chemmod a couple of years ago:

This reaction is the opposite of Problem 13.11d so to make the comparison easier I will flip this reaction (this does not change the chemistry). In that problem you are given the cell diagram to help in the selection and you then solve for the final reaction and Ecell0. In this problem you are given the final reaction, but need to (like Problem 13.11d) get the redox reactions and calculate Ecell0.

The solution for Problem 13.11d can be found here: https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=139&t=3354

To start we flip the given reaction (does not affect anything):

H2O (l) ---> H+(aq) + OH-

We see that we have one reactant and two products. Thus, for our oxidation reaction and reduction reaction H2O (l) will be used as the reactant in both.

So we now have the two reactions we are focused on:

H2O (l) ---> H+(aq)
H2O (l) ---> OH-

Now we want to find reduction half-reactions which correspond to these two reactions:

Reduction/Cathode: O2,H+/H2O: O2(g) + 4H+(aq) + 4e- ---> 2 H2O(l) E0=1.23
Oxidation/Anode: O2,H2O/OH-: O2(g) + 2H2O(l)l + 4e- ---> 4OH- E0=0.40

We assign the cathode and anode since we want this process to be spontaneous so we make the cathode reaction the one with the highest/most positive reduction potential.

Getting to this step probably the hardest part. After this we balance the two reactions:

Since our oxidation reaction is written as a reduction then we need to flip it and change the sign of E0=0.40.

Therefore:

Reduction/Cathode: O2(g) + 4H+(aq) + 4e- ---> 2 H2O(l) E0 = 1.23
Oxidation/Anode: 4OH- ---> O2(g) + 2H2O(l)l + 4e- E0 = -0.40

Now we balance and we get:

4 H+ (aq) + 4 OH- (aq) ---> 4 H2O (l)
H+(aq) + OH-(aq) ---> H2O (l)

Therefore:
Ecell0 = Eox0 + Ered0
Ecell0 = -0.40 V + 1.23 V
Ecell0 = +0.83 V

Here's the link to the original post https://lavelle.chem.ucla.edu/forum/viewtopic.php?t=3361

Yinhan_Liu_1D
Posts: 51
Joined: Sat Sep 24, 2016 3:00 am

Re: H2O->H+ + OH-

Postby Yinhan_Liu_1D » Thu Feb 09, 2017 3:53 pm

Navi Sidhu 1F wrote:This was a response by Chemmod a couple of years ago:

This reaction is the opposite of Problem 13.11d so to make the comparison easier I will flip this reaction (this does not change the chemistry). In that problem you are given the cell diagram to help in the selection and you then solve for the final reaction and Ecell0. In this problem you are given the final reaction, but need to (like Problem 13.11d) get the redox reactions and calculate Ecell0.

The solution for Problem 13.11d can be found here: https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=139&t=3354

To start we flip the given reaction (does not affect anything):

H2O (l) ---> H+(aq) + OH-

We see that we have one reactant and two products. Thus, for our oxidation reaction and reduction reaction H2O (l) will be used as the reactant in both.

So we now have the two reactions we are focused on:

H2O (l) ---> H+(aq)
H2O (l) ---> OH-

Now we want to find reduction half-reactions which correspond to these two reactions:

Reduction/Cathode: O2,H+/H2O: O2(g) + 4H+(aq) + 4e- ---> 2 H2O(l) E0=1.23
Oxidation/Anode: O2,H2O/OH-: O2(g) + 2H2O(l)l + 4e- ---> 4OH- E0=0.40

We assign the cathode and anode since we want this process to be spontaneous so we make the cathode reaction the one with the highest/most positive reduction potential.

Getting to this step probably the hardest part. After this we balance the two reactions:

Since our oxidation reaction is written as a reduction then we need to flip it and change the sign of E0=0.40.

Therefore:

Reduction/Cathode: O2(g) + 4H+(aq) + 4e- ---> 2 H2O(l) E0 = 1.23
Oxidation/Anode: 4OH- ---> O2(g) + 2H2O(l)l + 4e- E0 = -0.40

Now we balance and we get:

4 H+ (aq) + 4 OH- (aq) ---> 4 H2O (l)
H+(aq) + OH-(aq) ---> H2O (l)

Therefore:
Ecell0 = Eox0 + Ered0
Ecell0 = -0.40 V + 1.23 V
Ecell0 = +0.83 V

Here's the link to the original post https://lavelle.chem.ucla.edu/forum/viewtopic.php?t=3361



Thank you, it helps!


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