14.11

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Emily Glaser 1F
Posts: 156
Joined: Thu Jul 27, 2017 3:01 am

14.11

Postby Emily Glaser 1F » Mon Feb 19, 2018 5:25 pm

can someone do step by step of part d. I don't understand why, in the solutions manual, it says:

O2 +2H2O + 4e- --> 4OH- for the cathode (positive E value). OH- is listed before O2, so shouldn't the O2 be on the right side and OH on the left with a positive E value

Clara Rehmann 1K
Posts: 53
Joined: Fri Sep 29, 2017 7:03 am

Re: 14.11

Postby Clara Rehmann 1K » Thu Feb 22, 2018 2:07 pm

Yes, but the catch with this problem is that the battery does not have a positive voltage, as the Eº of the anode is greater than that of the cathode.

Pt(s) | O2(g) | H+(aq) || OH-(aq) | O2 (g) | Pt(s)

Overall reaction: 4H2O(l) → 4H+(aq) + 4OH-(aq)

Anode: O2(g) + 4H+(aq) → 2H2O(l) +1.23 V
Cathode: O2(g) + 2H2O(l) + 4e- → 4OH-(aq) +.4 V

Eº = Eºcathode - Eºanode = .4 V - 1.23 V = -.83 V

904676178
Posts: 5
Joined: Fri Jun 17, 2016 11:28 am

Re: 14.11

Postby 904676178 » Thu Feb 22, 2018 8:54 pm

Clara Rehmann 1K wrote:Yes, but the catch with this problem is that the battery does not have a positive voltage, as the Eº of the anode is greater than that of the cathode.

Pt(s) | O2(g) | H+(aq) || OH-(aq) | O2 (g) | Pt(s)

Overall reaction: 4H2O(l) → 4H+(aq) + 4OH-(aq)

Anode: O2(g) + 4H+(aq) → 2H2O(l) +1.23 V
Cathode: O2(g) + 2H2O(l) + 4e- → 4OH-(aq) +.4 V

Eº = Eºcathode - Eºanode = .4 V - 1.23 V = -.83 V

How did you know what the anode is? Isn't the cathode supposed to be the one with the greater positive Eº?


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