6L.3d

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705121606
Posts: 68
Joined: Wed Sep 18, 2019 12:17 am

6L.3d

Postby 705121606 » Wed Feb 26, 2020 8:15 pm

For this part we are given the cell:
Pt(s)|O2(g)|H+(aq)|| OH-(aq)|O2(g)|Pt(s)

How are we supposed to write the half reaction for the anode?
In the solutions manual e=-.83

Why would it be negative if we are talking about a galvanic cell?

Eileen Si 1G
Posts: 120
Joined: Fri Aug 30, 2019 12:17 am

Re: 6L.3d

Postby Eileen Si 1G » Thu Feb 27, 2020 12:01 am

You would write the half-reaction without Pt(s), since it simply serves as a solid electrode for the cell diagram. Because an oxidation reaction is occurring in the anode, there is a loss of electrons from water as it oxidizes to O2(g), H+(a), and electrons. So the anode half-reaction is: 2H2O(l) --> O2(g) + 4H+(aq) + 4e-.
Last edited by Eileen Si 1G on Mon Mar 02, 2020 6:40 pm, edited 1 time in total.

705121606
Posts: 68
Joined: Wed Sep 18, 2019 12:17 am

Re: 6L.3d

Postby 705121606 » Thu Feb 27, 2020 12:29 am

Eileen Si 1G wrote:You would write the half-reaction without Pt(s), since it simply serves as a solid electrode for the cell diagram. Because a reduction reaction is occurring in the anode, there is a gain of electrons added to O2(g) as it also reacts with water, H2O(l) to produce hydroxide, OH-(aq). So the anode half-reaction is: O2(g) + 2H2O(l) + 4e- --> 4OH-(aq).


are we taking H+ on the anode side to be water instead of H+?

Eileen Si 1G
Posts: 120
Joined: Fri Aug 30, 2019 12:17 am

Re: 6L.3d

Postby Eileen Si 1G » Mon Mar 02, 2020 6:42 pm

705121606 wrote:
Eileen Si 1G wrote:You would write the half-reaction without Pt(s), since it simply serves as a solid electrode for the cell diagram. Because a reduction reaction is occurring in the anode, there is a gain of electrons added to O2(g) as it also reacts with water, H2O(l) to produce hydroxide, OH-(aq). So the anode half-reaction is: O2(g) + 2H2O(l) + 4e- --> 4OH-(aq).


are we taking H+ on the anode side to be water instead of H+?


Sorry, I mistakenly wrote the half-reaction for the cathode side and not the anode side. I edited my original statement though, so hopefully that clarifies some things. Apologies again!!

ShastaB4C
Posts: 100
Joined: Thu Sep 26, 2019 12:18 am

Re: 6L.3d

Postby ShastaB4C » Mon Mar 02, 2020 9:29 pm

Eileen Si 1G wrote:
705121606 wrote:
Eileen Si 1G wrote:You would write the half-reaction without Pt(s), since it simply serves as a solid electrode for the cell diagram. Because a reduction reaction is occurring in the anode, there is a gain of electrons added to O2(g) as it also reacts with water, H2O(l) to produce hydroxide, OH-(aq). So the anode half-reaction is: O2(g) + 2H2O(l) + 4e- --> 4OH-(aq).


are we taking H+ on the anode side to be water instead of H+?


Sorry, I mistakenly wrote the half-reaction for the cathode side and not the anode side. I edited my original statement though, so hopefully that clarifies some things. Apologies again!!



Why did they write the cathode notation with OH- as if it was the product of the rxn?
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