14.11

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April P 1C
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Joined: Thu Jan 12, 2017 3:01 am

14.11

Postby April P 1C » Wed Feb 21, 2018 10:12 pm

14.11 Write the half reactions and the balanced equation for the cell reaction for each of the following galvanic cells:

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

I know that on the left of the || is anode and the right is cathode. Oxidation occurs at anode and reduction occurs at cathode.

How do I find the half reaction for this galvanic cell??
Also, why am I able to just ignore Pt(s)? Is it because it's not being reduced or oxidized?

Merzia Subhan 1L
Posts: 33
Joined: Thu Jul 13, 2017 3:00 am

Re: 14.11

Postby Merzia Subhan 1L » Thu Feb 22, 2018 1:10 am

I also have this question. In the solutions manual for the anode, they put O2(g) and H+(aq) on the same side of the equation and made H2O but I thought that they would be on opposite sides of each other in the chemical equation. Can anyone explain why this is?

Kyung_Jin_Kim_1H
Posts: 53
Joined: Thu Jul 27, 2017 3:00 am

Re: 14.11

Postby Kyung_Jin_Kim_1H » Thu Feb 22, 2018 1:24 am

Pt (Platinum) can be ignored because it's not losing or gaining electrons. It's acting as an inert electrode for electron transfer. You'll see Pt/inert electrodes pop up when both the oxidized and reduced species are in the solution OR it is a gas/ion electrode reaction.

April P 1C
Posts: 38
Joined: Thu Jan 12, 2017 3:01 am

Re: 14.11

Postby April P 1C » Thu Feb 22, 2018 1:52 pm

Someone correct me if I'm wrong but I think the reason why O2 and H+ are on the same side is because the reaction would not be oxidizing if it was either
H+ -> O2 or
O2 -> H+
The equation has to be H2O -> O2 + H+ because then the reaction will be oxidizing for H in H2O -> H+


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