6L.3 part d

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inlovewithchemistry
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Joined: Fri Sep 28, 2018 12:19 am

6L.3 part d

Postby inlovewithchemistry » Tue Feb 26, 2019 6:38 pm

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

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

I don't understand how to get the answer that is in the solutions manual or how to set up a half reaction with the cathode. Can someone explain what they did?

David Effio 1H
Posts: 38
Joined: Wed Feb 14, 2018 3:01 am

Re: 6L.3 part d

Postby David Effio 1H » Tue Feb 26, 2019 9:21 pm

So for this problem, there is a "hidden" component where water will appear in the reaction, creating an electric current that is then passed through a platinum cathode and anode.

The two half-reactions would look like:
2H2O (l) --> O2(g) + 4H+ (aq) + 4e- anode (oxidation)
O2 (g) + 2 H2O (l) + 4e- --> 4OH-(aq) cathode (reduction)

The combined reaction will be:
4 H2O (l) --> 4H+(aq) + 4OH- (aq)

Which can then be simplified to:
H2O (l) --> H+ (aq) + OH-(aq)

This is known as water dissociation, and it happens very often in water, though very briefly. This release and movement of protons in the presence of oxygen gas can actually create a functioning galvanic cell. I don't really know HOW it would work IRL, but trust that if it is moving electrons from an anode to a cathode, it is a galvanic cell of some sort.

As a general rule of thumb, any time that hydrogen appears in a cell seemingly out of nowhere, as in when none of the other elements or compounds in the cell diagram include hydrogen, it is more likely due to the presence of water, and it may really just be water dissociating or reacting with the cathode or anode in some way. I hope that makes sense!


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