Lecture 3/14

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veneziaramirez 3I
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Joined: Fri Sep 29, 2017 7:07 am

Lecture 3/14

Postby veneziaramirez 3I » Wed Mar 14, 2018 9:49 pm

For the first problem that we reviewed today, how do we know which reaction to choose to be oxidizing and reducing? Do we choose our oxidizing to be the reaction with the largest cell voltage?

DianaTrujillo2K
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Joined: Fri Sep 29, 2017 7:04 am

Re: Lecture 3/14

Postby DianaTrujillo2K » Wed Mar 14, 2018 9:50 pm

For a galvanic cell the over charge of the cell must be positive, so you determine which one to switch based on which one will make the charge positive.

Kyle Alves 3K
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Joined: Thu Jul 27, 2017 3:01 am

Re: Lecture 3/14

Postby Kyle Alves 3K » Sat Mar 17, 2018 5:20 pm

also, if it wasn't given that the cell is galvanic, you can determine anode and cathode by:
reversing the half reaction with the lower E as that is more likely to be oxidized - will be the anode
the higher E will be reduced - will be the cathode

Justin Bui 2L
Posts: 51
Joined: Fri Sep 29, 2017 7:06 am

Re: Lecture 3/14

Postby Justin Bui 2L » Sat Mar 17, 2018 7:15 pm

Yeah, you just generally want to try to make the Ecathode - Eanode a positive number because then it will proceed spontaneously and that's what you want your cells to do. So just see the V when looking at the reduction potentials and reverse the right equation so that you can try your best to keep Ecell positive.


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