Midterm Winter 2014 Q. 7 and 8


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Jeffrey Wang
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Joined: Fri Sep 25, 2015 3:00 am

Midterm Winter 2014 Q. 7 and 8

Postby Jeffrey Wang » Tue Feb 09, 2016 7:21 pm

I don't understand how to make which half reaction either part of the anode or the cathode. Like for #7, it makes sense that Zn 2+/Zn is the anode, as it needs to be oxidized, therefore making the cell potential overall a positive value and a favorable reaction. But for #8, what makes the first equation the anode? If you do that, your cell potential is negative, which is unfavorable. I guess what I'm asking is...

Given two half reactions, how do you determine which is the cathode and which is the anode (besides looking at the cell potential, because that does not seem to be a factor...)?

Michelle Pham 1G
Posts: 21
Joined: Fri Sep 25, 2015 3:00 am

Re: Midterm Winter 2014 Q. 7 and 8

Postby Michelle Pham 1G » Tue Feb 09, 2016 7:34 pm

If the question says that it is a galvanic cell, you would look at cell potential because E has to be positive for a galvanic cell. So you would make the cathode the more positive one and the anode the negative one.

If it does not say that it is a galvanic cell, you look at the balanced equation. In Q8 they ask for the Ka for HF. The only way to get the overall reaction
2HF 2H+ + 2F- is if F is being reduced. You know that F is the cathode because in order for the 2 half reactions to sum up to HF, F has to be reduced. This would mean that the other half reaction is being oxidized and thus the anode.

Shrita Pendekanti 4B
Posts: 22
Joined: Fri Sep 25, 2015 3:00 am

Re: Midterm Winter 2014 Q. 7 and 8

Postby Shrita Pendekanti 4B » Wed Feb 10, 2016 3:56 pm

I believe that for this specific question, you are able to figure out which is the anode because it asks for the Ka value. Ka is the acid dissociation constant and therefore, H+ must be a product. So, it is part of the anode reaction.


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