## 2014 midterm #8

Ivanna_Tang_3B
Posts: 22
Joined: Wed Sep 21, 2016 2:58 pm

### 2014 midterm #8

Using standard cell potentials:
F2+2H++2e--->2HF E=3.03V
F2+2e--->2F- E=2.87

How do we know which one will be the anode and which one will be the cathode, so we know which equation to flip?

question asks to calculate the Ka for HF.

Thank you!

Chem_Mod
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### Re: 2014 midterm #8

You want the overall reaction to be in the correct order (HF on the left to calculate Ka). Therefore, the first reaction should be flipped.

Jonathan Sarquiz 3F
Posts: 32
Joined: Fri Jul 15, 2016 3:00 am

### Re: 2014 midterm #8

This question is asking to calculate the Ka for HF. If we write the dissociation equation for HF, we get HF H+ + F-. This tells us we want HF as a reactant and H+ and F- as products. In order for us to match the overall equation, we must flip the first half reaction. When we flip it, we now know that the top reaction is the oxidation half reaction and the second equation (which we did not flip) is the reduction half reaction.

Belicia Tang 1B
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Joined: Wed Sep 21, 2016 3:00 pm

### Re: 2014 midterm #8

I was confused as to which reaction to flip, because I thought that usually you will flip the reaction with the lower reduction potential, which in this case, would be the second one. Flipping the first equation will give us a negative standard potential, which means the reaction is unfavorable.

Chem_Mod
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### Re: 2014 midterm #8

You are correct, the reaction is unfavorable, but this question isn't asking for the spontaneous reaction. Instead, it is asking for the dissociation Ka value of HF. Therefore, it is just testing your understanding of the connection between Ka values and anodes and cathodes. You can disregard whether it is spontaneous (for this question).