2014 Midterm #8 [ENDORSED]

Michelle_Nguyen_3F
Posts: 40
Joined: Wed Sep 21, 2016 2:59 pm

2014 Midterm #8

Hello!

The question is:

Using the standard cell potentials:
$F_{2} (g)+ 2H^{+} (aq) + 2 e^{-}\rightarrow 2 HF (aq)$ Enot = +3.03V
$F_{2} (g) + 2 e^{-} \rightarrow 2 F^{-}(aq)$ Enot = +2.87V

How do we determine which one is the cathode and which on is the anode because both written as reduction reactions?

Thank you!

Chem_Mod
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Re: 2014 Midterm #8  [ENDORSED]

The more positive the potential, the stronger the reaction is a oxidizing agent and therefore functions as the anode. As such, the one with the higher reduction potential would be at the anode and the lower reduction potential would be at the cathode.

janavi_patel_2K
Posts: 20
Joined: Wed Sep 21, 2016 2:59 pm

Re: 2014 Midterm #8

I am still a little confused. In the textbook and the solutions manual, it states that the one with the more positive reduction potential is where reduction occurs and, therefore, contains the oxidizing agent. This means that the one with the higher reduction potential would be at the cathode. However, this is opposite of what you said and what is in the answer to the midterm question. Can you please clarify?

stephanieyang_3F
Posts: 62
Joined: Wed Sep 21, 2016 2:55 pm

Re: 2014 Midterm #8

For this question, because we're trying to find out the value for Ka (K constant for acidic reactions), by definition, Ka is $\frac{[H+][F-]}{HF}$. The K value is always products/ reactants, so ideally this Ka value would represent a reaction that looks like this: HF $HF\rightleftharpoons H^+(aq)+F^-(aq)$. You use the given standard cell potentials to find a balanced reaction in that format. The reason why the more positive reduction potential is not at the cathode, rather the anode, is because making it the cathode would give the reverse of the reaction we want in the first place. Normally if we are trying to find the E cell of a galvanic cell, a spontaneous process, you'd want to maximize the E cell by making the lesser reduction potential the anode. But we are only trying to find the Ka value here. You'll find the overall E cell is actually negative, and that the dissociation of HF is actually a non spontaneous reaction (it's a weak acid so its not likely to dissociate naturally in the first place). Hope this makes sense.