Determining cathode and anode from two half-reactions...

Regina Chi 2K
Posts: 51
Joined: Fri Sep 26, 2014 2:02 pm

Determining cathode and anode from two half-reactions...

I have posted this problem before, but it was specific to a certain problem. However, if the cell is not defined to be galvanic, and we are given two half-reactions and their respective E values, how will we know which one we should set as the anode and which on as the cathode?

Chem_Mod
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Re: Determining cathode and anode from two half-reactions...

You would want your Standard Potential to be positive in order for your E(cell) to be positive and thus spontaneous. When you are given two half reactions and their respective potentials, have E(cathode)- E(anode) result in a positive E. For example if given the half reactions:
Cu2+(aq) + 2e- => Cu(s) E= 0.34V
Zn2+(aq) + 2e- => Zn(s) E=-0.76V

You would want to make Zn2+/Zn become the anode to maintain a positive E. (E= 0.34V - (-0.76V)= 1.10V)
Another way to think about it is the half reaction with the more positive E is more likely to be reduced.

Shelbyhamilton 3H
Posts: 37
Joined: Fri Sep 26, 2014 2:02 pm

Re: Determining cathode and anode from two half-reactions...

That is helpful. However, I have a problem where there are two positive cell potentials and the larger cell potential is actually subtracted from the smaller one creating a negative cell potential. this kind of contradicts what was said above and I'm really confused why.
So this is question 8 in the 2014 midterm.
using the standard cell potentials
f2 (g) + 2H+ (aq) + 2e- ---> 2HF (aq) Eo = +3.03 V
f2 (g) + 2e- ---> 2F- Eo= +2.87 V
calculate Ka for HF.

so then they write
HF--> H+ (aq) + F- (aq)
what confuses me is that they flip the first equation even though its the bigger one..... this results in e(Cathode)- e (anode) = 2.87-3.03. now our cell potential is -.16 v!!!
why is this??

Regina Chi 2K
Posts: 51
Joined: Fri Sep 26, 2014 2:02 pm

Re: Determining cathode and anode from two half-reactions...

This may be a little late, but it is because we are finding the Ka of the problem and Ka requires the acid (HF) to be in the denominator of the reaction quotient. So it must be a reactant and not a product.