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Electrochemistry Understanding Cell Potentials & Nernst EQN

Hi,

So I've been trying my best to understand this example my prof pulled up on the board one day:

Looking at the cell potentials, N2 reaction is 0.06 V, O2 reaction is 1.23 V.

From my understanding: the higher the cell potential of a reaction the higher the thermodynamic drive something has to be reduced. Therefore, If these two reactions were connected up into a cell, the o2 reaction would be reduced (as it has the more positive cell potential) and by proxy the N2 reaction would be oxidised (due it having lower cell potential out of the two). Reduction is gain of electrons, The cathode gives up electrons, so naturally reduction occurs at the cathode, therefore the forward reaction of o2 occurs at the cathode.

same reasoning for other electrode: Anode accepts electrons, oxidation happens at anode, therefore the backward reaction of N2 occurs at anode.

Cell potential: Potential at anode - Potential at cathode
0.06 - 1.23 = -1.17 V

...right?

Apparently not, as shown by the next slide.

I got the overall potential correct, but.
I've apparently got the redox reactions completely the opposite way around, the backward reaction happens to the O2, not the N2

Where have I gone wrong in my thought process?

EDIT: I also do not understand why there's 6 for the number of electrons in the nernst eqn, I get theres 6 e in the N2 reaction, but theres 4e in the o2 reaction, why do you have to choose 6??

Thank you in advance : )

(edited 2 years ago)

Reply 1

Seonsem Leo

In an electrolytic cell, cathode gains e and anode loses e. The one of smaller electrode potential, that is N₂, should be the cathode, for E(ec)=E(ca)-E(an)<0.

And the other question, in this reaction N₂ + 3H₂O → 2NH₃ +1.5O₂, I guess you forgot to multiply the coefficient 1.5.