[Big-Daddy]

Given that E°=+0.16 for the electrode half-cell with equilibrium Cu²⁺ (aq) + e^-  Cu⁺ (aq) and E°=0.52 for the electrode which establishes equilibrium Cu⁺ (aq) + e^-  Cu (s), can we deduce E° for the electrode half-cell with equilibrium 1/2 Cu²⁺ (aq) + e^-  Cu (s)?

My current method would be:

E°[1/2 Cu²⁺ (aq) + e^-  Cu (s)]=(0.16 · 0.52)^1/2=0.289.

Can someone confirm this or say how I should add them properly? E° is defined from standard chemical potentials which I'm sure cancel out so this must be possible...

[Big-Daddy]

Can someone help with this? I'm actually no longer sure it is possible.

[Corribus]

You can't add cell potentials like this.

Check out

Fe(s) Fe³⁺ + 3e^-   E° = +0.04 V

and

Fe(s) Fe²⁺ + 2e^-  E° = +0.44 V
Fe²⁺ Fe³⁺ + e^-   E° = -0.77 V

Summing the latter two equations does not equal the upper equation.  If you want to add half reactions, you have to convert to Gibbs energy.

See here:

http://www.chem1.com/acad/webtext/elchem/ec3.html

[Big-Daddy]

ΔG° appears to cancel out?

ΔG°[final]=(1/2)*(ΔG°₁+ΔG°₂)
ΔG°[final]=(1/2)*(-F*E°₁-F*E°₂)
ΔG°[final]=-(1/2)*F*(E°₁+E°₂)
-F*E°[final]=-(1/2)*F*(E°₁+E°₂)
E°[final]=(1/2)*(E°₁+E°₂)=(1/2)*(0.16+0.52)=0.34

Which, unless this is a coincidence, would suggest that E° is additive over half-reactions in the same way that ΔG° is.

I suppose this might not work if the number of electrons were not the same in each of these reactions (as ΔG°=-nFE°). That's not a problem for me, because the book I work from suggests that it is always good practice to divide through in each reaction until the coefficient on the electron is 1. And I think that the E° value will always be the same for a half-reaction even if you multiply or divide that half-reaction? (Because then by definition the value of n multiplies or divides by the same amount.)

I'd also like to note that it would seem we can calculate an equilibrium constant for the half-cell reaction using -R*T*log_e(K)=-FE°. And if you multiply the number of electrons by 3, you will cube K (both the numerator at the denominator of the expression for K are cubed).

Is this all right?

[Corribus]

Check out example 2 in the link I pasted above.  You will see that simply adding the potentials gives quite a different answer than converting to Gibbs energy, adding, then converting back to potentials.

You can see this pretty easy from algebra:

ΔG_tot = ΔG¹ + ΔG²

since ΔG_x = - n_xFE°_x

ΔG_tot = -n₁FE°₁ - n₂FE°₂

Also,

ΔG_tot = -n_totFE°_tot = -(n₁+n₂)FE°_tot

Combining the last two equations gives

n₁E°₁ + n₂E°₂ = (n₁+n₂)E°_tot

And that

E°_tot = (n₁E°₁ + n₂E°₂) / (n₁+n₂)

It is pretty obvious now that the overall potential change is in general not just the sum of the potential changes for the two half reactions.  The "n" terms do not divide out.

The reason for all of this is because standard potentials by definition refer to transfer of 1 mole of electrons.  When adding half reactions that deal with different numbers of electrons, you're messing up your "electron accounting".  Converting to Gibbs energy first will always get you the right answer, because the accounting is implicity incorporated in the nFE definition.  (Converting to Gibbs energy is a good practice anyway - because it is a state function, you know you can always add it without fear of doing something wrong.)

[Big-Daddy]

Sorry if my previous post was a bit long. I have already said this in there: that I convert the half-reactions each to one mole of electrons per mole of reaction by dividing through by n from the start. The E° is defined for this reaction, whereas obviously due to being additive the ΔG° has to be multiplied by number of electrons.

If I first convert each half-reaction to E° format, i.e. 1 electron per reaction equation, as my book suggests, then E° should be additive in the same way ΔG° is.

e.g. Fe(s) Fe²⁺ + 2e^-  E° = +0.44 V
Fe²⁺ Fe³⁺ + e^-   E° = -0.77 V

(Though I thought protocol was to write the equilibria with electrons in the reactants rather than products?)

Change the first reaction to have only one electron:
(1/2) Fe(s) (1/2) Fe²⁺ + e^- (Rxn 1)
Fe²⁺ Fe³⁺ + e^- (Rxn 2)

Now we will use Rxn 1+(1/2)*Rxn 2. This is (1/2) Fe (s)  (1/2) Fe³⁺ + (3/2) e^-.

Now we need to divide the whole thing by (3/2) to reduce the coefficient on electrons to 1, (1/3) Fe (s)  Fe³⁺ + e^-.

So our final transformation is E°_desired=(3/2)*(E°₁+(1/2)*E°₂)=(3/2)*(0.44+(1/2)*(-0.77))=+0.0825. Why is this different from the true answer you quoted?

I know you've suggested using Gibbs' free energy changes but honestly as I proved in my last post I cannot see why, having reduced all reaction equations to one mole of electrons per mole of reaction, and ensuring that for our desired reaction we also produce one mole of electrons, E° should not be directly additive. I understand the principle behind ΔG° being used instead of E° but not why this doesn't work.

[Big-Daddy]

I have a slightly different question now. I think I understand the above topic and am ok with it.

The equation ΔG=-nFE_cell, I understand.

I also understand that, at unity activities of all species, Q=1 so R*T*ln(Q)=R*T*ln(1)=0. And therefore E_cell=E°_cell and ΔG=ΔG°.

However, surely we cannot then write that ΔG°=-nFE°_cell=-R*T*ln(K), since ΔG° is measured at unity activity when Q=1, rather than at K? In other words, ΔG°=-nFE°_cell requires that the species be at unity activity (i.e. Q=1) so that it can come from ΔG=-nFE_cell. But surely if Q=1 and K≠Q (if K does equal Q then, as we discussed some time ago, nothing is measured because equilibrium has already been reached!), we cannot write ΔG=0, so we cannot write ΔG°=-R*T*ln(K). So we should not be able to make the connection -R*T*ln(K)=-nFE°_cell as far as I can see, because ΔG°=-nFE°_cell is only true when Q=1 and ΔG°=-R*T*ln(K) is only true when Q=K and thus ΔG=0, and we in general assume that K≠1 (so that we are not at equilibrium to start with, meaning there is some spontaneous reaction that enables us to measure potential).

[Big-Daddy]

And do the redox equilibria have the same standard electrode potential regardless of the direction (i.e. whether the forward reaction is the reduction or oxidation)?

By the way I have figured out the post 2 above this one so I don't need help with that. But I do need help with the post right above this one, which asks how we can write -nFE=-RT*ln(K).