# Chemical Forums

## Chemistry Forums for Students => High School Chemistry Forum => Chemistry Olympiad and other competitions => Topic started by: Big-Daddy on February 05, 2014, 04:12:01 PM

Title: Latimer calculation
Post by: Big-Daddy on February 05, 2014, 04:12:01 PM
How far will VO2+ be reduced by Zn  :rarrow: Zn2+ + 2e- reaction (E°=-0.76 V for the reduction, E°=0.76 V for the oxidation), given that I have calculated E° for these reactions to be:

E°(VO2+ + 2H+ + e-  :rarrow: VO2+ + H2O) = 1.00 V
E°(VO2+ + 4H+ + 2e-  :rarrow: V3+ + 2H2O) = 0.68 V
E°(VO2+ + 4H+ + 3e-  :rarrow: V2+ + 2H2O) = 0.3683333 V
E°(VO2+ + 4H+ + 5e-  :rarrow: V + 2H2O) = -0.231 V

It is tempting to hypothesize that the last reaction with positive E° will be the last that occurs, i.e. it will be reduced to +2 oxidation state but no further - but this appears to have no dependence on E°(oxidation) of the reducing agent, which can't be right. So how do we work out?
Title: Re: Latimer calculation
Post by: Big-Daddy on February 06, 2014, 12:18:28 PM
The original source of this problem is International Chemistry Olympiad 2008 Preparatory Problem 13, http://www.icho.hu/files/prep_problems_icho40.pdf , if someone wants to check that I've done E° calculations right.

I considered calculating ΔG° for the overall redox reaction, but I don't think it is fair to say that the most negative ΔG° will show how far VO2+ is reduced, because then we could just multiply one of the (less) negative (but still negative) reactions by some large number through all its stoichiometric coefficients and say "now this reaction has most negative ΔG°".
Title: Re: Latimer calculation
Post by: Rutherford on February 06, 2014, 12:27:14 PM
Convert the voltages to Gibbs' energy and calculate the Gibbs' energy of the overall reactions for each case and see whether the reaction is spontaneous.
Title: Re: Latimer calculation
Post by: Big-Daddy on February 06, 2014, 12:37:44 PM
Convert the voltages to Gibbs' energy and calculate the Gibbs' energy of the overall reactions for each case and see whether the reaction is spontaneous.

Well we can't use ΔG since we don't have enough information about concentrations, but we can use ΔG°. My results are

ΔG°(2VO2+ + 4H+ + Zn :rarrow: 2VO2+ + 2H2O + Zn2+) = -339.6 kJ/mol
ΔG°(VO2+ + 4H+ + Zn :rarrow: V3+ + 2H2O + Zn2+) = -277.9 kJ/mol
ΔG°(2VO2+ + 8H+ + 3Zn :rarrow: 2V2+ + 4H2O + 3Zn2+) = -653.2 kJ/mol
ΔG°(2VO2+ + 8H+ + 5Zn :rarrow: 2V + 4H2O + 5Zn2+) = -510.4 kJ/mol

So you'd think then that VO2+ is reduced all the way to V, since every reaction on the line has (strongly) negative ΔG°. But this isn't the correct answer... instead it stops at V2+. Most negative ΔG° argument, refer to my post above?
Title: Re: Latimer calculation
Post by: Rutherford on February 06, 2014, 12:48:02 PM
No, calculate the Gibbs' energy for zinc oxidation and then calculate the Gibbs energy for the overall reaction in each of the four cases (just as using Hess law).
Title: Re: Latimer calculation
Post by: Big-Daddy on February 06, 2014, 01:04:53 PM
No, calculate the Gibbs' energy for zinc oxidation and then calculate the Gibbs energy for the overall reaction in each of the four cases (just as using Hess law).

Didn't I just calculate the (standard) Gibbs' energy for the overall reaction in each of the four cases?
Title: Re: Latimer calculation
Post by: Rutherford on February 06, 2014, 01:56:27 PM
Oops. Sorry, didn't see. Most negative ΔG° argument is correct, you can prove it. Calculate the Gibbs energy for V2+ to V reduction. It should be positive, so as the V2+ production has the most negative Gibbs energy, and V2+ isn't reduced further to V, it can be concluded that VO2+ will be mostly reduced to V2+.
Title: Re: Latimer calculation
Post by: Big-Daddy on February 06, 2014, 02:24:06 PM
Oops. Sorry, didn't see. Most negative ΔG° argument is correct, you can prove it. Calculate the Gibbs energy for V2+ to V reduction. It should be positive, so as the V2+ production has the most negative Gibbs energy, and V2+ isn't reduced further to V, it can be concluded that VO2+ will be mostly reduced to V2+.

I see the logic. One point - if we consider VO2+ to V3+ reduction, do we have to use ΔG°(2VO2+ + 8H+ + 2Zn :rarrow: 2V3+ + 4H2O + 2Zn2+) = -277.9*2 = -555.8 kJ/mol to consider it fairly alongside the others? So if, e.g. the ΔG° for the reduction by Zn to V2+ was -520 kJ/mol rather than -653 kJ/mol, then we'd get reduction to V3+ and no further?
Title: Re: Latimer calculation
Post by: Rutherford on February 06, 2014, 02:52:39 PM
Yes.
Title: Re: Latimer calculation
Post by: Big-Daddy on February 06, 2014, 03:12:22 PM
Ok. Now let's say we make it more complicated and use Cu metal instead of Zn. Cu can be oxidized to +1 or +2 oxidation states or not at all (http://hyperphysics.phy-astr.gsu.edu/hbase/tables/electpot.html). Would you still know how to solve it? (by solve it, I mean find the main reaction of the system)

I think mastering Latimer diagrams would be good for this year's IChO.
Title: Re: Latimer calculation
Post by: Rutherford on February 07, 2014, 08:14:40 AM
Then you would have 8 reactions plus 4 more to compare the copper(I) oxidation to copper(II) in each case.
Title: Re: Latimer calculation
Post by: Big-Daddy on February 07, 2014, 08:47:51 AM
Why plus 4 more? We are starting with Cu, so isn't it true that either Cu remains Cu (no reaction occurs, the VO2+ does not get to form a redox equilibrium at all), or it gets oxidized to Cu+ (by one of the 4 reductions), or it gets oxidized to Cu2+ (by one of the 4 reductions)?

So then, in this case, the most negative ΔG° approach is still going to be for the main reaction (which shows how far Cu gets oxidized and VO2+ gets reduced)?

Note: if Cu remains Cu, we could call this a null reaction with K=1 and ΔG°=0, meaning that if all the possible reactions in our system had ΔG°>0, the null reaction would 'occur' and nothing will happen.
Title: Re: Latimer calculation
Post by: Rutherford on February 07, 2014, 08:51:05 AM
Maybe Cu+ is even easier oxidized to Cu2+ than Cu. I am not sure exactly if it is, but this should be checked.
Title: Re: Latimer calculation
Post by: Big-Daddy on February 07, 2014, 11:27:43 AM
Maybe Cu+ is even easier oxidized to Cu2+ than Cu. I am not sure exactly if it is, but this should be checked.

From general knowledge I know that Cu+ tends to disproportionate to Cu2+ and Cu. But I would think that all these possibilities are included when you calculate ΔG° for all 8 reactions you mentioned before? I do not think "the main reaction" in this question could start with Cu+ because there is no Cu+ initially present. And if there was, it would just disproportionate.

So if we have a set of the reactions that can happen starting from the initial reactants present (here, seems to be 8 - starting from Cu and VO2+) are you sure most negative ΔG° is the right way to find the main reaction?
Title: Re: Latimer calculation
Post by: Rutherford on February 07, 2014, 11:41:04 AM
I don't see why not. I didn't say that the main reaction would start with Cu(I). If Cu(I) disproportionates easily, then in the overall reaction Cu(II) would be produced even if you previously get the most negative Gibbs energy for an overall reaction with Cu(I) being produced (by checking the 8 reactions).

Title: Re: Latimer calculation
Post by: Big-Daddy on February 07, 2014, 12:45:36 PM
Ok. I've learnt something valuable today.

I didn't say that the main reaction would start with Cu(I). If Cu(I) disproportionates easily, then in the overall reaction Cu(II) would be produced even if you previously get the most negative Gibbs energy for an overall reaction with Cu(I) being produced (by checking the 8 reactions).

This is just Hess' law (now with ΔG°). The route Cu -> Cu(I) -> Cu(II) will have the same ΔG° as the route Cu -> Cu(II), so long as the reaction equation is the same, which it is.

I did a proof of this fact: that if Cu(I) disproportionates to Cu and Cu(II), it is guaranteed that you will get the most negative ΔG° for a reaction with Cu(II) being produced. You can say this because E°(Cu+/Cu)>E°(Cu2+/Cu) etc. But that's just mathematical semantics. If Cu+ is unstable and disproportionates to form Cu(II), it's not hard to predict that Cu will go straight to Cu(II).

Edit: It should be noticed that we have treated the reduced species as needing to have the same stoichiometric coefficient for us to compare fairly between values of ΔG°, but the oxidized species then does not have the same stoichiometric coefficient from reaction to reaction as we compare ΔG°. Why the reason for this asymmetry?
Title: Re: Latimer calculation
Post by: Rutherford on February 07, 2014, 03:02:43 PM
Because we know which oxidation state it will obtain, but we wanted to see how far will vanadium(V) be reduced.
Title: Re: Latimer calculation
Post by: Big-Daddy on February 07, 2014, 03:40:06 PM
True, but the same logic cannot be applied to the Cu case since we don't know (beforehand) whether it will go to Cu(II) or Cu(I), nor do we know how far the VO2+ will go. How do we then make sure that our comparison of ΔG° values is fair in terms of stoichiometric coefficients on VO2+ or Cu?
Title: Re: Latimer calculation
Post by: Rutherford on February 08, 2014, 02:13:32 AM
Then apply it to both copper and vanadium separately.