Find n for Reaction Equations: General Method

[OmniReader]

how can the number of electrons, n, in a reaction equation be found generally? for example:

Tl³⁺ + 4I^– → TlI₄^–
I₂(s) + I^–(aq) → I₃^–(aq)
Tl⁺ + I₃^- → Tl³⁺ + 3I^-

How to work out standard electrode potential from K or vice versa depends on value of n. for these 3 I know they are n=1, n=2 and n=2 respectively. (thus this is not a homework question) but why does n take these values for this reactions, and how to work out n for general reactions?

 

[Simon Bridge]

It's related to the fine detail of the atomic structure - why atoms form particular ions at all.
A chemistry course short-cuts the underlying physics by just getting you to learn the emerging rules.

 

[OmniReader]

So then can you explain it to me or give me a link?
Even the emerging rules would be good enough, just so long as I can work out n for reactions as I see them. understanding the physics is ideal ofc

 

[DrDu]

Maybe you could state more clearly what you want and how you define n.
Your first and second reactions aren't even redox reactions, so presumably n=0.
If you were to perform reaction 3 in a electrolytic cell (seems to be what you are interested in),
then you have two half reactions:
<br /> Tl^+ \rightarrow Tl^{3+} +2e^-
and
I_3^- +2 e^- \rightarrow 3 I^-
The number of electrons on both sides is fixed by the condition of charge conservation (and, of course, atom number conservation).

 

[OmniReader]

Thanks for the method for reaction 3. For non-redox reactions there are problems where you get equilibrium constant for these reactions, e.g. 1 and/or 2, and have to use in conjunction with standard potential to work out standard potential of another couple. manipulating standard potentials additively requires you to convert each to 1-electron equations (dividing by coefficient on e^-). thus we need to know n, as in dG=-nFE, to manipulate even non-redox reaction to reach "standard electrode potential" for these reactions just to combine these potentials with others, even if they have no physical sense.

if we go by route of converting all potentials to constants for redox reaction (n known) and using the constants for the non-redox, we run into wall of, once all equations with the known constants are combined to form the desired equation for which we need E, we know K from transforming the known K values in the right way, but do not know n for this desired reaction so cannot find E. so, a value of 'n' is needed for even non-redox reactions purely for accounting purposes

 

[DrDu]

Please specify, using your first reaction as an example, what you want to calculate.
The equilibrium constant for that reaction?

 

[OmniReader]

Ok how about this then:

Tl+(aq) + e– → Tl(s) Eº1 = – 0.336 V
Tl3+(aq) + 3e– → Tl(s) Eº2 = + 0.728 V
I2(s) + 2e- → 2I–(aq) Eº3 = + 0.540 V
The equilibrium constant for the reaction I2(s) + I–(aq) → I3–(aq): K1 = 0.459.
The cumulative formation constant for the reaction Tl3+ + 4I– → TlI4–: K2=10^35.7

Write the reaction that takes place when a solution of the more stable isomer of thallium iodide (TlI₃) is treated with an excess of KI. Calculate the equilibrium constant for this reaction, and its standard electrode potential if possible.

(And I already have the answer for equilibrium constant so it's the explanation that could be really useful!) Difficult part is knowing how to use the two non-redox reactions.

 

[DrDu]

I take your word that this is no homework question!
I don't think you can state a standard electrode potential for that reaction as it is not a half reaction, ie. oxidation of Tl and reduction of I3- take place simultaneously.

You may translate the first three equations for the emf into equilibrium constants using Nernst.
Then just eliminate the ions/compounds which don't show up in the equilibrium constant.
That is:
half reaction 1: K1'=1/ ([Tl+] ) ; here n=1
half reaction 2: K2'=1/ ([Tl3+]) ; here n=3
half reaction 3: K3'=1/ ([I-]^2): here n=2
K1=[I3-]/[I-]
K2=[TlI4-]/[Tl3+][I-]^4

For the reaction (I suppose TlI3 is completely soluble)
TlI3+I- -> TlI4-
the equilibrium constant is
K=[TlI4-]/[TlI3][I-]=[TlI4-]/([Tl+][I3-][I-])
You should be able to express this in terms of all the K's given.

 

[OmniReader]

thanks, this was in fact a problem on my exam, I solved it fine for equilibrium constant K but lost the next mark because I said there was standard electrode potential but failed to find the "correct" value.

see post below

 

[OmniReader]

is the complete robust method to not use standard potentials but rather convert all potentials and equilibrium constants into standard Gibbs' free energy using the Delta Gº = -n * F * Eº and Delta Gº = -R * T * ln(K) relations after which the Delta Gº values are completely additive, and then once I have it for required reaction I can then convert back to Eº or K for that reaction with same relationship? alternatively you could convert all values to equilibrium constants using Delta Gº = -R * T * ln(K) and K = exp(n * F * Eº / (R * T)), then combine multiplicatively, till you reach required reaction which you can easily shift back to Eº or Delta Gº if wanted. forget about combining in Eº itself, just convert to Delta Gº for convenience in all cases. (or you could always convert to K if you want) am I right?

question becomes slightly different then. now, I have to ask, how to tell at a glance how many electrons involved in desired reaction like Tl+ + I3- + I- -> TlI4-, purely to convert from Delta Gº for this reaction to Eº? if Eº exists for this, if not then forget about it and thanks for help, if you can just confirm the above thing about Delta Gº always being additive once I use those reliable transformations, I'd be grateful.

 

[DrDu]

Converting everything into Delta G and using its additivity is certainly a valid approach.

Note that to state E^0 you need one or two half reactions. In these half reactions, the electrons show up explicitly, so you can directly read off their number.

 

[OmniReader]

Thanks, I have another problem. I want to know how to deduce reliably from galvanic cell representation, the half-cell reactions, for any cell, or vice versa.

Pt(s) | H2(g) | HCl(aq) | Cl2(g) | Pt(s)
Pb(s) | PbCl2(s) | HCl(aq) | H2(g) | Pt(s)
Pb(s) | PbSO4(s) | K2SO4(aq) || KNO3(aq) || KCl(aq) | PbCl2(s) | Pb(s)

have seen the basic guidelines on how to write cells knowing the reaction, but they do not explain why in these representations are so many salts rather than ions. nor is it clear how to approach deducing half-cell reactions from the cells above...