[brooke23l]

There is something about these redox titrations that I am just not getting. If anyone has any insight into this problem I would greatly appreciate the assistance

15-5. Consider the titration of 25.0 mL of 0.050 0 M Sn2 with
0.100 M Fe3 in 1 M HCl to give Fe2 and Sn4, using Pt and
calomel electrodes.
(a) Write a balanced titration reaction.
(b) Write two half-reactions for the indicator electrode.
(c) Write two Nernst equations for the cell voltage.
(d) Calculate E at the following volumes of Fe3: 1.0, 12.5, 24.0,
25.0, 26.0, and 30.0 mL. Sketch the titration curve.

I'm stuck on part D. This is what I have so far:

a) 2Fe³⁺ + Sn²⁺ -> 2Fe²⁺ + Sn⁴⁺

b)Fe³⁺ + e^- <-> Fe²⁺ ... E_o =.732
Sn²⁺ <-> Sn⁴⁺ + 2e^- ... E_o = .139

c)E_- = .732 -(.05916/2)log([Fe²⁺]/[Fe³⁺])
E₊ = .139 -(.05916/2)log([Sn⁴⁺]/[Sn²⁺])

d)at 1.0 ml Fe³⁺...
1 mol Fe³⁺ = 1 mol Fe²⁺ + .5 mol Sn⁴⁺
(.1 M Fe³⁺)(.001 L) = .0001 mol Fe³⁺... so there is .0001 mol Fe²⁺ and .00005 mol Sn⁴⁺
(.050M Sn²⁺)(.025 L) = .00125 mol Sn²⁺
E = .139 -(.05916/2)log([Sn⁴⁺]/[Sn²⁺) - .268 (calomel electrode)
E = .139 -(.05916/2)log(.00005/.00125) - .268 (calomel electrode) = -.0876 V
Yet the correct answer is -.143...

[Borek]

E_- = .732 -(.05916/2)log([Fe²⁺]/[Fe³⁺])

Why /2?

1 mol Fe³⁺ = 1 mol Fe²⁺ + .5 mol Sn⁴⁺

I wouldn't put it this way, but I guess you refer to the stoichiometry.

.00005 mol Sn⁴⁺

Yes, that's the amount of Sn⁴⁺ produced.

(.050M Sn²⁺)(.025 L) = .00125 mol Sn²⁺

That was the initial amount of Sn²⁺, part of it already reacted. (Edit: typo corrected)

Not that it changes your answer by much. Either the book answer is wrong, or I need more coffee.

[Hunter2]

That was the initial amount of S²⁺, part of it already reacted.

one "n" is missing for tin

[mjc123]

E+ = .139 -(.05916/2)log([Sn²⁺]/[Sn⁴⁺])
How much Sn²⁺ is there after reaction with the Fe³⁺?
Where do you get the value 0.268? Saturated calomel electrode is 0.244V https://en.wikipedia.org/wiki/Saturated_calomel_electrode
With these corrections I get E = -0.146V.

[brooke23l]

E+ = .139 -(.05916/2)log([Sn²⁺]/[Sn⁴⁺])
How much Sn²⁺ is there after reaction with the Fe³⁺?
Where do you get the value 0.268? Saturated calomel electrode is 0.244V https://en.wikipedia.org/wiki/Saturated_calomel_electrode
With these corrections I get E = -0.146V.

Oops, I was looking at the unsaturated calomel electrode, and I forgot that some of the Sn²⁺ is used up (duh!). Thank you so much, you're a lifesaver!

[brooke23l]

I'm having some trouble with the 25.00 ml titration. This is the equivalence point, correct? So we add both nernst equations...

2E₊ = .732 +.139 -(.05916/2)log1 - (.05916/2)log1.... log1 because [Fe²⁺]=[Fe³⁺] and [Sn²⁺] = [Sn⁴⁺] at V₂
E₊ = .4355

E= E₊ + E_calomel  = .4355 - .241 = .1954

The correct answer is .096 V

[mjc123]

[Fe2+]=[Fe3+] and [Sn2+] = [Sn4+] at V2

These statements are not simultaneously true at any point, and neither is true at equivalence.

[brooke23l]

[Fe2+]=[Fe3+] and [Sn2+] = [Sn4+] at V2

These statements are not simultaneously true at any point, and neither is true at equivalence.

Sorry, I meant that 2[Fe²⁺] = [Sn⁴⁺] and 2[Fe³⁺] = [Sn²⁺] at 2V
This is the equivalent point, right? The textbook doesnt really explain how to find it.
I've been looking at this problem for so long, I just keep confusing myself...
Here's the relevant section of the text....

[mjc123]

At the equivalence point, all the Sn²⁺ has supposedly reacted with the stoichiometric amount of Fe³⁺ and you have 0.00125 mol Sn⁴⁺ and 0.0025 mol Fe²⁺. In reality you will have a small amount of Fe³⁺ and Sn²⁺ - let us say x mol Sn²⁺ and 2x mol Fe³⁺. Let us suppose - we should check this at the end of the calculation - that x << 0.00125 mol. Then (as volume cancels out)
[Sn²⁺]/[Sn⁴⁺] = x/0.00125 = 2x/0.0025 = [Fe³⁺]/[Fe²⁺] = Q
E+ = 0.732 + 0.05916 logQ (note, don't divide by 2, as Borek said - why?)
E+ = 0.139 - 0.05196/2 logQ
These equations are both true, so by weighted addition (eq. 1 + 2* eq 2) 3E+ = 0.732 + 0.278 = 1.010 hence E+ = 0.337 V
E = 0.337 - 0.241 = 0.096 V
Subtracting eq 2 from eq 1 we get
0 = 0.593 + 0.05916*3/2 logQ
logQ = -6.68
Q = 2 x 10^-7
so our assumption that x is very small was correct.
Note: I have used your figure of 0.732 as it gives what you say is the correct answer. But is not E°(Fe³⁺/Fe²⁺) 0.77V? Isn't that what it says in the extract you posted?