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### Topic: Redox Titration Help  (Read 5883 times)

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#### brooke23l

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##### Redox Titration Help
« on: November 10, 2015, 09:55:36 PM »
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+ ... Eo =.732
Sn2+ <-> Sn4+ + 2e- ... Eo = .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...
« Last Edit: November 10, 2015, 10:23:25 PM by brooke23l »

#### Borek

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##### Re: Redox Titration Help
« Reply #1 on: November 11, 2015, 03:46:37 AM »
E- = .732 -(.05916/2)log([Fe2+]/[Fe3+])

Why /2?

Quote
1 mol Fe3+ = 1 mol Fe2+ + .5 mol Sn4+

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

Quote
.00005 mol Sn4+

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

Quote
(.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.
« Last Edit: November 12, 2015, 02:41:24 AM by Borek »
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#### Hunter2

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##### Re: Redox Titration Help
« Reply #2 on: November 11, 2015, 04:36:46 AM »
Quote
That was the initial amount of S2+, part of it already reacted.

one "n" is missing for tin

#### mjc123

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##### Re: Redox Titration Help
« Reply #3 on: November 11, 2015, 09:16:57 AM »
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

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##### Re: Redox Titration Help
« Reply #4 on: November 11, 2015, 10:57:43 AM »
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

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##### Re: Redox Titration Help
« Reply #5 on: November 11, 2015, 02:15:26 PM »
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+ + Ecalomel  = .4355 - .241 = .1954

The correct answer is .096 V

#### mjc123

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##### Re: Redox Titration Help
« Reply #6 on: November 11, 2015, 06:50:23 PM »
Quote
[Fe2+]=[Fe3+] and [Sn2+] = [Sn4+] at V2
These statements are not simultaneously true at any point, and neither is true at equivalence.

#### brooke23l

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##### Re: Redox Titration Help
« Reply #7 on: November 11, 2015, 08:35:23 PM »
Quote
[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

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##### Re: Redox Titration Help
« Reply #8 on: November 12, 2015, 06:13:17 AM »
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?