# Redox titrations questionWatch

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#1
So I found this experiment: http://alevelchem.com/aqa_a_level_ch....6/inorg01.htm
and conducted it for my Internal Assessment. The experiment itself went well, but would someone care to explain to me the calculations based on the results? I'm pretty sure there is something wrong with the ones that can be found here and I'm not exactly sure how to proceed.
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3 years ago
#2
(Original post by catcheroni)
So I found this experiment: http://alevelchem.com/aqa_a_level_ch....6/inorg01.htm
and conducted it for my Internal Assessment. The experiment itself went well, but would someone care to explain to me the calculations based on the results? I'm pretty sure there is something wrong with the ones that can be found here and I'm not exactly sure how to proceed.
What kind of something do you think is wrong?
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#3
It's all good, I managed to figure it out during the night, thanks for the help!
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#4
(Original post by catcheroni)
It's all good, I managed to figure it out during the night, thanks for the help!
Actually:
Therefore, fromthe reaction stoichiometry, moles of iron(II) = 5 x 0.0001 = 1.00 x 10-4 mol
This is in 27.90ml, therefore in 250 ml there is (250/27.9) x 1.00 x 10-4 = 8.96 x 10-4 moles of iron

Isn't this wrong? How is 5 x 0.0001 = 1.00 x 10^-4??
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3 years ago
#5
(Original post by catcheroni)
Actually:
Therefore, fromthe reaction stoichiometry, moles of iron(II) = 5 x 0.0001 = 1.00 x 10-4 mol
This is in 27.90ml, therefore in 250 ml there is (250/27.9) x 1.00 x 10-4 = 8.96 x 10-4 moles of iron

Isn't this wrong? How is 5 x 0.0001 = 1.00 x 10^-4??
Yes, there was an error in the calculation. It is now corrected.
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#6
(Original post by charco)
Yes, there was an error in the calculation. It is now corrected.
So for all further calculations I should use the value 5 x 10^-4? And in this example the amount of iron should be 250mg, am I correct?
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3 years ago
#7
(Original post by catcheroni)
So for all further calculations I should use the value 5 x 10^-4? And in this example the amount of iron should be 250mg, am I correct?
yes
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#8
Wow, are you the author of that other site? Because as I entered it today, the mistakes are corrected 0
3 years ago
#9
(Original post by catcheroni)
Wow, are you the author of that other site? Because as I entered it today, the mistakes are corrected I told you they were corrected in my earlier post...
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3 years ago
#10
(Original post by charco)
I told you they were corrected in my earlier post...
Hi,

Using the information below, how would you actually derive the half equations and the full equations? In other words how would you know what is produced for the half equations? It says that Fe2+ will be oxidised to Fe3+ on reacting with manganate, however, it doesn't say what will happen to manganate, which is the oxidising agent?
3.00 g of a lawn sand containing an iron (II) salt was shaken with dilute H2SO4. The resulting solution required 25.00 cm3 of 0.0200 mol dm-3 potassium manganate (VII) to oxidise the Fe2+ ions in the solution to Fe3+ ions. Use this to calculate the percentage by mass of Fe2+ ions in this sample of lawn sand.
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3 years ago
#11
(Original post by SANTR)
Hi,

Using the information below, how would you actually derive the half equations and the full equations? In other words how would you know what is produced for the half equations? It says that Fe2+ will be oxidised to Fe3+ on reacting with manganate, however, it doesn't say what will happen to manganate, which is the oxidising agent?
The oxidising agent takes the electrons and is itself reduced.

In this case manganate(VII) ions are reduced to manganese(II) ions

3.00 g of a lawn sand containing an iron (II) salt was shaken with dilute H2SO4. The resulting solution required 25.00 cm3 of 0.0200 mol dm-3 potassium manganate (VII) to oxidise the Fe2+ ions in the solution to Fe3+ ions. Use this to calculate the percentage by mass of Fe2+ ions in this sample of lawn sand.
Calculate moles of manganate(VII) needed
Use this to calculate moles of iron(II) reacted
Use mol = mass/Ar to calculate mass of iron
Percentage = 100 * mass/total mass
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3 years ago
#12
(Original post by charco)
The oxidising agent takes the electrons and is itself reduced.

In this case manganate(VII) ions are reduced to manganese(II) ions

Calculate moles of manganate(VII) needed
Use this to calculate moles of iron(II) reacted
Use mol = mass/Ar to calculate mass of iron
Percentage = 100 * mass/total mass
If iron (Fe2+) loses one electron to become Fe3+, then isn't that one electron gained by manganate, so manganate 7+ becomes manganate 6+?
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3 years ago
#13
(Original post by SANTR)
If iron (Fe2+) loses one electron to become Fe3+, then isn't that one electron gained by manganate, so manganate 7+ becomes manganate 6+?
This is why you need to use the redox half equations:

Fe2+ --> Fe3+ + 1e

BUT

MnO4- + 8H+ + 5e --> Mn2+ + 4H2O

So, 1 manganate(VII) ion reacts with 5 iron(II) ions
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3 years ago
#14
(Original post by charco)
This is why you need to use the redox half equations:

Fe2+ --> Fe3+ + 1e

BUT

MnO4- + 8H+ + 5e --> Mn2+ + 4H2O

So, 1 manganate(VII) ion reacts with 5 iron(II) ions
In redox questions, if they ask you to give a couple of reasons why emf is less than calculated in practice, what valid points can you make? I was thinking saying something about standard conditions not being used, but not sure what variation on standard conditions would specifically give a lower emf. Is there anything else I could say?
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3 years ago
#15
(Original post by Funky_Giraffe)
In redox questions, if they ask you to give a couple of reasons why emf is less than calculated in practice, what valid points can you make? I was thinking saying something about standard conditions not being used, but not sure what variation on standard conditions would specifically give a lower emf. Is there anything else I could say?
Better to say that concentrations are not standard.

Even better to say whether they're too high or low.

Problem with plan B is that if the MS wanted "non-standard concs" and you went higher rather than lower it might CON the mark.
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