Redox titrations question

So I found this experiment: http://alevelchem.com/aqa_a_level_chemistry/unit3.6/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.

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??

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?

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...

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.

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+?

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?