[eesux]

I am in desperate need of help. I have to write a 4000 word report on my experiment which is due in 2 days. I've already finished doing most of my calculations, and have about 3,000 words all done, I just realized this problem right now. I know this is a lot of reading to do for you guys but please, PLEASE help me!!!

So I'm doing an experiment to answer this question: What is the effect of different cooking length on the iron(II) content in spinach.

Basically what I'm doing is boiling 10g of chopped up spinach in 20mL of distilled water at 100˚C. Then I remove the spinach chunks, and mix sulfuric acid (H2SO4) to the water I just boiled the spinach with. This will make FeSO4 because some of the iron(II) will be in the water. Then, I'm going to titrate this FeSO4 with potassium permanganate KMnO4 to see how much Fe(II) was lost from the iron.

The problem is though, when I add H2SO4 to the boiled water (that used to have spinach), since I obviously don't know how much iron is in it, I might add excess H2SO4 right? That would change the titrating equation from

FeSO4 + KMnO4 -> KMnSO4  + FeO4  to
FeSO4 + H2SO4 + KMnO4 ->?
since I'll have excess H2SO4.
Would this effect my data on the redox titration?!

I'm going to thank you guys before hand: THANK YOU A MILLION TIMES!!

[UG]

No, it wouldn't matter since you still have the same number of moles of Fe(II), the concentration will be diluted but it will not affect your calculations. Btw...FeSO4 + KMnO4 -> KMnSO4  + FeO4 is incorrect, iron is oxidised to Fe(III)

[eesux]

UG,

Thank you so much.
Oh right because it's a redox reaction right? I'm usually good at chemistry but this reduction-oxidation throws me off a little. So KMnO4 which is a good oxidizing agent will cause the Fe(II) in FeSO4 to go through oxidation, which means it will turn into Fe(III)...does that sound good?

I know this is a stupid question....but which one is getting reduced?

[UG]

MnO₄^- is reduced.
MnO₄^- + 5e^- + 8H⁺  Mn²⁺ + 4H₂O

[eesux]

MnO₄^- is reduced.
MnO₄^- + 5e^- + 8H⁺  Mn²⁺ + 4H₂O

8 H2SO4 + 2 KMnO4 + 10 FeSO4 = 5 Fe2(SO4)3 + 2 MnSO4 + K2SO4 + 8H2O

does this look like a good equation? Though, I don't really understand why the H2SO4 needs to be there...

[UG]

Well, the sulfuric acid is there as a catalyst, to provide the H⁺ ions. It doesn't need to be written in the equation, all you need to do is write a net ionic equation. Just add the oxidation of iron together with the reduction half-reaction.

[eesux]

UG you're a life saver!

However...I'm sorry I'm not quite sure how to include oxidation and reduction in this equation. Do I just show the charges or...?

[UG]

Okay, so the reduction goes:
MnO₄^- + 5e^- + 8H⁺  Mn²⁺ + 4H₂O

and the oxidation goes

Fe²⁺  Fe³⁺ + e^-

Now in order to combine the two, you gotta make the electrons cancel out on both sides, this is done by multiplying the oxidation reaction by 5. So now you can cancel out the electrons and put the two together 
i.e, 5Fe²⁺ + MnO₄^- + etc..

[eesux]

Alright!! Thank you so much

I might have some more questions later on...hopefully you'll be here to help me

[Borek]

Well, the sulfuric acid is there as a catalyst, to provide the H⁺ ions.

It is not a catalyst - H⁺ are being consumed! But SO₄^2- is a spectator.

http://www.titrations.info/permanganate-titration

http://www.titrations.info/permanganate-titration-iron

[UG]

Yes, you're right, my mistake.  Thanks for the correction.

[eesux]

It is not a catalyst - H⁺ are being consumed! But SO₄^2- is a spectator.

Borek,

Thank you! I just have one little question...I've been googling this but I still haven't gotten a clear answer. When I boil spinach in normal, distilled water, since Fe(II) is soluble, it dissolves into the water right? But in what form does it stay? Does it stay as simply Fe+2 ions floating around in the water, or does it combine with water to make a certain compound such as FeOH or Fe(H2O) or something?