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Iron(ii) and Iron (iii) Seperation

  • Thread starter RE-com
  • Start date
1. The problem statement, all variables and given/known data
You must devise a volumetric procedure to determine the percentage or Iron(ii) and Iron (iii) in a mixture containing both.

You are provided with 200cm[tex]^{}3[/tex] of a solution containing between 1.1g and 1.3g of iron ions as a mixture of Fe[tex]^{}2+[/tex] and Fe[tex]^{}3+[/tex](aq). You may assume that each of the two ions is present to at least 30% by mass. You have the normal apparatus required for volumteric analysis but a balance is not available.

3. The attempt at a solution

The max and min. mass range = 1.1x0.3=0.33g min. and 1.3x0.7=0.91g


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Try to put in a little more effort in your attempt at a solution. Do you know of any tests that are specific for either iron (II) or iron (III)? Know of any way to convert one quantitatively into the other?
I do have some ideas I've already jotted down but whether or not they will work I have no idea:

Fe[tex]^{}2+[/tex] - e [tex]\rightarrow[/tex] Fe[tex]^{}3+[/tex]

Using MnO[tex]_{}4[/tex][tex]^{}-[/tex]+8H[tex]^{}+[/tex] + 5e [tex]\rightarrow[/tex]Mn[tex]^{}2+[/tex]+4H[tex]_{}2[/tex]O

Work that out ?

Then (just tell me to stop if this is completely wrong):

Fe[tex]^{}3+[/tex]+ e [tex]\rightarrow[/tex]Fe[tex]^{}2+[/tex]

Then use a reducing agent so :

2I[tex]^{}-[/tex] - 2e[tex]\rightarrow[/tex] I[tex]_{}2[/tex]

Using S[tex]_{}2[/tex]O[tex]_{}3[/tex][tex]^{}2-[/tex] for getting rid of the excess Iodide that would be left

so bascally measure the Fe[tex]^{}2+[/tex] with the MnO[tex]_{}4[/tex][tex]^{}-[/tex] then convert Fe[tex]^{}3+[/tex] to Fe[tex]^{}2+[/tex]
Measure it again, then difference is the amount of Fe[tex]^{}3+[/tex]

but it says justify concentrations etc, I don't have a clue where to start on that front

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