How Can You Determine the Percentage of Iron(ii) and Iron(iii) in a Mixture?

[RE-com]

Homework Statement

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^{}3 of a solution containing between 1.1g and 1.3g of iron ions as a mixture of Fe^{}2+ and Fe^{}3+(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.

The Attempt at a Solution

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

 

[chemisttree]

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?

 

[RE-com]

I do have some ideas I've already jotted down but whether or not they will work I have no idea:

Fe^{}2+ - e \rightarrow Fe^{}3+

Using MnO_{}4^{}-+8H^{}+ + 5e \rightarrowMn^{}2++4H_{}2O

Work that out ?

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

Fe^{}3++ e \rightarrowFe^{}2+

Then use a reducing agent so :

2I^{}- - 2e\rightarrow I_{}2

Using S_{}2O_{}3^{}2- for getting rid of the excess Iodide that would be left

so bascally measure the Fe^{}2+ with the MnO_{}4^{}- then convert Fe^{}3+ to Fe^{}2+
Measure it again, then difference is the amount of Fe^{}3+

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