# Iron(ii) and Iron (iii) Seperation

#### RE-com

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$$^{}3$$ 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.

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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#### chemisttree

Homework Helper
Gold Member
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 $$\rightarrow$$Mn$$^{}2+$$+4H$$_{}2$$O

Work that out ?

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

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

Then use a reducing agent so :

2I$$^{}-$$ - 2e$$\rightarrow$$ I$$_{}2$$

Using S$$_{}2$$O$$_{}3$$$$^{}2-$$ 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

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