[Lou Reed]

Hi! I'm having trouble with this question, i'm really only having problems with what actually happens in the reaction, any help very welcome.

"25.0cm3 amount of a solution containing both iron (II) and iron (III) sulphates required 21.9cm3 of 0.02M potassium dichromate solution for oxidation. A further 25.0cm of the same solution after reduction required 36.6cm3 of the same dichromate. Calculate the concentration of the Fe(2+) and Fe(3+) in g/dm3."

I think i understand the first bit of the question; Fe(2+) is oxidised to Fe(3+) and Cr2O7(2-) is reduced to Cr(3+) giving the half equations:

Fe(2+) --> Fe(3+) + e(-)
Cr2O7(2-) + 14H(+) + 6e(-) --> 2Cr(3+) + 7H2O

combined gives...

6Fe(2+) + Cr2O7(2-) + 14H(+) --> 6Fe(3+) + 2Cr(3+) + 7H2O

Ok! I think thats right (if not please correct me!) but then what happens to the iron (IIII) sulphate? Is it also oxidised?

And then we get onto the second part of the question. I think that there are only Fe(3+) ions in the solution now because all the Fe(2+) have been oxidised, but then what happens when you add more dichromate? Does it form...Fe2O3?

Help really appreciated!

[Borek]

iron (IIII) sulphate?

Huh?

In the second titration you have REDUCED all Fe³⁺ present to Fe²⁺ before titration.

[Lou Reed]

iron (IIII) sulphate?

Huh?

meant iron III sorry!

In the second titration you have REDUCED all Fe³⁺ present to Fe²⁺ before titration.

Ohhhh...so...25.0cm3 of Fe(2+) requires 36.6cm3 0.02M potassium dichromate to titrate it?

What about the first step then? Does only the Fe(2+) react (oxidise) does the Fe(3+) not do anything?

Thanks!

[Lou Reed]

Anyways i got concentrations to be:

Fe(2+): 5.87gdm-3
Fe(3+): 3.94gdm-3

to 3 sig figs

If anyone can be bothered checking id be really grateful!

[Borek]

Looks OK.

Checked with stoichiometry calculator