I need to work out the equation for the reaction between potassium dichromate(VI) and ammonium iron(II) sulphate, but I just don't know where to start...any pointers?

Do you know what the products are or what sort of reaction this is?

I know that the potassium dichromate(VI) oxides the iron(II) to iron(III) but that's all :/

It's a good start.

Do you know what the products are?

Still a bit clueless on the products. I've managed to find out this though, when potassium dichromate is reduced, this is the half equation:

Cr2O72+ + 14H+ + 6e- --> 2Cr3+ + 7H2O

And I think the H+ comes from the acidified solution that the potassium dichromate is in.

 

And I think the formula for ammonium iron(II) sulphate is:

Fe(NH4)2(SO4)2

Right so far.

I assume this reaction was done in dilute sulphuric acid which,as you say, supplies the extra H+ ions.

If I say the other products are potasssium sulphate, ammonium sulphate Cr(III) sulphate, and Fe(III) sulphate does that help?

Thanks, that's a great help, I have guessed then that the (unbalanced) reaction is:

K2Cr2O7 + Fe(NH4)2(SO4)2 + 7H2SO4 → K2SO4 + (NH4)2SO4 + Cr2(SO4)3 + Fe2(SO4)3 + 7H2O

 

But I have been fiddling around with it pretty much since your post and I can't balance it (feeble brain huh). I can manage to get the ammoniums andand everything else to balance, but never the sulphates. I couldn't seem to get enough of them on the left hand side, so I changed the H+ to H2SO4, but even so, I couldn't balance it..does it involve some pretty large numbers?

Each Cr(VI) takes 3 electrons to become Cr(III)

It only takes the removal of one to turn an Fe(II) into Fe(III)

Each dichromate has 2 Cr atoms.

How many Fe(II) does it take to supply the elctrons needed to reduce a molecule of K2Cr2O7?

So 6 iron(II) ions would be needed.

 

I get this:

Cr2O72- + 14H+ + 6Fe2+ --> 2Cr3+ + 7H2O + 6Fe3+