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[UNIT 5] Hard question on finding mole ratio of Iron half cell

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A solution is prepared for use in the standard half cell Fe3+(aq), Fe2+(aq)|Pt. What is the mole ratio of the solids iron(II) sulfate, FeSO4, and iron(III) sulfate, Fe2(SO4)3, which should be dissolved to make the solution for this cell?

A 2:1

B 1:1

C 1:2

D 3:2

Answer is A

EDIT:
Look at FeSO4:Fe2(SO4)3 in a 1:1 mole ratio
In FeSO4 = we have 1 mole of Fe2+

In Fe2(SO4)3 = we have 2 moles of Fe2+

We have to have equal amount of moles of Fe ions for this cell to work.
We need to double the FeSO4 to get 2 moles of Fe2+ which is exactly the same amount of moles of Fe3+, which is 2.

And therefore the mole ratio FeSO4:Fe2(SO4)3 = 2:1

(edited 6 years ago)

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