Electrochemistry

https://imgur.com/a/S3SJGJA

Solution A contains both iron(II) and iron(III) ions
Q:
A student suggested that, for standard conditions, solution A should be
made by mixing equal volumes of 1 moldm−3 iron(II) sulfate solution and 1 moldm−3 iron(III) sulfate solution.
This mixture is not suitable for measuring the standard electrode potential of this cell.
State how this mixture should be changed. Justify your answer.

Answer:
Iron(II) sulfate should be 2 mol dm—3 (Mark 1)
Mixture should be 1 mol dm—3 with respect to each iron ion (in standard
electrode) (mark 2)

Can some explain how to approach this question and how to get the answers mentioned above?
Thank you!

ps - I added a link with the cell diagram if necessary

Standard conditions are 1 mol dm-3 for the Fe2+ and Fe3+ ions, rather than the conc of the compounds.

1 mol of Fe2(SO4)3 i.e. iron(III) sulfate contains 2 mol of Fe3+ ions. 1 mol dm-3 Fe2(SO4)3 therefore contains 2 mol dm-3 of Fe3+ ions.

bro could you please explain your answer?

ok so basically, for a standard solution to be formed, the cell solution must be 1mol dm^-3 with respect to both Fe 2+ and Fe 3+ ions...

not the compounds feso4, fe2(SO4)3 etc

hence we have to find the indivdual ion concentrations from the compound itself...

to do that we will simply look at the mole ratio.

1 mol of FeSO4 produces 1 mol of Fe2+

1 mol of Fe2(SO4)3 produces 2 moles of Fe 3+ ....

The mole ratio we just found works for conc too

eg 1 moldm^-3 of fe2(so4)3 gives 2 moldm^3 of Fe 3+


note that when volume is doubled... conc is halved...

Surely if you double concentration of FeSO4 that would make it 2moldm-3 and Fe2(SO4)3 is also 2 moldm-3. But, doesn't it have to be 1 moldm-3.

So why don't you half concentration of Fe2(SO4)3 so they are both 1 moldm-3?