Question

What is the colour change when aqueous iron(II) ions are treated with an acidified solution of dichromate(VI) ?

Answer 1

The colour change is from pale blue/green to yellow/brown.

Explanation:

Dichromate(VI) is a powerful oxidising agent and takes in electrons according to the 1/2 equation:

`Cr_2O_(7(aq))^(2-)+14H_((aq))^(+)+6erarr2Cr_((aq))^(3+)+7H_2O_((l))" "color(red)((1))`

The iron(II) ions give up their electrons according to:

`Fe_((aq))^(2+)rarrFe_((aq))^(3+)+e" "color(red)((2))`

To get the electrons to balance you can see that we need to multiply `color(red)((2))` by 6 then add this to equation `color(red)((1))rArr`

`Cr_2O_(7(aq))^(2-)+14H_((aq))^(+)+cancel(6e)+6Fe_((aqa))^(2+)rarr6Fe_((aq))^(3+)+cancel(6e)+2Cr_((aq))^(3+)+7H_2O_((l))"`

Iron(II) solutions contain the hexaaquoiron(II) ion `[Fe(H_2O)_(6)]^(2+)` which is a pale blue/green colour.

In the solid, hydrated state they look like this:

In aqueous conditions a solution of iron(III) contains the hexaaaquoiron(III) ion `[Fe(H_2O)_6]^(3+)]`. The solution looks a yellow/brown colour:

The colour is actually yellow due to the hydrolysis of the ion in water:

`[Fe(H_2O)_6]^(3+)+H_2O_((l))rightleftharpoons[Fe(H_2O)_5(OH)]^(2+)+H_3O^(+)`

It is the `[Fe(H_2O)_5(OH)]^(2+)` that gives the yellow/brown colour.

Pure iron(III) ions are actually pale violet. This is what they look like in the solid state:

However, these colours will be masked when you carry out the reaction.

Potassium chromate(VI) is orange in colour and that's what you see before the reaction. These are converted to green `Cr^(3+)` ions so you will see a mixture of these and the iron(III) species after the reaction.