Question

What is the normality of the following solution?

Consider the following unbalanced redox reaction:
`Cr_2O_7^(2-)+Fe^(2+)+H^+rarr Cr^(3+)+Fe^(3+) +H_2O`

What is the normality of a `K_2Cr_2O_7` solution, 35.0 `cm^3` of which contains 3.87g of `K_2Cr_2O_7`
I am struggling to work out the equivalent weight. I have no idea how to get it.

Answer 1

Well, by definition, `"molarity"="moles of solute"/"volume of solution"`

Explanation:

And so...`((3.87*g)/(294.18*g*mol^-1))/(35.0*cm^-3xx10^-3*L*cm^-3)=0.376*mol`

And so `n_(Cr_2O_7^(2-))=0.376*mol`

As regards equivalence, we need to write the appropriate redox reaction, for which we use the method of half-equations...

Orange dichromate, `Cr(VI+)`, is reduced to green `Cr^(3+)`:

`Cr_2O_7^(2-)+14H^+ +6e^(-) rarr 2Cr^(3+)+7H_2O(l)` `(i)`

And `"ferrous ion"` is OXIDIZED to `"ferric ion"`...

`Fe^(2+) rarr Fe^(3+) + e^(-)` `(ii)`

And clearly, to balance, we take `(i)+6xx(ii)`...and cancel common reagents....we add the equations in these mulitples in order to remove the electrons are virtual particles of convenience....

`Cr_2O_7^(2-)+6Fe^(2+)+14H^+ +6e^(-) rarr 6Fe^(3+) + 2Cr^(3+)+7H_2O(l)+6e^(-)`

`Cr_2O_7^(2-)+6Fe^(2+)+14H^+ rarr 6Fe^(3+) + 2Cr^(3+)+7H_2O(l)`

Now you have the molar equivalence of `Fe(II)`....`1/6*"equiv"*Cr_2O_7^(2-)-=1*"equiv"*Fe^(2+)`.

Does this help you?