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.

1 Answer
Mar 22, 2018

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?