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?