Question #eaf9b

2 Answers
Aug 9, 2016

Answer:

#sf(30color(white)(x)ml)#

Explanation:

Start with the 1/2 equations:

#sf(Fe^(2+)rarrFe^(3+)+e)" "color(red)((1))#

#sf(Cr_2O_7^(2-)+14H^(+)+6erarr2Cr^(3+)+7H_2O" "color(red)((2)))#

In order to get the electrons to balance, you can see that we need to x #color(red)((1))# by 6 then add to #color(red)((2))rArr#

#sf(Cr_2O_7^(2-)+14H^(+)+cancel(6e)+6Fe^(2+)rarr2Cr^(3+)+7H_2O+6Fe^(3+)+cancel(6e))#

This tells us that #sf(1"mol"color(white)(x)Cr_2O_7^(2-)-=6"mol"color(white)(x)Fe^(2+))#

#sf(c=n/v)#

#:.##sf(n=cxxv)#

#:.##sf(nCr(VI)=0.02xx25/1000=0.0005)#

#:.##sf(nFe(II)=0.0005xx6=0.003)#

For the manganate(VII) titration the 1/2 equation is :

#sf(MnO_4^(-)+8H^(+)+5erarrMn^(2+)+4H_2O" "color(red)((3)))#

To get the electrons to balance, we need to x #color(red)((1))# by 5 then add to #color(red)((3))rArr#

#sf(MnO_4^(-)+8H^(+)+cancel(5e)+5Fe^(2+)rarrMn^(2+)+4H_2O+5Fe^(3+)+cancel(5e))# .

This tells us that #sf(1"mol"color(white)(x)MnO_4^(-)-=5"mol"color(white)(x)Fe^(2+))#

#:.##sf(nMnO_4^(-)=0.003/5=0.0006)#

#sf(c=n/v)#

#:.##sf(v=n/c=0.0006/0.02=0.03color(white)(x)L)#

#sf(=30color(white)(x)ml)#

Aug 9, 2016

The reduction half reaction of #Cr_2O_7^"2-"# ion in acid medium

#Cr_2O_7^"2-" +14H^+ +6e->2Cr^"3+"+7H_2O....(1)#

And the oxidation half reaction of #Fe^"2+"#

#Fe^"2+"-e->Fe^"3+"......(2)#

The reduction half reaction of #MnO_4^"-"# ion in acid medium

#MnO_4^"-" +8H^+ +5e->Mn^"2+"+4H_2O....(3)#

Multiplying (2) by 6 and then adding with (1) the balanced equation of redox reaction by dichhromate becomes

#Cr_2O_7^"2-" +14H^+ +6Fe^"2+"->2Cr^"3+"+7H_2O+6Fe^"3+"..(4)#

This equation (4)reveals that

#1"mol "Fe^"2+"equiv 1/6 "mol "Cr_2O_7^"2-"color(red)(......(4A))#

Multiplying (2) by 5 and then adding with (3) the balanced equation of redox reaction by permanganate becomes

#MnO_4^"-" +8H^+ +5Fe^"2+"->Mn^"2+"+4H_2O+5Fe^"3+"..(5)#

This equation (5)reveals that

#1"mol "Fe^"2+"equiv 1/5 "mol "MnO_4^"-"color(red)(......(5A))#

So by combining (4A) and (5A) we can say

# 1/6 "mol "Cr_2O_7^"2-"equiv1/5" mol "MnO_4^"-"#

#:. 1 "mol "Cr_2O_7^"2-"equiv6/5" mol "MnO_4^"-"#

Now
#50"ml "Fe^"2+"equiv25ml" "0.02M" "Cr_2O_7^"2-"#

#=25xx0.02xx10^-3"mol "Cr_2O_7^"2-"#

#equiv6/5xx25xx0.02xx10^-3"mol "MnO_4^"-"#

#equiv30xx0.02xx10^-3"mol "MnO_4^"-"#

#equiv30"ml "0.02M" "MnO_4^"-"#

So if titration is carried out with 0.02 M #MnO_4^-# solution instead of 0.02M # Cr_2O_7^"2-"#
solution then 30ml 0.02M permanganate solution will be required.