Question

Question #eaf9b

Answer 1

`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)`

Answer 2

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.