Question

KMnO4 oxidises `X^n+` to `XO^(3-)`, itself changing to Mn2+ in acidic medium. 2.68x10^-3 mol of X^n+ require 1.61x10^-3 mol of KmnO4. The value of n is?

Answer 1

n = 2

Explanation:

`sf(X^(n+))` is oxidised by the `sf(MnO_4^-)`:

`sf(X^(n+)rarrXO_3^-)`

The change in oxidation state is:

`sf(+nrarr+5)`

The `sf(MnO_4^-)` is reduced:

`sf(MnO_4^(-)+8H^(+)+5erarrMn^(2+)+4H_2O" "color(red)((1)))`

The change in oxidation state is:

`sf(+7rarr+2)`

We find the ratio in moles by which they react:

`sf((nX^(n+))/(nMnO_4^-)=(2.68x10^(-3))/(1.61xx10^(-3))=1.66=5/3)`

`:.` `sf(nX^(n+):nMnO_4^(-)=5:3)`

We can build up the stoiciometric equation:

`sf(5X^(n+)color(white)(xxxx)+color(white)(xxxx)3MnO_4^(-)rarrcolor(white)(xxx)5XO_3^(-)color(white)(xxx)+3Mn^(2+)`

`sf(5xx(+n)color(white)(xxxxxx)3xx+7color(white)(xxxxxx)5xx+5color(white)(xxx)3xx+2)`

`sf(+5ncolor(white)(xxxxxxxxxxx)+21color(white)(xxxxxxxx)+25color(white)(xxxxxx)+6)`

`color(white)(xxxxxxxxxxxxxxx)stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(rarr)`

`color(white)(xxxx)stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(red)(rarr)`

You can see that the net oxidation state of Mn has gone down from +21 to +6.

To effect this change will require 21 - 6 = 15 moles of electrons.

These will be provided by `sf(5X^(n+))` going up to `sf(5XO_3^-)`

`:.` we can say: `sf(25-5n=15)`

From which `sf(n=2)`

The 1/2 equation for the oxidation becomes:

`sf(X^(2+)+3H_2OrarrXO_3^(-)+6H^++3e" "color(red)((2)))`

To get the electrons to balance we multiply `sf(color(red)((1))` by 3 and `sf(color(red)((2))` by 5 then add`rArr`

`sf(5X^(2+)+cancel(15H_2O)+3MnO_4^(-)+cancel(24H^+)+cancel(15)erarr5XO_3^(-)+cancel(30H^(+))+cancel(15e)+3Mn^(2+)+cancel(12H_2O))`

This cancels down to:

`sf(5X^(2+)+3H_2O+3MnO_4^(-)rarr5XO_3^(-)+6H^(+)+3Mn^(2+))`

As you can see, charge and mass is conserved.