#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.
