In the presence of sodium fluoride, #25cm^3# of 0.4moldm-3 aqueous iron(‖)sulphate requires #25cm^3# of 0.10moldm-3 aqueous potassium manganate(Ⅶ)for complete reaction.What is the final oxidation state of manganese?

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Answer 1 · 2018-03-28T18:37:42

+3

Manganate(VII) is an oxidising agent and oxidises iron(II) to iron(III):

#sf(Fe^(2+)rarrFe^(3+)+e)#

The number of moles of Fe(II) is given by:

#sf(n_(Fe^(2+))=cxxv=0.4xx25/1000=0.01)#

The number of moles of manganate(VII) is given by:

#sf(n_(MnO_4^-)=cxxv=0.1xx25/1000=0.0025)#

#:.# 0.0025 mol #sf(MnO_4^-) -=#0.01 mol #sf(Fe^(2+))#

#:.# 1 mol #sf(MnO_4^-)-=# #sf(0.01/0.0025=4)# mol #sf(Fe^(2+)#

This means that:

#sf(4Fe^(2+)rarr4Fe^(3+)+4e)#

Since the oxidation number of Mn in #sf(MnO_4^(-))# is +7 this means that adding 4 electrons must reduce its oxidation number to +3:

#sf(MnO_4^(-)+8H^(+)+4erarrMn^(3+)+4H_2O)#

In reality the usual product in the titration is #sf(Mn^(2+))#.

Perhaps the NaF has something to do with this?