In the presence of sodium fluoride, $25 c m^{3}$ $25cm^3$ of 0.4moldm-3 aqueous iron(‖)sulphate requires $25 c m^{3}$ $25cm^3$ 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?

In the presence of sodium fluoride, 25cm^325cm325cm^3 of 0.4moldm-3 aqueous iron(‖)sulphate requires 25cm^325cm325cm^3 of 0.10moldm-3 aqueous potassium manganate(Ⅶ)for complete reaction.What is the final oxidation state of manganese?