A 1.0 g sample of Fe_(2)O_(3) solid of 55.2% purity is dissolved in acid and reduced by heating the solution with zinc dust. The resultant solution is cooled and made up to 100.0 mL. An aliquot of 25.0 mL of this solution requires 17.0 mL of 0.0167 M solution of an oxidant for titration. Calculate the number of electrons taken up by the oxidant in the reaction of the above titration.

A 1.0 g sample of Fe_(2)O_(3) solid of 55.2% purity is dissolved in ac

1.	A 1.0 g sample of Fe_(2)O_(3) solid of 55.2% purity is dissolved in acid and reduced by heating the solution with zinc dust. The resultant solution is cooled and made up to 100.0 mL. An aliquot of 25.0 mL of this solution requires 17.0 mL of 0.0167 M solution of an oxidant for titration. Calculate the number of electrons taken up by the oxidant in the reaction of the above titration.
Answer» Solution :The redox changes are : For reduction of `Fe_(2)O_(3)` by zinc dust `Fe_(2)^(6+) + 2e^(-) rarr 2 FE^(2+)` `Fe^(2+) rarr Fe^(3+) + e^(-)` oxidant + ne rarr reductant Meq. of `Fe_(2)O_(3)` in25 mL = Meq. of `Fe^(2+)` formed = Meq. of oxidant used to oxidize `Fe^(2+)` again Meq. of `Fe_(2)O_(3)` in 25 mL = Meq. of oxidant = `17 XX 0.0167 xx n` Where n is the number of ELECTRONS gained by 1 molecule of oxidant Meq. of `Fe_(2)O_(3)`in 100 mL = `17 xx 0.0167 xx n xx 100/25` `:. 1 xx 55.2 xx 1000/(100 xx M/2) = 17 xx 0.0167 xx n xx 4` Molecule wt. of `Fe_(2)O_(3)` = 160 `n = 1 xx 55.2 xx 2 xx 1000/(100 xx 160 xx 17 xx 0.0167 xx 4) = 6` Hence, number of electrons gained by one molecule of oxidant = 6