1.0 g of `Fe_2O_3` solid of `55.2%` purity is dissolved in acid. The solution is reduced by heating with Zn dust. The resultant solution is cooled and made up to 100 mL. An aliquot of 25 " mL of " this solution requires 17 " mL of " 0.0167 M solution of an oxidant. Calculate the number of electrons taken up by oxidant in the above titration.

Explanation

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)` in 25 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