A `1 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 upto `100 mL`. An aliquot of `25 mL` of this solution requires `17 mL` of `0.0167M` solution of an oxidant for titration. Calculate no.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
`{:(2e+,Fe_(2)^(3)rarr,2Fe^(2+),,),( ,Fe^(2+)rarr,Fe^(3+)+e,,),("Oxidant","ne"rarr,"Reductant",,):}`
Meq.of `Fe_(2)O_(3)` in `25 mL=` Meq.of oxidant
`=17xx0.0167xxn`
where `n` is no. of electrons gained by `1` molecule of oxidant.
`therefore` Meq.of `Fe_(2)O_(3)` in `100mL=17xx0.0167xxnxx(100)/(25)`
`:. (1xx55.2xx1000)/(100xxM//2)=17xx0.0167xxnxx4`
`( because M.wt."of" Fe_(2)O_(3)=160)`
`:.n =(1xx55.2xx2xx1000)/(100xx160xx17xx0.0167xx4)=6`
`:.` No. of electrons gained by one molecule of oxidant `=6`