Question

A $1$ g sample of $F{e}_2{O}_3$ 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 $0.0167$ M solution of an oxidant for titration. Calculate the number of electrons taken up by oxidant in the above titration.

Answer 1

The redox changes are as follows:
For $F{e}_2{O}_3,$ $2\left(F{e}^{3+}\right)+2e^{-}⟶2F{e}^{2+}$ (by zinc dust)
Again, $F{e}^{2+}⟶F{e}^{3+}+e^{-}$ (by oxidant)
$Oxidant+n{e}^{-}⟶Reductant$, where $n$ is the number of electrons gained by 1 molecule of oxidant.
$\therefore$ Meq. of $F{e}_2{O}_3$ in 25 mL $=$ Meq. of $F{e}^{3+}$ in $F{e}_2{O}_3$ $=$ Meq. of $F{e}^{2+}$ formed $=$ Meq. of oxidant used for $F{e}^{2+}$
or, Meq. of $F{e}_2{O}_3$ in 25 mL $=$ Meq. of oxidant $=17\times 0.0167\times n$
$\therefore$ Meq. of $F{e}_2{O}_3$ in 100 mL $=17\times 0.0167\times n\times \left(\frac{100}{25}\right)$
or, $\frac{1\times 55.2\times 1000}{10\times \left(\frac{160}{2}\right)}=17\times 0.0167\times n\times \frac{100}{25}$ $(\because MolarmassofF{e}_2{O}_3=160)$
$\therefore n=6$ (integer)
Therefore, the number of electrons gained by one molecule of oxidant $=6$.