Question

1gm of $F{e}_2{O}_3$ solid of 55.2% purity is dissolved in acid and reduced by heating the solution with Zn 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. Calculate the number of electrons taken up by oxidant in the above reaction.

Answer 1

$F{e}_2{O}_3+Zn⟶F{e}^{+2}⟶100ml$ solution
25 ml sample $\equiv$ 17 ml of 0.0167 M of oxidant.

Let 'n' be the number of electrons taken up by oxidant. Now meq. of $F{e}^{+2}$ in 25 ml = meq of oxidant,
$=[0.0167\times n]\times 17$

meq. of $F{e}^{+2}$ in 100 ml $=[0.0167\times n\times 17]\times 4$

Also meq of $F{e}_2{O}_3=$ meq of $F{e}^{+2}$ is 100 ml

meq of $F{e}_2{O}_3=\frac{0.552}{E}\times 1000=(0.0167\times 68)n$

$E_{F{e}_2{O}_3}=?$

1 mol of $F{e}_2{O}_3\equiv 2F{e}^{+3}$;

$F{e}^{+3}+1e^{-}⟶F{e}^{+2}$

$\Rightarrow x=2$ for 1 mol of $F{e}_2{O}_3$

$E_{F{e}_2{O}_3}=\frac{160}{2}=80$

$n=(\frac{0.552}{80}\times 1000)÷(0.0167\times 68)=6$

Electrons taken by oxidant = 6