Some Basic Concepts of Chemistry 2 Question 41
41. A $1.0 \mathrm{~g}$ sample of $\mathrm{Fe}{2} \mathrm{O}{3}$ 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 $100.0 \mathrm{~mL}$. An aliquot of $25.0 \mathrm{~mL}$ of this solution requires for titration. Calculate the number of electrons taken up by the oxidant in the reaction of the above titration.
(1991, 4M)
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Solution:
- Mass of $\mathrm{Fe}{2} \mathrm{O}{3}=0.552 \mathrm{~g}$
millimol of $\mathrm{Fe}{2} \mathrm{O}{3}=\frac{0.552}{160} \times 1000=3.45$
During treatment with Zn-dust, all $\mathrm{Fe}^{3+}$ is reduced to $\mathrm{Fe}^{2+}$, hence
millimol of $\mathrm{Fe}^{2+}($ in $100 \mathrm{~mL})=3.45 \times 2=6.90$
$\Rightarrow$ In $25 \mathrm{~mL}$ aliquot, $\frac{6.90}{4}=1.725$ millimol $\mathrm{Fe}^{2+}$ ion.
Finally $\mathrm{Fe}^{2+}$ is oxidised to $\mathrm{Fe}^{3+}$, liberating one electron per $\mathrm{Fe}^{2+}$ ion. Therefore, total electrons taken up by oxidant.
$$ \begin{aligned} & =1.725 \times 10^{-3} \times 6.023 \times 10^{23} \ & =1.04 \times 10^{21} \end{aligned} $$
