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1gm of $Fe_2O_3$ 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 up to 100ml,25ml of this solution requires 17ml of 0.0167M solution of an oxidant.Calculate the number of electrons taken up by the oxidant in the above reaction.


1 Answer

$Fe_2O_3+Zn\rightarrow Fe^{2+}\rightarrow 100ml$ solution
25ml sample=17ml of 0.0167M of an oxidant.
Let 'n' be the no of electrons taken up by oxidant.
Now Meq of $Fe^{2+}$ in25ml=Meq of oxidant
$\Rightarrow [0.0167\times n]\times 17$
Meq of $Fe^{2+}$ in 100ml=$[0.0167\times n\times 17]\times 4$
Also meq of $Fe_2O_3$=Meq of $Fe^{2+}$ in 100ml
$\Rightarrow 0.0167\times 68]n$
Meq of $Fe_2O_3=\large\frac{0.552}{E}$$\times 1000$
$\Rightarrow (0.0167\times 68)n$
$E_{\large Fe_2O_3}=?$
1 mole of $Fe_2O_3=2Fe^{3+};Fe^{3+}+1e^-\rightarrow Fe^{2+}$
$X=2$ for 1 mole of $Fe_2O_3$
$\Rightarrow E_{Fe_2O_3}=\large\frac{160}{2}$$=80$
$n=\big[\large\frac{0.552}{80}$$\times 1000\big]/[0.0167\times 68]=6$
$\Rightarrow$ Electrons taken by Oxidant=6
Hence (d) is the correct answer.
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