Question

25.0 cm${}^3$ of 0.0497 M solution of $N{H}_{2}OH$ was boiled with excess of ferric alum solutuion in $H_{2}S{O}_4$. The Fe(II) formed rquired 24.7 cm${}^3$ of 0.1007 $N{K}_{2}C{r}_2{)}_7$ solution. Identify the product formed by oxidation of $N{H}_{2}OH$.

• A. $N_2{O}_4$
• B. $N_{2}O$
• C. $N_2{O}_2$
• D. $NO$

Answer 1

The correct option is B $N_{2}O$

$F{e}^{3+}$ is reduced to $F{e}^{2+}$ which in turn is oxidised to $F{e}^{3+}$ by $C{r}_2{O}_7^{2-}$ in acidic medium.
$N{H}_{2}OH+F{e}^{3+}⟶F{e}^{2+}+\left(oxidationproduct\right)$
$6F{e}^{2+}+C{r}_2{O}_7^{2-}+14H^{+}⟶6F{e}^{3+}+2C{r}^{3+}+7H_{2}O$
Equivalent of $C{r}_2{O}_7^{2-}=\frac{24.7\times 0.1007}{1000}$
$=0.0025$
Equivalent of $F{e}^{3+}$ or $F{e}^{2+}=0.0025$
Equivalent of $N{H}_{2}OH=0.0025$
Moles of $N{H}_{2}OH$ (taken) $=\frac{25\times 0.0497}{1000}$
$=0.00125$
Moles $\left(N{H}_{2}OH\right)=$ Change in oxidation number $\times$ Equivalent of $N{H}_{2}OH$
$\therefore$ Change in oxidiation number of $N{H}_{2}OH=\frac{0.0025}{0.00125}$
$=2$
Oxidation number of N is $N{H}_{2}OH=-1$
Oxidation number of N in oxidised product $=-1+2=+1$
so product is $N_{2}O$
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