$Cr _{2} O _{7}^{2-}+14 H ^{+}+6 e ^{-} \rightarrow 2 Cr ^{3+}+7 H _{2} O$
The amount of $Cr ^{3+}$ obtained was $0.104\, g$. The
efficiency of the process(in\%) is
(Take : $F =96000\, C$, At. mass of chromium $=52$ )
Using stoichiometry; theoritically
$\frac{ n _{ e ^{\ominus}} used }{6}=$$\frac{ n _{ cr ^{+3}} \text {produced }}{2}$
$\Rightarrow n _{ cr ^{+3}}$ produced $=\frac{2}{6} \times \frac{8 \times 60 \times 2}{96000}$
$=\frac{0.02}{6}$
$\Rightarrow wt _{ cr ^{+3}}$ theoritically produced
$=\left(\frac{0.02}{6} \times 52\right) g$
$\Rightarrow \%$ efficiency $=\frac{0.104 g }{\left(\frac{0.02 \times 52}{6}\right) g } \times 100$
$=60 \%$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(A)$ Dinitrogen is a diatomic gas which acts like an inert gas at room temperature.
$(B)$ The common oxidation states of these elements are $-3,+3$ and +5 .
$(C)$ Nitrogen has unique ability to form $\mathrm{p} \pi-\mathrm{p} \pi$ multiple bonds.
$(D)$ The stability of +$5$ oxidation states increases down the group.
$(E)$ Nitrogen shows a maximum covalency of $6$.
Choose the correct answer from the options given below.
$[P]\xrightarrow{\begin{subarray}{l}
(i)\,NAN{O_2}/HCl,\,0 - {5^o}C \\
(ii)\,\beta - napthol/NaOH
\end{subarray} }Colored\,\,Solid$
$[P]\xrightarrow{{B{r_2}/{H_2}O}}{C_7}{H_6}NB{r_3}$
The compound $[P]$ is