Electrochemistry - Result Question 22
####23. $Al_2 O_3$ is reduced by electrolysis at low potentials and high currents. If $4.0 \times 10^{4}$ amperes of current is passed through molten $Al_2 O_3$ for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency and At. mass of $Al=27 g mol^{-1}$ )
[2009]
(a) $8.1 \times 10^{4} g$
(c) $1.3 \times 10^{4} g$
(b) $2.4 \times 10^{5} g$
(d) $9.0 \times 10^{3} g$
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Solution:
- (a) As $Q=i \times t$
$ \begin{aligned} \therefore Q & =4.0 \times 10^{4} \times 6 \times 60 \times 60 C \\ & =8.64 \times 10^{8} C \end{aligned} $
Now since $96500 C$ liberates $9 g$ of Al
$ \begin{aligned} & 8.64 \times 10^{8} C \text{ liberates } \frac{9}{96500} \times 8.64 \times 10^{8} gAl \\ & =8.1 \times 10^{4} g \text{ of Al } \end{aligned} $
24 .
(d) No. of gram equivalent of $H^{+}=$No. of eq. of $Al^{3+}=\frac{4.5}{9}=0.5$
Eq. wt of $Al^{3+}=\frac{27}{3}=9$
No. of gm eq. of $H^{+}=$no. of mole of $H^{+}$
Hence, Mass of $H^{+}=0.5 \times 1 g=0.5 g$
We know that, $2 g H_2$ at $STP=22.4 L$
$ \therefore 0.5 g H_2 \text{ at } STP=\frac{22.4}{2}, 0.5=5.6 L $
