Electrochemistry - Result Question 95
43. An acidic solution of $Cu^{2+}$ salt containing $0.4 g$ of $Cu^{2+}$ is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of solution kept at $100 $ $mL$ and the current at $1.2 A$. Calculate the volume of gases evolved at NTP during the entire electrolysis.
$(1989,5 M)$
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Solution:
- If the salt is $CuSO _4$
During deposition of $Cu$ at cathode, $O _2(g)$ will evolve at anode gram-equivalent of $Cu$ deposited $=\frac{0.4 \times 2}{63.5}=0.0126$
Volume of $O _2$ liberated at NTP at anode
$=0.0126 \times 5600 mL=70.56 $ $mL$
In the next $7 min, H _2$ at cathode and $O _2$ at anode would be produced.
Faraday’s passed $=\frac{1.2 \times 7 \times 60}{96500}=5.22 \times 10^{-3}$
$\Rightarrow$ Volume of $H _2($ at NTP $)=5.22 \times 10^{-3} \times 11200 $ $mL$
$=58.46 $ $mL$
Volume of $O _2($ at NTP $)=5.22 \times 10^{-3} \times 5600 mL=29.23 $ $mL$
Therefore, $O _2(g)$ at $NTP=70.56+29.23=99.79 $ $mL$
$ H _2(g) \text { at } NTP=58.46 $ $mL $
BETA
