Electrochemistry 1 Question 44
43. An acidic solution of $\mathrm{Cu}^{2+}$ salt containing $0.4 \mathrm{~g}$ of $\mathrm{Cu}^{2+}$ is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of solution kept at $100 \mathrm{~mL}$ and the current at $1.2 \mathrm{~A}$. Calculate the volume of gases evolved at NTP during the entire electrolysis.
$(1989,5 \mathrm{M})$
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Answer:
Correct Answer: 43. $(1.172 \mathrm{M})$
Solution:
- If the salt is $\mathrm{CuSO}_{4}$
During deposition of $\mathrm{Cu}$ at cathode, $\mathrm{O}_{2}(\mathrm{~g})$ will evolve at anode gram-equivalent of $\mathrm{Cu}$ deposited $=\frac{0.4 \times 2}{63.5}=0.0126$
Volume of $\mathrm{O}_{2}$ liberated at NTP at anode
$=0.0126 \times 5600 \mathrm{~mL}=70.56 \mathrm{~mL}$
In the next $7 \mathrm{~min}, \mathrm{H}{2}$ at cathode and $\mathrm{O}{2}$ at anode would be produced.
Faraday’s passed $=\frac{1.2 \times 7 \times 60}{96500}=5.22 \times 10^{-3}$
$\Rightarrow$ Volume of $\mathrm{H}_{2}($ at NTP $)=5.22 \times 10^{-3} \times 11200 \mathrm{~mL}$
$=58.46 \mathrm{~mL}$
Volume of $\mathrm{O}_{2}($ at NTP $)=5.22 \times 10^{-3} \times 5600 \mathrm{~mL}=29.23 \mathrm{~mL}$
Therefore, $\mathrm{O}_{2}(g)$ at $\mathrm{NTP}=70.56+29.23=99.79 \mathrm{~mL}$
$$ \mathrm{H}_{2}(g) \text { at } \mathrm{NTP}=58.46 \mathrm{~mL} $$
