Question

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.

Answer 1

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 $NTP =70.56+29.23=99.79 \,mL$
$H _{2}(g)$ at $NTP =58.46 \,mL$