Question

An acidic solution of $C{u}^{2+}$ salt containing $0.4g$ of $C{u}^{2+}$ 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 $1.2$ amp. Find the volume of gases evolved at anode and Cathode at NTP during the entire electrolysis.

• A. $78.20ml,99.78ml$
• B. $58.48ml,78.20ml$
• C. $98.56ml,58.24ml$
• D. $78.20ml,58.48ml$

Answer 1

The correct option is C $98.56ml,58.24ml$

Solution:- (C) $98.56mL,58.24mL$

Assuming ${Cu}^{+2}$ salt to be $CuS{O}_4$, the reactions occuring at the electrodes would be-

At anode:-

$H_{2}O⟶2H^{+}+\frac{1}{2}{O}_2+2e^{-}$

At cathode:-

${Cu}^{+2}+2e^{-}⟶Cu$

Equivalent weight of ${Cu}^{+2}=31.5g$

$0.4g$ of ${Cu}^{+2}=\frac{0.4}{31.5}=0.0127g$ equivalent

At the same time the oxygen deposited at anode.

Equivalent weight of oxygen $=8gm$

Mass of oxygen deposited $=0.0127\times \frac{8}{32}=0.0031\text{mol}$

After the complete deposition of copper, the reactions would be-

At anode:-

$H_{2}O⟶2H^{+}+\frac{1}{2}{O}_2+2e^{-}$

At cathode:-

$2H_{2}O+2e^{-}⟶H_2+2{OH}^{-}$

Given:-

$I=1.2A$

$t=7\text{min}=7\times 60=420s$

Amount of charge passed $=I\times t=1.2\times 7\times 60=504C$

Therefore,

Amount of oxygen liberated $=\frac{1}{96500}\times 504=0.00523g$ equivalent $=\frac{8}{32}\times 0.00523=0.0013\text{mol}$

Amount of hydrogen liberated $=0.00523g$ equivalent $=\frac{1}{2}\times 0.00523=0.0026\text{mol}$

Now,

Gas evolved at anode $=O_2$

Total no. of moles of $O_2$ evolved $=0.0031+0.0013=0.0044\text{mol}$

$\therefore$ Volume of gas evolved at anode $=0.00447\times 22400=98.56mL$

Gas evolved at cathode $=H_2$

Total no. of moles of $H_2$ evolved $=0.0026\text{mol}$

$\therefore$ Volume of gas evolved at cathode $=0.0026\times 22400=58.24mL$

Hence the volume of gases evolved at anode and cathode at NTP during the entire electrolysis $98.56mL$ and $58.24mL$ respectively.