Question

An acidified solution of $C{u}^{2+}$ salt containing 0.636 g of $C{u}^{2+}$ is electrolysed until all the copper is deposited. The electrolysis is continued for another 16 min 5 sec with the volume of the solution kept at 100 mL and current at 1.2 amps. Calculate the volume of gases evolved during the entire electrolysis. (atomic wt of Cu = 63.6)

Answer 1

When $C{u}^{2+}$ is present, the reactions at the cathode and anode are as follows.
Cathode
$C{u}^{2+}+2e^{-}\to Cu$
Anode:
$2H_{2}O\left(l\right)\to O_2\left(g\right)+4H^{+}\left(aq\right)+4e^{-}$
After all of the $C{u}^{2+}$ is deposited as Cu, the reaction at the cathode and anode are
Cathode
$2H_{2}O+2e^{-}\to H_2+2O{H}^{-}$
Anode:
$2H_{2}O\left(l\right)\to O_2\left(g\right)+4H^{+}\left(aq\right)+4e^{-}$
Step 3 : Equivalents of Cu deposited $=\frac{\text{Mass of copper}}{\text{Eq. wt of copper}}=\frac{0.636}{\frac{63.6}{2}}$ g
This must equal the equivalents of $O_2$ evolved.
Weight of $O_2$ evolved = equivalents of $O_2\times$ eq. wt. of $O_2$
$=\frac{0.636\times 2}{63.6}\times \frac{32}{4}$
Volume of $O_2$ evolved $=\frac{0.636\times 2}{63.6}\times \frac{32}{4}\times \frac{22.4}{32}=0.112$ litres
In the extra 16 min and 5 sec of electrolyses, the total charge passed
$=Q=It=1.2\times (16\times 60+5)=1158C$
96500 C of charge will release 1 gram eqvt of $O_2$& $H_2$
$\therefore 1158C$ will release $\frac{1158}{96500}=12\times {10}^{-3}$ Gram equivalents
Mass of $H_2$ evolved $=12\times {10}^{-3}\times \frac{2}{2}=12\times {10}^{-3}$ g
Volume of $H_2$ evolved $=12\times {10}^{-3}\times \frac{22.4}{2}=0.1344$ litres
Volume of $O_2$ evolved $=12\times {10}^{-3}\times \frac{22.4}{4}=0.0672$ litres
Total volume of gas evolved $=0.112+0.1344+0.0672=0.3136l$