Question

An acidic solution of $C{u}^{2+}$ salts containing 0.4 g of $C{u}^{2+}$ is electrolyzed 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. The volume (in mL) of gases evolved at STP during the entire electrolysis is : (write your answer to nearest integer)

Answer 1

For first part of electrolysis :

At anode : $2H_{2}O\to 4H^{\oplus }+O_2+4e^{-}$

At cathode :

$C{u}^{2+}+2e^{-}\to Cu$

$\therefore$ Equivalent of $O_2$ formed = equivalent of Cu =

$\frac{0.4\times 2}{63.6}=12.58\times {10}^{-3}$

For second part of electrolysis :

Since $C{u}^{2+}$ ions are discharged completely and thus further passage of current through solution will lead to following changes:

At anode :

$2H_{2}O\to 4H^{\oplus }+O_2+4e^{-}$

At cathode :
$2H_{2}O+2e^{-}\to H_2+2\stackrel{\ominus }{O}H$

Thus, equivalent of $H_2$ =
equivalent of $O_2=\frac{It}{96500}$

=$\frac{1.2\times 7\times 60}{96500}=5.22\times {10}^{-3}$

$\therefore$ Total equivalent of $O_2$ = Equivalent of $O_2$ for first part + Equivalent of $O_2$ for second part of electrolysis.

= $5.22\times {10}^{-3}+12.58\times {10}^{-3}$

= $17.8\times {10}^{-3}$

$\therefore$ 4 equivalent of $O_2$ at STP = 22.4 L

$\therefore 17.8\times {10}^{-3}$ equivalent of $O_2$ at STP.

$=\frac{22.4\times 17.8\times {10}^{-3}}{4}$ = 99.68 mL

Now, equivalent of $H_2=5.22\times {10}^{-3}$

$\because 17.8\times {10}^{-3}$ equivalent of $O_2$ at STP

$=\frac{22.1\times 17.8\times {10}^{-}3}{4}=99.68mL$

$\therefore$ Total volume of $O_2+H_2$ = 99.68 + 58.46 = $158.14mL$