An acidic acid of $C u^{2 +}$ salt containing 0.4 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 amp. Calculate the volume of $O_{2}$ gases evolved at NTP during the entire electrolysis., in ml. (write the value to the nearest integer).

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Solution

The number of moles of copper electrolyzed $= \frac{0.4 g}{63.5 g / m o l} = 6.3 \times 10^{- 3} m o l$
2 moles of electrons corresponds to 1 mole of copper.
Hence, $6.3 \times 10^{- 3} m o l$ moles of copper corresponds to $2 \times 6.3 \times 10^{- 3} m o l = 12.6 \times 10^{- 3} m o l$ of electrons.
The current at 1.2 amp is passed for additional 7 minutes.
Hence, the additional moles of electrons passed $= \frac{1.2 \times 7 \times 60}{96500} = 5.22 \times 10^{- 3}$ .
Hence, total number of moles of electrons passed $12.6 \times 10^{- 3} m o l + 5.22 \times 10^{- 3} m o l = 17.82 \times 10^{- 3} m o l$ .
The volume of oxygen liberated at NTP $\frac{17.82 \times 10^{- 3} m o l}{4} \times 22400 = 99.68 m l$

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