An acidic solution of Cu 2+ alt containing 0.4 g of Cu 2+ s 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 ampere. Calculate the volume of gases evolved at NTP during the entire electrolysis.
Text Solution
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Sol. For first part of electrolysis
At anode : 2H2O ⎯→ 4H + + O 2 + 4e – At cathode : Cu
2+ + 2e – ⎯→ Cu
∴ equivalents of O 2 formed = equivalent of Cu
= 
= 12.58 × 10 –3 For second part of electrolysis :
Since Cu
2+ ions are discharged completely and thus further passage of current through solution will lead to following changes.
At anode : 2H 2 O ⎯→ 4H + + O 2 + 4e – At cathode : 2H 2 O + 2e
– ⎯→ H 2 + 2OH – Thus, equivalent of H 2 = equivalent of
O 2 = 
= 
= 5.22 × 10
–3 ∴ Total equivalent of O 2 = equivalent of O 2 for Ist part + equivalent of O 2 for II part electrolysis
= 5.22 × 10
–3 + 12.58 × 10 –3 = 17.8 × 10
–3 ∴ 4 equivalent of O 2 at NTP = 22.4 L
∴ 17.8 × 10
–3 equivalent of O 2 at NTP
= 
= 99.68 mL
Now, equivalent of H 2 = 5.22 × 10 –3 2 Equivalent of H 2 at NTP = 22.4 L
∴ 5.22 × 10
–3 equivalent of H 2 at NTP
= 
= 58.46 mL
∴ Total volume of O 2 + H 2 = 99.68 + 58.46
= 158.14 mL
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