Electricity is passed through an acidic solution of Cu2+ till all the Cu2+ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is mL. (Nearest integer) [Given: Cu2+(aq) + 2e- Cu(s); E red = +0.34 V O2(g) + 4H+ + 4e- 2H2O; E red = +1.23 V Molar mass of Cu = 63.54 g mol-1 Molar mass of O2 = 32 g mol-1 Faraday Constant = 96500 C mol-1 Molar volume at STP = 22.4 L]
Answer
111
Solution
At the anode water is oxidised throughout: 2H2O O2 + 4H+ + 4e-, i.e. 1 mol O2 per 4 mol electrons.
Step 1 (deposition of Cu):
nCu = 0.30063.54 = 4.721 10-3 mol
electrons = 2 4.721 10-3 = 9.443 10-3 mol
nO2 = 9.443 10-34 = 2.361 10-3 mol
Step 2 (after Cu2+ is exhausted, H2 at cathode, O2 still at anode):
Q = It = 0.600 28 60 = 1008 C
electrons = 100896500 = 1.0446 10-2 mol
nO2 = 1.0446 10-24 = 2.611 10-3 mol
Total nO2 = 2.361 10-3 + 2.611 10-3 = 4.972 10-3 mol
VO2 = 4.972 10-3 22400 = 111.4 mL 111 mL
Answer: 111