Eq of Cu = Eq of O2
63.54300×10−3×2=nO2×4
2.36×10−3=nO2
When current is further passed
nO2×4=96500×1000600×28×60
nO2=2.611×10−3
Total O2 released
=[10−3×(2.36+2.611)]×22400 ml
=111.35 ml
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)Ered o=+0.34 V
O2( g)+4H++4e−→2H2OEred o=+1.23 V
Molar mass of Cu=63.54 g mol−1
Molar mass of O2=32 g mol−1
Faraday Constant =96500Cmol−1
Molar volume at STP=22.4 L]
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