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a current of 0.75 A is passed through acidic solution of CuSO4 for 10 minutes. The volume of oxygen at anode (at STP) will be

swetha , 6 Years ago
Grade 12
anser 1 Answers
vanshith

Last Activity: 6 Years ago

2cu+2e- ---->Cu
4OH-+4e------>2H2O+O2  See Q=0.75×10×60=450C
Now 4×96500c deposite 32g of Oxygen 450c will deposite=32×450/4×96500=0.0373g O2 at STP 32g of oxygen has volume 22.4L thus 0.373g will have 22.4×0.373/32=0.261L=26.1ml
 

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