What is the volume of `O_(2` liberated at anode at `STP` in the electrolysis of `CdSO_(4)` solution when a current of `2 A` is passed for `8 m i n ?`

Explanation

Aqueous `CdSO_(4)overset(el ectrolysis)rarr Cd^(2+)+SO_(4)^(2-)`
Number of Faradays `=(Ixxts)/(96500C)=(2Axx8xx60s)/(96500C)`
`=9.0099~~0.001 F`
In aqueous solution, oxidation of `H_(2)O` takes place than that of `SO_(4)^(2-)` ions, since oxidation potential of `H_(2)Ogt` oxidation potential of `SO_(4)^(2-)`. So oxidation of `H_(2)O` at anode occurs.
`2H_(2)O rarr O_(2)+4H^(o+)+4e^(-)`
Frist method
`4e^(-)=4F-= 1 mol O_(2)=22.4 L `at `STP`
`:. 0.001F=(22.4Lxx0.001F)/(4F)`
`=0.056L of O_(2)`
Second method
`1F=1` equivalent of `O_(2)=(22.4L)/(4)O_(2)`
`[{:(Equivalent of O_(2)),(=(Volume of 1 mol of a gas)/(n fact o r)),(=(22.4)/(4)=5.6LO_(2)),(n fact o r f o r O_(2)=4) :}]`
`2O^(2-) rarr O_(2)+4e^(-)`
`:. 1 F =5.6 L `or `O_(2)`at `STP`
`0.001 F=5.6 xx 0.001 =0.056 L O_(2)`