In the electrolysis of `7.2L` aqueous solution of `CuSO_(4)`, a current of `9.65A` passed for 2 hours.
`a.` Calculate the weight of `Cu` deposited at cathode.
`b.` Calculate the volume of `O_(2)` produced at anode at `27^(@)C` and `1 atm` pressure.

Explanation

First method
`(CuSO_(4) rarr Cu^(2+)+SO_(4)^(2-))`
At cathode`:` `Cu^(2+)+2e^(-) rarr Cu(s)`
At anode `:` `2H_(2)O rarr O_(2)+4H^(o+)+4e^(-)`
`(` Since the oxidation potential of `H_(2)Ogt` oxidation potential of `SO_(4)^(2-)`, so oxidation of `H_(2)O` occurs at anode and `SO_(4)^(2-)` and `H^(o+)` ions remain in the solution to give an acidic solution `).`
Number of faradays `=(Ixxt)/(96500C)=(9.65A xx 2 xx 3600 s )/(96500C mol^(-1))`
`a.` Now from cathode `: (Cu^(2+)+2e^(-)rarr Cu)`
`2e^(-) =2F=1` mol of `Cu=63 g ` of `Cu`
`0.72F=(63g xx 0.72F)/(2)=22.68g`
`b.` Now from anode `: (2H_(2)Orarr O_(2)+4H^(o+)+4e^(-))`.
`4e^(-)=4-=1 mo l `of `o_(2)`
`0.72F=(1xx0.72)/(4)=0.18 mol`
Now using gas equation `: PV=nRT`
`V=(nRT)/(P)=(0.18xx0.082xx300)/(1)=4.428L` of `O_(2)`
`c.` Now from anode` (2H_(2)O rarr O_(2)+4H^(o+)+4e^(-))`
`4e^(-)=4F=4 mol `of `H^(o+)`
`[H^(o+)]=("Mol")/("Volume in L")=(0.72 mol)/(7.2L)=10^(-1)M`
`:. pH=-log [10^(-1)]=1`
`A.` At cathode`:`
`1F=1` equivalent of `Cu=(63)/(2) `of `Cu`
`0.7F=(63)/(2)xx0.72=22.68g` of `Cu` deposited
`b.` At anode `:`
`1F=1` equivalent of `O_(2)`
`0.72F=0.72` equivalent of `O_(2)=(0.72)/(4)=0.18 mol ` of `O_(2)`
`[{:(Since mol=("Equivalent")/("n factor")`or `("Equivalent")/("charge"),=,(0.72)/(4)),(2O^(-2) rarrO_(2)+4e^(-1)(n fact o r=4),=,):}]`
Now for volume of `O_(2)`, process as in first method.
`c." "1F=1` equivalent of `H^(o+)` ions.
`0.72F=0.72` equivalent of `H^(o+)=0.72mol `of `H^(o+)`
`[` Since` n` factor for `H^(o+)=1]`
`:. [H^(o+)]=(0.72 mol)/(7.2L)=10^(-1)M`
`:. pH=1`