In the electrolysis of a `40L CuSO_(4)` solution, there are two possible reactions at anode `:`
`i. Cu rarr Cu^(2+)+2e^(-)`
`ii. 2H_(2)O rarr 4H^(o+)+4e^(-)+O_(2)`
A current of `1.07 A ` is passed for 2 hours. The loss in the mass of `Cu` at anode was `1.27 g`.
`(` Atomic weight of `Cu =63.5g mol ^(1))`.
Which of the following statement`(s)` is `//` are correct ?

Options

A. `0.08 mol` of electrons are passed through the solution during entire electrolysis
B. `224mL` of `O_(2)(g)` is liberated at `STP` at anode.
C. Fraction of current in the production of `Cu^(2+)` ions `=0.5.`
D. `pH` drops to 3

Answer

a,b,c,d

Explanation

`a.` Statement `(a)` is correct
Amound of `Cu` converted to `Cu^(2+)` ions `=(1.27g)/(63.5g mol^(-1))`
`[2F-=1 mol `of `Cu^(2+)]`
Amount of electrons involved `=2xx0.02=0.04 mol`.
Amount of electrons actually passed through solution
`=(1.07xx2xx60xx60)/(96500)`
`0.08 mol`
`b.` Statement `(b)` is correct
Amount of electrons involved is converting `H_(2)O` to `O_(2)`
`=0.08-0.04`
`=0.04 mol`
Amount of `O_(2)` liberated at anode `=(1)/(4)xx0.04`
`=0.01mol`
`:.` Volume of `O_(2)` produced at `STP` at anode
`=0.01xx22400mL=224 mL`
`c.` Statement `(c)` is correct
Fraction of current used in the production of `Cu^(2+)` ions `=(0.04)/(.08)=0.5`
`d.` Statement `(d)` is correct `:`
`:gt `Mol of `H^(o+)` ions produced at andoe `=0.04 mol`
`:. [H^(o+)]_(anode)=(0.04 mol)/(40L)=10^(-3)mol L^(-1)`
`:. pH=-log(10^(-3))=3`