What will happen during the electrolysis of aqueous solution of `CuSO_(4)` by using platinum electrodes ?

Options

A. Copper will deposit at cathode.
B. Copper will deposit at anode.
C. Oxygen will be released at anode.
D. Copper will dissolve at anode.

Answer

C

Explanation

`CuSO_(4)hArr Cu^(2+)+SO_(4)^(2-)`
`H_(2)OHarrH^(+)+OH^(-)`
At cathode,
`Cu^(2+)+2e^(-) to Cu," "E_(Cell)^(Theta)=0.34V`
`H^(+)+e^(-) to (1)/(2)H_(2)," "E_(Cell)^(Theta)=0.00V`
This reaction will take place due to higher reduction potential.
At anode,
`2SO_(4)^(2-)+2e^(-)toS_(2)O_(5)^(2-)," "E_(Cell)^(Theta)=1.96V`
`2H_(2)O to O_(2)+4H^(+)+4e^(-)," "E_(Cell)^(Theta)=1.23V`
The reaction with lower value of `E^(@)` will be preferred at anode, hence `O_(2)` is released at anode.