During the electrolysis of aqueous `CuSO_(4)` solutions using `Pt` elecrodes

Options

A. `Cu` is deposited at the anode
B. `O_(2)(g)` is liberated at cathode
C. `Cu` is deposited at the cathode
D. `H_(2)(g)` is liberated at the cathode

Answer

C

Explanation

The species we should consider for half reactions are `Cu^(2+)(aq.), SO_(4)^(2-)(aq.)`, and `H_(2)O`
Possible cathode half-reactions are
`Cu^(2+)(aq.)+2e^(-) rarr Cu(s), E^(@) = 0.34 V`
`2H_(2)O(l)+2e^(-) rarr H_(2)(g)+2OH^(-)(aq.): E^(@) = -0.83 V`
Because the copper electrode potential is much larger than the reduction potential of water, `Cu^(2+)` is reduced and `Cu` is deposited at the cathode.
`2SO_(4)^(2-)(aq.) rarr S_(2)O_(8)^(2-)(aq.)+2e^(-) E^(@) = 2.01 V`
`2H_(2)O(l) rarr O_(2)(g)+4H^(+)(aq.)+4e^(-), E^(@) = 1.23V`
Because the second electrode potential is lower, `H_(2)O` is oxidized and `O_(2)(g)` is liberated at the anode.
However, if electrolysis is carried out using copper electrodes then another possible anode reaction is
`Cu(s) rarr Cu^(2+)(aq.)+2e^(-) E^(@) = 0.34 V`
Thus, copper anode will dissolve or `Cu` anode will loose mass.