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During the electrolysis of aqueous `CuSO_(4)` solutions using `Pt` elecrodesA. `Cu` is deposited at the anodeB. `O_(2)(g)` is liberated at cathodeC. `Cu` is deposited at the cathodeD. `H_(2)(g)` is liberated at the cathode |
Answer» Correct Answer - C 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. |
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