1.

How does aqueous and acidified `CuSO_(4)` dissociate on the passage of current ? Also write the electrode reactions at the respective electrodes

Answer» Aqueous acidified copper sulphate dissociates as shown below.
`CuSO_(4) rarr Cu^(2+) + SO_(4)^(-2) " , " H_(2)O rarr H^(+) + OH^(-) " , " H_(2) SO_(4) rarr 2H^(+) + SO_(4)^(2-)`
Reaction at cathode: Copper gets discharged at the cathode because of lower discharge potential than that of hydrogen.
`Cu^(2+) + 2e^(-) rarr Cu`
Reaction at anode
(i) If anode is platinum or graphite (inactive anode) `OH^(-)` is discharged at the anode because its discharge potential is lower than that of `SO_(4)^(2-)` ion. `4OH^(-) rarr 2H_(2)O + O_(2) + 4e^(-)`
(ii) If the anode is copper itself (active anode)
`Cu - 2e^(-) rarr Cu^(2+)`
No product is formed at anode.


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