An aqueous solution of `CuSO_(4)` is electrolyzed using `Pt` electrodes in one case and `Cu` electrodes in another case. What are the products of electrolysis in both the cases ?

Explanation

In case of platinum electrodes, the products are Cu on cathode and `O_(2)` gas on anode. This because out of `Cu^(2+) and OH^(+)` ions (produced from `H_(2)O`), `Cu^(2+)` ions have lower discharge potential. Out of `SO_(4)^(2-)` and `OH^(-)` ions, `OH^(-)` ions have lower discharge potential `(OH^(-)toOH+e^(-),CutoCu^(2+)+e^(-))`
In case of copper electrodes, product at cathode is the same, viz, Cu but at anode, out of the three possible reactions, viz., `OH^(-)toOH+e,SO_(4)^(2-)toSO_(4)+2e^-,CutoCu^(2+)+2e^(-)`
Cu is more easiliy oxidized than `OH^(-) and SO_(4)^(2-)` ions. hence, anode of Cu dissolves into the solution to form `Cu^(2+)` ions.