Predict the products of electrolysis in each of the following:
(i) An aqueous solution of `AgNO_3` with silver electrodes.
(ii) An aqueous solution of `AgNO_3` with platinum electrodes.
(iii) A dilute solution of `H_2SO_4` with platinum electrodes.
(iv) An aqueous solution of `CuCl_(2)` , with platinum electrodes.

Explanation

Electrolysis of aqueous solution of `AgNO_(3)` with silver electrodes:
The ions produced by the dissociation are as under:
`AgNO_(3) (s) + aq to Ag^(+) + NO_(3)^(-3) (aq)`
`H_(2)O At cathode : `Ag^+` ions have lower discharge potential than H+ ions. Hence, `Ag^+` ions will be deposited in preference to `H^+` ions.
At anode : As Ag anode is attacked by `NO_(3)^(-)` ions, Ag of the anode will dissolve to form `Ag^+` ions in the solution.
`Ag to Ag^(+) + e^(-)`
(ii) Electrolysis of aqueous solution of `AgNO_3` using platinum electrodes : At cathode : Same as in (i).
At anode : Out of `OH^(-)` and `NO_3^(-)` ions, `OH^(-)` ions have lower discharge potential. Hence, `OH^(-)` ions will be discharged in preference to `NO_3^(-)` ions. OH ions decompose to give out `O_(2)` as under:
`OH^(-)(aq) to OH + e^(-), 4OH to 2H_(2)O (l) + O_(2)(g)`
(iii) Electrolysis of dilute `H_(2)SO_(4)` with platinum electrodes:
`H_(2)SO_(4)` and `H_(2)O` ionise as under:
`H_(2)SO_(4)(aq) to 2H^(+) (aq) + SO_(4)^(2-) (aq)`
`H_(2)O At cathode: `H^(+) + e^(-) to H, H + H to H_(2)(g)`
At cathode: `OH^(-) to OH +e^(-), 4OH to 2H_(2)O + O_(2)(g)` (because of lower discharge potential of `OH^(-)`)
(iv) Electrolysis of aqueous solution of `CuCl_2` with platinum electrodes :
`CuCl_(2)(s) + aq to Cu^(2+) (aq) + 2Cl^(-)(aq)`
`H_(2) O At cathode: `Cu^(2+)` ions will be reduced in preference to `H^(+)` ions
`Cu^(2+) + 2e^(-) to Cu`
At anode: `Cl^(-)` ions will be oxidised in preference to `OH^(-)` ions.
`Cl^(-) to Cl + e^(-), Cl + Cl to Cl_(2)(g)`
Thus, Cu will be deposited on the cathode and `Cl_2` will be liberated at the anode.