Predict the products of electrolysis in each of the following `:`
`a.` An aqueous solution of `AgNO_(3)` with silver electrodes.
`b.` An aqeous solution of `AgNO_(3)` with platinum electrodes,
`c.` A dilute solution of `H_(2)SO_(4)` with platinum electrodes.
`d.` An aqueous solution of `CuCl_(2)` with platinum electrodes.

Explanation

(i). `AgNO_(3)(s)aqtoAg^(+)(aq)+NO_(3)^(-)(aq)`
`H_(2)OhArrH^(+)+OH^(-)`
At cathode: `Ag^(+)` ions have lower discharge potential than `H^(+)` ions. Hence, `Ag^(+)` ions will be deposited as Ag I preference to `H^(+)` ions
`Ag^(+)(aq)+e^(-)toAg(s)`
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(s)toAg^(+)(aq)+e^(-)`
(ii). At cathode: `Ag^(+)` ions have lowr discharge potential than `H^(+)` ions. hence, `Ag^(+)` ions will be deposited as Ag in preference to `H^(+)` ions.
Ag anode: As anode is not attackable, 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, which then decompose to give out `O_(2)`.
`OH^(-)(aq)toOH+e^(-)`
`4OHto2H_(2)O,(l)+O_(2)(g)`
(iii). `H_(2)SO_(4)(aq)to2H^(+)(aq)+SO_(4)^(2-)(aq)`
`H_(2)OhArr H^(+)+OH^(-)`
At cathode: `H^(+)+e^(-)toH`
`H+HtoH_(2)(g)`
At anode: `OH^(-)toOH+e^(-)`
`4OHto2H_(2)O+O_(2)(g)`
Thus, `H_(2)` gas is liberated at the cathode
and `O_(2)` gas at the anode.
(iv). `CuCl_(2)(s)+aqtoCu^(2+)(aq)+2Cl^(-)(aq)`
`H_(2)OhArrH^(+)+OH^(-)`
At cathode: `Cu^(2+)` ions will be reduced in preference to `H^+` ions and copper will be deposited at cathode.
`Cu^(2+)+2e^(-)toCu`
At anode: `Cl^(-)` ions will be discharged in preference to `OH^(-)` ions which remains in solution
`Cl^(-)toCl+e^(-)`
`Cl+Cl toCl_(2)(g)`
Thus, Cu will be deposited on the cathode and `Cl_(2)` gas will be liberated at the anode.