Predict the products of electrolysis 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) An aqueous solution of `H_2SO_4` with platinum electrodes
(ii) An aqueous solution of `CuCl_2` with platinum electrodes

Explanation

(i) Electrolysis of aqueous solution of `AgNO_3` using silver electrodes :
`AgNO_3(s)+nH_2O rarr Ag^(+)(aq) +NO_3^(-) (aq)`
`" "H_2O hArr H^(+) +OH^(-)`
At cathode : `Ag^+` ions have lower discharge potential than `H^+` ions .
Hence `Ag^+` ions will be deposited as silver (in preference to `H^+` ions )
At anode : Since silver electrode is attacked by `NO_3^(-)` ions, Ag anode will dissolve to form `Ag^+` ions in the solution .
`Ag rarr Ag^(+) +e^(-)`
(ii) Electrolysis of aqueous solution of `AgNO_3` using platinum electrodes :
(ii) At cathode : same as above
At anode : Since silver is not attackable , out of `OH^- and NO_3^(-)` ions, `OH^-` ions have lower discharge potential and hence `OH^-` ions will be discharged in preference to `NO_3^-` . The `OH^-` will decompose to give `O_2`.
`OH^(-)(aq) rarr OH+e^(-)`
`4OH^(-) (aq) rarr 2H_2O(l) +O_2(g)`
(iii) Electrolysis of `H_2SO_4` with Pt electrodes :
`H_2SO_4(aq) rarr 2H^(+) (aq) +SO_4^(2-) (aq)`
`H_2OhArr H^(+) +OH^(-)`
At cathode : `H^(+) + e^(-) rarr H`
`H+H rarr H_2(g)`
At anode : `OH^(-) rarr OH+e^(-)`
`4OH rarr 2H_2O+O_2(g)`
(iv) Electrolysis of aqueous solution of `CuCl_2` with platinum electrodes :
`CuCl_2(s) +(aq) rarr Cu^(2+)(aq) +2Cl^(-) (aq)`
`H_2O hArr H^(+) +OH^(-)`
At cathode : `Cu^(+)` will be reduced in preference to `H^+` is
`Cu^(2+)+2e^(-) rarr Cu`
At anode : `Cl^(-)` ions will be oxidised in preference to `OH^-` ions
`Cl^(-) rarr Cl+e^(-)`
`Cl+Cl rarr Cl_2`
Thus copper will be deposited on the cathode and `Cl_2` will be liberated at anode .