Predict the products of electrolysis in each of the following:
(i) An aqueous solution of `AgNO_(3)` with silver electrodes.
(ii) An aqueous solution `AgNO_(3)` with platinum electrodes.
(iii) A dilute solution of `H_(2)SO_(4)` with platinum electrodes.
(iv) An aqueous solution of `CuCl_(2)` with platinum electrodes.

Explanation

(i) Aqueous solution of `AgNO_(3),Ag_((aq))^(+)andNO_(3(aq))^(-)` ions are formed.
`AgNO_(3(aq))toAg_((aq))^(+)+NO_(3(aq))^(-)`
During electrolysis `Ag^(+)orH_(2)O` gets reduced at cathode. But reduction potential of `Ag^(+)` is more than that of `H_(2)O`.
`Ag_((aq))^(+)+e^(-)toAg_((s))" "E^(@)=+0.80"volt"`
`2H_(2)O_((l))+2e^(-)toH_(2)+2OH_((aq))^(-)" "E^(@)=-0.83"volt"`
Therefore, `Ag^(+)` gets reduced at cathode.
As same, Ag metal or `H_(2)O` atoms oxidises at anode. But oxidation potential of Ag is more. than `H_(2)O`, therefore Ag gets oxidised at anode.
`Ag_((s))toAg_((aq))^(+)+e^(-)," "E^(@)=-0.80"volt"`
`2H_(2)O_((l))toO_(2(g))+4H_((aq))^(+)+4e^(-)," "E^(@)=-1.23"volt"`
(ii) Pt do not gets oxidized therefore at anode water gets reduced and gathered at cathode.
(iii) Aqueous solution of `H_(2)SO_(4)` on electrolysis `H^(+)andSO_(4)^(-2)` ions are formed.
`H_(2)SO_(4(aq))to2H_((aq))^(+)+SO_(4(aq))^(-2)`
During electrolysis `H^(+)orH_(2)O` gets reduced at cathode. But reduction potential of `H^(+)` is more then `H_(2)O`, therefore `H^(+)` is reduced at cathode and `H_(2)` gas is liberated.
`2H_((aq))^(+)+2e^(-)toH_(2(g))" "E^(@)=-0.0"volt"`
`2H_(2)O_((aq))+2e^(-)toH_(2)+2OH_((aq))^(-)" "E^(@)=-0.63"volt"`
At anode, `SO_(4)^(-2)orH_(2)O` gets oxidized but for the oxidation of `SO_(4)^(-2)`, more energy is used for the breaking the bond than `H_(2)O`.
Therefore, oxidation potential of `SO_(4)^(-2)` is less than `H_(2)O`. At anode, their is oxidation of `H_(2)O` and oxygen gas is released.
(iv) Aqueous solution of `CuCl_(2)` gives `Cu^(+2)andCl^(-)` ions.
`CuCl_(2(aq))toCu_((aq))^(+2)+2Cl_((aq))^(-)`
During electrolysis, `Cu^(+2)` ion of `H_(2)O` is reduced but electrode potential of `Cu^(+2)` is more than that of `H_(2)O`, therefore `Cu^(+2)` is reduced and gathered at cathode.
`Cu_((aq))^(+2)+2e^(-2)toCu_((aq))" "E^(@)=+0.34"volt"`
`H_(2)O_((l))+2e^(-)toH_(2)+2OH^(-)" "E^(@)=-0.83"volt"`
Some as, at anode `Cl^(-)orH_(2)O` is oxidised. But oxidation potential of `H_(2)O` is more than `Cl^(-)`.
`2Cl_((aq))^(-)toCl_(2(g))+2e^(-)" "E^(@)=-1.36"volt"`
`2H_(2)O_((l))toO_(2(g))+4H_((aq))^(+)+4e^(-)" "E^(@)=-1.23"volt"`
For the oxidation of `H_(2)O,O_(2)` gas is released when their is more potential is applied. Therefore, `Cl^(-)` is oxidised at anode and `Cl_(2)` gas is evolved.