Predict the products of electrolysis in eaCHM 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)SSO_(4)` with platinum electrodes.
`d.` An aqueous solution of `CuCl_(2)` with platinum electrodes.

Explanation

(i) In aqueous solution `AgNO_(3)` ionises ot give `Ag^(+)(Aq)and NO_(3)^(-)(aq)`ions
`AgNO_(3)(aq)rarrAg^(+)(aq)+NO_(3_^(-) (aq)`
thus when electricity is passed `Ag^(+) (aq)+e^(-)rarrAg(s), E^(@)=+_0.80 v`
`2H_(2)O(l)+2e^(-)rarrH_(2)(g)+2O^(-)(aq),e^(@)=-0.83 v`
since the elctrode potential i.e reductin potential of `Af^(+)` (aq) ions is higher than that of `H_(2)O` molecules therefore at the cathode it is the `Ag^(+)` (aq) ions (rather than `H_(2)O` molecules ) which are reduced similary at the naode either ag metal of the anode ofr `H_(2)O` molecules may be oxidised their electrode
thus when an queous soluton of `AgNO_(3)` is electrolysed As from As anode dissolved while `Ag^(+)` (aq) ins present in the solution get reduced and get depostied on the cathode
(iii) if howver elctrolysis of `AgNO_(3)` is electrolysed Ag from Ag anode dissolves wile `Ag^(+)` ions present in the solutoin get reduced and get depostited on the cathode
since the elctron potential of `H^(+)` (aq) ions is higher than that of `H_(2)O` therefore at the cathode it is `H^(+)` (aq) ioons which are reduced to evolve `H_(2)O` gas of `SO_(4)^(2-)` is expected to be much lower than that of `H_(2)O` molecuels therefore at the anode it is `H_(2)O` mlecules which are oxidised to evolve `O_(2)` gas
from the abvoe discussion it follow that during electrolysis of an aqueous solution of `H_(2)O` only the electrolysis of `H_(2)O` occurs liberating `H_(2)` at the cathode and `O_(2)` at the anode
thus as cathode either `Cu^(2+)`(aq) or `H_(2)O` molecuels are reduced their electrode potential are
`Cu^(2+)(aq)+2e^(-)rarrCu(s) , =+ 0.34 V`
`H_(2)O(l)+2e^(-)rarrH_(2)(g)+2OH^(-), E^(@)=-0.83 V`
since the electrode potential of `Cu^(2+)`(aq) ions is much higher than that of `H_(2)O` therefore at the cathode it is `Cu^(2+)` (aq) ions whihc are reduced and not 1 `H_(2)O` molecules
similary at the anode either `CI^(-)` (aQ) ions or `H_(32)O` molecules are oxidised their oxidation potentials are
`2CI^(-)(aq)rarrCI_(2)(g)+2e^(-), E^(@)=-1.36 v`
`2H_(2)O(l)rarrO_(2)+4H^(+)(aq)+4e^(-),E^(@)=-1.23 v`
Although oxidation potential of `H_(2)O` molecules is higher than of `CI^(-)` ions neverthelss oxidation of `CI^(-)(aq)` ions occurs is prefrence to `H_(2)O` since due to `"overvoltage"^(*)` of `O_(2)` much lower potential than than -1.36 v is needed for the oxidation of `H_(2)O` molecules
thus when aqueous solution of `CuCI_(2)` is electrolysed Cu metal is liberated at the cathode while `CI_(2)` gas is evolved at the anode