Predict the products of electrolysis in each of the following `:`
`a.` An aqueous solution of `AgNO_(3)` with silver electrodes.
`b.` An aqueous 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) An aqueous solution of `AgNO_(3)` using silver electrodes :
Both `AgNO_(3)` and water will ionise in aqueous solution
`AgNO_(3)(s)overset((aq))rarrAg^(+)(aq)+NO_(3)^(-)(aq)`
`H_(2)O overset((aq)) harr H^(+)(aq)+HO^(-)(aq)`
At cathode : `Ag^(+)` ions with less discharge potential are reduced in preference to `H^(+)` ions which will remain in solution. As a result, silver will be deposited at cathode.
`Ag^(+)(aq)+e^(-) to Ag` (deposited) ltbr. At anode : An equivalent amount of silver will be oxidised to `Ag^(+)` ions by releasing electrons.
`Ag(s)+e^(-)overset((aq))to Ag^(+)(aq)+e^(-)`
(ii) An aqueous solution of `AgNO_(3)` using platinum electrodes :
In this case, the platinum electrodes are the non-attackable electrons. On passing current the following changes will occur at the electrodes.
At cathode : `Ag^(+)` ions will be reduced to Ag which will get deposited at the cathode.
At anode : Both `NO_(3)^(-)` and `OH^(-)` ions will migrate. But `OH^(-)`ions with less discharge potential will be oxidised in preference to `NO_(3)^(@)` ions which will remain in solution.
`OH^(-)(aq)toOH+e^(-) ,4OH to 2H_(2)O(l)+O_(2)(g)`
Thus, as a result of electrolysis, silver is deposited on the cathode while `O_(2)` is evolved at the anode. The solution will be acidic due to the presence of `HNO_(3)`.
(iii) A dilute solution of `H_(2)SO_(4)` using platinum electrodes :
On passing current, both acid and water will ionise as follows :
`H_(2)SO_(4)(l)overset((aq))rarr2H^(+)(aq)+SO_(4)^(2-)(aq)`
`H_(2)Ooverset((aq))harrH^(+)(aq)+OH^(-)(aq)`
A cathode : `H^(+)` (aq) ions will migrate to the cathode and will be reduced to `H_(2)`.
`H^(+)(aq)+e^(-)toH , H+H to H_(2)(g)`
Thus, `H_(2)`(g) will be evolved at cathode.
At anode : `OH^(-)` ions will be released in preference to `SO_(4)^(2-)` ions because their discharge potential is less. They will be oxidised as follows :
`OH^(-)(aq)toOH+e^(-) , 4OHto2H_(2)O(l)+O_(2)(g)`
Thus, `O_(2)`(g) will be evolved at anode. The solution will be acidic and will contain `H_(2)SO_(4)`.
(iv) An aqueous solution of `CuCl_(2)` using platinum electrodes :
The electrolysis proceeds in the same manner as discussed in the case of `AgNO_(3)` solution. Both `CuCl_(2)` and `H_(2)O`will ionise as follows :
`CuCl_(2)overset((aq))rarrCu^(2+)(aq)+2Cl^(-)(aq)`
`H_(2)O overset((aq))harrH^(+)(aq)+OH^(-)(aq)`
At cathode : `Cu^(2+)` ions will be reduced in preference to `H^(+)` ions and copper will be deposited at cathode
`Cu^(2+)(aq)+2e^(-)to Cu` ("deposited")
A anode : `Cl^(-)` ions will be discharged in preference to `OH^(-)` ions which will remain in solution.
`Cl^(-)toCl^(-)+e^(-) , Cl+Cl to Cl_(2)(g)`
Thus, `Cl_(2)` will be evolved at anode.