Match the following columns.
 

	Column I (Substance)		Column II (Product after electrolysis)
A.	Aqueous solution of AgNO3 using Ag electrodes	p.	Oxygen is produced at anode
B.	Aqueous solution of AgNO3 using Pt electrodes	q.	Hydrogen is produced at cathode
C.	Dilute solution of H2SO4 using Pt electrodes	r..	Silver is deposited at cathode
D.	Aqueous solution of CuCl2 using Pt electrodes	s.	Neither O2 nor H2 is produced

$\mathrm{A}\to \mathrm{r},\mathrm{s};\mathrm{B}\to \mathrm{r},\mathrm{p};\mathrm{C}\to \mathrm{p},\mathrm{q};\mathrm{D}\to \mathrm{s}$
A. An aqueous  solution of AgNO₃ with silver electrodes. In aqueous solution, ionisation of AgNO₃ and H₂O takes place.
       $\begin{array}{r}{\mathrm{AgNO}}_3\left(\mathrm{s}\right)\stackrel{\left(\mathrm{aq}\right)}{\rightleftharpoons }{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{NO}}_3^{-}\left(\mathrm{aq}\right)\\ {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\rightleftharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\end{array}$
At cathode : Ag⁺ ions has less discharge potential than H⁺ ions so silver will be deposited at cathode.
                        ${\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}⟶\mathrm{Ag}\left(\mathrm{s}\right)$
At anode : An equivalent amount of silver will be oxidised to Ag⁺ ions by releasing electrons.
                       $\mathrm{Ag}\left(\mathrm{s}\right)⟶{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}$
            Ag anode is attacked by ${\mathrm{NO}}_3^{-}$ ions, so it will also produce Ago in the solution.
B. An aqueous solution of AgNO₃ with platinum electrodes.
    In aqueous solution, ionisation of AgNO₃ and H₂O both occur.
                   $\begin{array}{r}{\mathrm{AgNO}}_3\left(\mathrm{s}\right)\stackrel{\left(\mathrm{aq}\right)}{⟶}{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{NO}}_3^{-}\left(\mathrm{aq}\right)\\ {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\rightleftharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\end{array}$
      As platinum electrodes are non-attackable electrodes, they will not be reacted upon by ${\mathrm{NO}}_3^{-}$ ions.
At cathode Ag will be deposited at cathode.
            ${\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}⟶\mathrm{Ag}\left(\mathrm{s}\right)$
At anode Out of ${\mathrm{NO}}_3^{-}$ and OH^- ions, only OH- ions will be oxidised (due to less discharge potential) preferentially and ${\mathrm{NO}}_3^{-}$ ions will remain in the solution.
             $\begin{array}{l}{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)⟶\mathrm{OH}+{\mathrm{e}}^{-}\\ 4\mathrm{OH}⟶2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\end{array}$
So, oxygen gas is produced at anode. The solution remains acidic due to the Presence of HNO₃.
           ${\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{NO}}_3^{-}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{HNO}}_3\left(\mathrm{aq}\right)$
C.A dilute solution of H₂SO₄ with platinum electrodes. Both H₂SO₄ and water ionise in the solution.
        $\begin{array}{r}{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\rightleftharpoons 2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)\\ {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\rightleftharpoons 2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\end{array}$
At cathode H⁺ ions will be reduced and hydrogen gas is produced at cathode.
          $\begin{array}{r}{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}⟶\mathrm{H}\left(\mathrm{g}\right)\\ \mathrm{H}\left(\mathrm{g}\right)+\mathrm{H}\left(\mathrm{g}\right)⟶{\mathrm{H}}_2\left(\mathrm{g}\right)\end{array}$
At anode OH^- ions will be released preferentially and not ${\mathrm{SO}}_4^{-}$ions due to less discharge potential.
           $\begin{array}{r}{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)⟶\mathrm{OH}+{\mathrm{e}}^{-}\\ 4\mathrm{OH}⟶2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\end{array}$
         Oxygen gas is Produced at anode. Solution will be acidic and will contain H₂SO₄.
D. An aqueous solution of CuCl₂ with platinum electrodes. Both CuCl₂ and water ionise as usual.
             $\begin{array}{r}{\mathrm{CuCl}}_2\stackrel{\left(\mathrm{aq}\right)}{\rightleftharpoons }{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)\\ {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\rightleftharpoons 2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\end{array}$
     At cathode Cu²⁺ ions will be reduced preferentially due to less discharge potential than H⁺ ions.
              ${\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}⟶\mathrm{Cu}\left(\mathrm{s}\right)$
      Copper metal is deposited at cathode.   
   At anode Cl^- ions will be discharged in preference to OH^- ions and chlorine gas is produced at anode.
          $\begin{array}{r}{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)⟶\mathrm{Cl}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}\\ \mathrm{Cl}\left(\mathrm{g}\right)+\mathrm{Cl}\left(\mathrm{g}\right)⟶{\mathrm{Cl}}_2\left(\mathrm{g}\right)\end{array}$