Question

Match the following columns.

Column I(Substance)		Column II(Product after electrolysis)
A	Aqueous solution of $AgN{O}_3$ using $Ag$ electrodes	p	Oxygen is produced at anode
B	Aqueous solution of $AgN{O}_3$ using $Pt$ electrodes	q	Hydrogen is produced at cathode
C	Dilute solution of $H_{2}S{O}_4$ at using $Pt$ electrodes	r	Silver is deposited cathode
D	Aqueous solution of $CuC{l}_2$ is using $Pt$ electrodes	s	Neither $O_2$ nor $H_2$ produced

• A. $A-(r,q),B-(p,q),C-(q,s),D-(p,s)$
• B. $A-(r,q),B-(p,q,s),C-(q,s),D-(q,r,s)$
• C. $A-(p),B-(q),C-(r),D-(s)$
• D. $A-(r,s),B-(r,p),C-(p,q),D-(s)$

Answer 1

Answer: (D)

$A\to r,s;B\to r,p;C\to p,q;D\to s$
A. An aqueous solution of $AgN{O}_3$ with silver electrodes. In aqueous solution, ionisation of $AgN{O}_3$ and $H_{2}O$ takes place.
$AgN{O}_3(s)\stackrel{(aq)}{\rightleftharpoons }A{g}^{+}(aq)+N{O}_3^{-}(aq)$
$H_{2}O(l)\rightleftharpoons H^{+}(aq)+O{H}^{-}(aq)$
At cathode $A{g}^{+}$ions has less discharge potential than $H^{+}$ ions so silver will be deposited at cathode.
$A{g}^{+}(aq)+e^{-}⟶Ag(s)$
At anode An equivalent amount of silver will be oxidised to $A{g}^{+}$ions by releasing electrons.
$Ag(s)⟶A{g}^{+}(aq)+e^{-}$
Ag anode is attacked by $N{O}_3^{-}$ions, so it will also produce $A{g}^{+}$in the solution.
B. An aqueous solution of $AgN{O}_3$ with platinum electrodes. In aqueous solution, ionisation of $AgN{O}_3$ and $H_{2}O$ both occur.
$AgN{O}_3(s)\stackrel{(aq)}{\rightleftharpoons }A{g}^{+}(aq)+N{O}_3^{-}(aq)$
$H_{2}O(l)\rightleftharpoons H^{+}(aq)+O{H}^{-}(aq)$
As platinum electrodes are non-attackable electrodes, they will not be reacted upon by $N{O}_3^{-}$ions.
At cathode Ag will be deposited at cathode.
$A{g}^{+}(aq)+e^{-}⟶Ag(s)$
At anode Out of $N{O}_3^{-}$and $O{H}^{-}$ions, only $O{H}^{-}$ions will be oxidised (due to less discharge potential) preferentially and $N{O}_3^{-}$ions will remain in the solution.
$O{H}^{-}(aq)⟶OH+e^{-}$
$4OH⟶2H_{2}O(l)+O_2(g)$
So, oxygen gas is produced at anode. The solution remains acidic due to the presence of $HN{O}_3$.
$H^{+}(aq)+N{O}_3^{-}(aq)\rightleftharpoons HN{O}_3(aq)$
C. A dilute solution of $H_{2}S{O}_4$ with platinum electrodes. Both $H_{2}S{O}_4$ and water ionise in the solution.
$H_{2}S{O}_4(aq)\rightleftharpoons 2H^{+}(aq)+{\mathrm{SO}}_4^{2-}(aq)$
$H_{2}O(l)\rightleftharpoons 2H^{+}(aq)+O{H}^{-}(aq)$
At cathode $H^{+}$ions will be reduced and hydrogen gas is produced at cathode.
$H^{+}(aq)+e^{-}⟶H(g)$
$H(g)+H(g)⟶H_2(g)$
At anode $O{H}^{-}$ions will be released preferentially and not $S{O}_4^{-}$ions due to less discharge potential.
$O{H}^{-}(aq)⟶OH+e^{-}$
$4OH⟶2H_{2}O(l)+O_2(g)$
Oxygen gas is produced at anode.
Solution will be acidic and will contain $H_{2}S{O}_4$.
D. An aqueous solution of $CuC{l}_2$ with platinum electrodes. Both $CuC{l}_2$ and water ionise as usual.
$CuC{l}_2\stackrel{(aq)}{\rightleftharpoons }C{u}^{2+}(aq)+2C{l}^{-}(aq)$
$H_{2}O(l)\rightleftharpoons 2H^{+}(aq)+O{H}^{-}(aq)$
At cathode $C{u}^{2+}$ ions will be reduced preferentially due to less discharge potential than $H^{+}$ions.
$C{u}^{2+}(aq)+2e^{-}⟶Cu(s)$
Copper metal is deposited at cathode.
At anode $C{l}^{-}$ions will be discharged in preference to $O{H}^{-}$ions and chlorine gas is produced at anode.
$C{l}^{-}(aq)⟶Cl(g)+e^{-}$
$Cl(g)+Cl(g)⟶C{l}_2(g)$