Question

List -I	List -II
(i) Dilute solution of $HCl$	(P) $O_2$ evolved at anode
(ii) Dilute solution of $NaCl$	(Q)$H_2$ evolved at cathode
(iii) Concentrate solution of $NaCl$	(R) $C{l}_2$ evolved at anode
(iv) Fairly concentrate solution of $AgN{O}_3$	(S) $Ag$ deposition at cathode

Which of the following option has the incorrect combination considering List-I and List-II

• A. $i-S$
• B. $ii-Q$
• C. $iii-R$
• D. $iv-P$

Answer 1

The correct option is A $i-S$

During electrolysis of $dil.HCl(ordil.NaCl)$ solution, the water dissociates to form hydrogen and oxygen. Hydrogen is evolved at Cathode and oxygen is evolved at the anode.
At cathode: $2H^{+}\left(aq\right)+2e^{-}\to H_2\left(g\right)$
At anode: $4O{H}^{-}\left(aq\right)\to O_2\left(g\right)+2H_{2}O\left(l\right)+4e^{-}$
​​​​​​During electrolysis of concentrated $NaCl$ solution at the cathode, hydrogen is obtained from water and at the anode, chlorine is obtained from $NaCl$.
At cathode: $2H^{+}\left(aq\right)+2e^{-}\to H_2\left(g\right)$
At anode: $2C{l}^{-}\left(aq\right)\to C{l}_2\left(g\right)+2e^{-}$ ( $O{H}^{-}$ left in the solution)
During electrolysis of fairly concentrated silver nitrate ($AgN{O}_3$) at the anode, oxygen is evolved from water and at the cathode, Silver is deposited.
At cathode: $A{g}^{+}+e^{-}\to Ag\downarrow$
At anode: $2H_{2}O\to 4H^{+}+O_2+4e^{-}$
$I–(Q,P),II–(P,Q),III–(Q,R),IV–(P,S)$