gas will be evolved as a product of electrolysis of: (A) an aqueous solution of using silver electrodes. (B) an aqueous solution of using platinum electrodes. (C) a dilute solution of using platinum electrodes. (D) a high concentration solution of us

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Solution

Q.
$O_{2}$ gas will be evolved as a product of electrolysis of:

(A) an aqueous solution of ${AgNO}_{3}$ using silver electrodes.
(B) an aqueous solution of ${AgNO}_{3}$ using platinum electrodes.
(C) a dilute solution of $H_{2} {SO}_{4}$ using platinum electrodes.
(D) a high concentration solution of $H_{2} {SO}_{4}$ using platinum electrodes.

Choose the correct answer from the options given below: [2025]

1 (B) and (D) only

2 (A) and (C) only

3 (A) and (D) only

4 (B) and (C) only

Ans.
(4)

(A) Ions present:

$A g^{+}, N O_{3}^{-}, H^{+}, O H^{-}$

At anode:

$N O_{3}^{-}$ and $O H^{-}$ move towards anode for oxidation. Anode is of silver. Oxidation potential of Ag is more than that of $N O_{3}^{-}$ and $O H^{-}$ , hence Ag is oxidised in preference to $N O_{3}^{-}$ and $O H^{-}$ .

$A g (s) \to A g^{+} (a q) + e^{-}$

At cathode:

$A g^{+}$ and $H^{+}$ move towards cathode for reduction. Reduction of $A g^{+}$ is preferred over $H^{+}$ as reduction potential of Ag is more than that of $H^{+}$ .

$A g^{+} (a q) + e^{-} \to A g (s)$

Net reaction:
$A g (a n o d e) \to A g (c a t h o d e)$

Thus, silver is dissolved at anode and deposited at cathode.

(B) Ions present: $A g^{+}, N O_{3}^{-}, H^{+}, O H^{-}$

At anode:

$N O_{3}^{-}$ and $O H^{-}$ move towards anode for oxidation. $O H^{-}$ is oxidised in preference to $N O_{3}^{-}$ as it has more oxidation potential than that of $N O_{3}^{-}$ .

$4 O H^{-} (a q) \to O_{2} (g) + 2 H_{2} O (l) + 4 e^{-}$

At cathode:

${Ag}^{+}$ and $H^{+}$ move towards cathode for reduction. Reduction of ${Ag}^{+}$ is preferred over $H^{+}$ as reduction potential of Ag is more than that of $H^{+}$ .

$A g^{+} (a q) + e^{-} \to A g (s)$

Net reaction:
$4 A g^{+} (a q) + 4 O H^{-} (a q) \to 4 A g (s) + O_{2} (g) + 2 H_{2} O (l)$

(C) Ions present: $S O_{4}^{2 -}, H^{+}, O H^{-}$

At anode:

$S O_{4}^{2 -}$ and $O H^{-}$ move towards anode for oxidation. In dilute solution, $O H^{-}$ is oxidised in preference to $S O_{4}^{2 -}$ as $O H^{-}$ has more oxidation potential.

$4 O H^{-} (a q) \to O_{2} (g) + 2 H_{2} O (l) + 4 e^{-}$

At cathode:

$H^{+}$ move towards cathode for reduction and is reduced.

$4 H^{+} (a q) + 4 e^{-} \to 2 H_{2} (g)$

Net reaction:

$4 O H^{-} (a q) + 4 H^{+} (a q) \to O_{2} (g) + 2 H_{2} O (l) + 2 H_{2} (g)$

$4 H_{2} O (l) \to O_{2} (g) + 2 H_{2} O (l) + 2 H_{2} (g)$

$2 H_{2} O (l) \to O_{2} (g) + 2 H_{2} (g)$

Thus net reaction is electrolysis of water.

(D) In concentrated solution of sulphuric acid, oxidation of sulphate ion is preferred.

$2 S O_{4}^{2 -} (a q) \to S_{2} O_{8}^{2 -} (a q) + 2 e^{-}$

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