$\mathrm{O}_2$ gas will be evolved as a product of electrolysis of :

(A) an aqueous solution of $\mathrm{AgNO}_3$ using silver electrodes.

(B) an aqueous solution of $\mathrm{AgNO}_3$ using platinum electrodes.

(C) a dilute solution of $\mathrm{H}_2 \mathrm{SO}_4$ using platinum electrodes.

(D) a high concentration solution of $\mathrm{H}_2 \mathrm{SO}_4$ using platinum electrodes.

Choose the correct answer from the options given below :

<p>(A) An aqueous solution of AgNO$_3$ using silver electrodes.</p> <p>Cathode - Reduction - $Ag_{(aq)}^ + + {e^ - } \to Ag(s)$</p> <p>Anode - Oxidation - $Ag(s) \to Ag_{(aq)}^ + + {e^ - }$</p> <p>Solid silver will be deposited at the cathode. Solid anode (silver) will dissolve, releasing silver ions into the solution.</p> <p>So, there is no formation of O$_2$ gas in this electrolysis.</p> <p>(B) An aqueous solution of AgNO$_3$ using platinum electrodes.</p> <p>Cathode - Reduction - $Ag_{(aq)}^ + + {e^ - } \to Ag(s)$</p> <p>Anode - Oxidation - $2{H_2}O \to 4{H^ + } + {O_2} + 4{e^ - }$</p> <p>When platinum electrodes are used, Ag$^+$ from solution is reduced and deposited at cathode whereas O$_2$ is produced at the anode.</p> <p>(C) A dilute soution of H$_2$SO$_4$ using platinum electrodes.</p> <p>Cathode - Reduction - $2{H^ + } + 2{e^ - } \to {H_2}$</p> <p>Anode - Oxidation - $2{H_2}O \to {O_2} + 4{H^ + } + 4{e^ - }$</p> <p>H$_2$ gas is produded at the cathode and O$_2$ gas is produced at the anode.</p> <p>(D) a high concentration solution of H$_2$SO$_4$ using platinum electrodes.</p> <p>O$_2$ gas is not formed in this case.</p> <p>Cathode - Reduction - The substance formed is H$_2$ gas.</p> <p>Anode - Oxidation - The substance formed is not O$_2$ gas.</p> <p>So, statements (B) and (C) are correct.</p>