$$O_{2}$$ 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 :

Let us analyze each case for $$O_2$$ evolution at the anode during electrolysis:

(A) Aqueous $$AgNO_3$$ with silver electrodes: At the anode, the silver electrode dissolves ($$Ag \to Ag^+ + e^-$$) since silver is oxidized preferentially. No $$O_2$$ evolved.

(B) Aqueous $$AgNO_3$$ with platinum electrodes: At the inert Pt anode, water is oxidized: $$2H_2O \to O_2 + 4H^+ + 4e^-$$. $$O_2$$ is evolved. ✓

(C) Dilute $$H_2SO_4$$ with platinum electrodes: At the Pt anode, water is oxidized to give $$O_2$$. ✓

(D) Concentrated $$H_2SO_4$$ with platinum electrodes: At high concentrations, $$SO_4^{2-}$$ or $$HSO_4^-$$ may get oxidized to form peroxodisulphate ($$S_2O_8^{2-}$$) at the anode instead of $$O_2$$. So $$O_2$$ is not the primary product.

The correct answer is Option 2: (B) and (C) only.