O₂ gas will be evolved as a product of electrolysis of :
(A) an aqueous solution of AgNO₃ using silver electrodes.
(B) an aqueous solution of AgNO₃ using platinum electrodes.
(C) a dilute solution of H₂ SO₄ using platinum electrodes.
(D) a high concentration solution of H₂ SO₄ using platinum electrodes.
Choose the correct answer from the options given below :

Solution

Chemistry ma’am · 3 steps

Look at the anode first: if anything there is easier to oxidise than water, oxygen never forms.

1. Where oxygen forms. O₂ can only come from the anode, by oxidising water: 2H₂O → O₂ + 4H⁺ + 4e⁻. So for each case, ask whether something at the anode is oxidised more easily than water.

   Case	What is oxidised at the anode	Anode product
   (A) AgNO₃, Ag electrodes	The silver anode itself: Ag → Ag⁺ + e⁻	Ag⁺ goes into solution, no gas
   (B) AgNO₃, Pt electrodes	Water, because Pt is inert and NO₃⁻ resists oxidation	O₂
   (C) dilute H₂SO₄, Pt electrodes	Water, which is in large excess	O₂
   (D) concentrated H₂SO₄, Pt electrodes	HSO₄⁻: 2HSO₄⁻ → S₂O₈²⁻ + 2H⁺ + 2e⁻	Peroxodisulphate, not O₂

2. A and D fail. A silver anode is oxidised before water, so it dissolves instead of giving gas. In concentrated H₂SO₄ there is little free water and plenty of HSO₄⁻, so the anode makes peroxodisulphuric acid, H₂S₂O₈.
3. So. Oxygen is evolved only in (B) and (C), which is option (a).

The idea to keep. At the anode, an active metal electrode beats water, and water beats nitrate and sulphate. Concentration matters too: dilute H₂SO₄ gives O₂, concentrated H₂SO₄ gives S₂O₈²⁻.

Answer from the NTA final answer key. Solution by PrepWiser.