Question

O\(_2\) 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 H2SO4 using platinum electrodes. 
(D) a high concentration solution of H2SO4 using platinum electrodes. 
Choose the correct answer from the options given below :

• (B) and (C) only
• (A) and (D) only
• (B) and (D) only
• (A) and (C) only

Hint:

Oxygen gas is evolved at the anode during electrolysis when water or dilute sulfuric acid is electrolyzed. At high concentrations of sulfuric acid, hydrogen gas evolution is favored at the anode.

Answer 1

The Correct Option is A

In the electrolysis of water or aqueous solutions, oxygen gas (O\(_2\)) is produced at the anode under certain conditions. 

(A) An aqueous solution of AgNO\(_3\) using silver electrodes: In this case, silver ions are reduced at the cathode to form silver metal. Oxygen gas is not produced at the anode because the silver electrode undergoes oxidation to form silver ions. Therefore, no O\(_2\) is evolved. 

(B) An aqueous solution of AgNO\(_3\) using platinum electrodes: Platinum is an inert electrode, and when AgNO\(_3\) is electrolyzed, oxygen gas will evolve at the anode due to the oxidation of water. This is correct. 

(C) A dilute solution of H\(_2\)SO\(_4\) using platinum electrodes: In this case, water is the main electrolyte, and oxygen gas will evolve at the anode during electrolysis of the dilute sulfuric acid solution. This is correct.

(D) A high concentration solution of H\(_2\)SO\(_4\) using platinum electrodes: At high concentrations of H\(_2\)SO\(_4\), oxygen evolution is suppressed, and hydrogen gas is more likely to evolve at the anode. This is not correct for oxygen evolution. 

Thus, the correct answer is (1) (B) and (C) only.

Answer 2

Step 1: Understand the question.
We are asked to determine in which of the following electrolysis cases oxygen gas (\( O_2 \)) will be evolved as a product. Oxygen gas is typically produced at the anode during the oxidation of water or hydroxide ions, depending on the nature of the electrolyte and the electrodes used.

Step 2: Analyze each option.
(A) Aqueous solution of AgNO₃ using silver electrodes
In this case, both the electrolyte and the electrode contain silver. The reactions are:
At the cathode: \( Ag^+ + e^- \rightarrow Ag \) (silver deposition)
At the anode: \( Ag \rightarrow Ag^+ + e^- \) (silver dissolves)
Since oxidation and reduction involve only silver, no \( O_2 \) is released. Hence, (A) is false.

(B) Aqueous solution of AgNO₃ using platinum electrodes
Platinum is an inert electrode, so it does not participate in the reaction. The reactions are:
At the cathode: \( Ag^+ + e^- \rightarrow Ag \)
At the anode: Water is oxidized because platinum cannot dissolve.
\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \] Hence, \( O_2 \) is evolved. Therefore, (B) is true.

(C) A dilute solution of H₂SO₄ using platinum electrodes
For dilute \( H_2SO_4 \), the major ions are \( H^+ \) and \( SO_4^{2-} \).
At the cathode: \( 2H^+ + 2e^- \rightarrow H_2 \) (hydrogen gas evolves)
At the anode: \( 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \) (oxygen gas evolves)
Thus, \( O_2 \) is produced at the anode. Therefore, (C) is true.

(D) A high concentration solution of H₂SO₄ using platinum electrodes
In concentrated \( H_2SO_4 \), sulfate ions are oxidized to peroxydisulfate ions (\( S_2O_8^{2-} \)) instead of oxygen gas:
\[ 2SO_4^{2-} \rightarrow S_2O_8^{2-} + 2e^- \] Hence, no \( O_2 \) gas is evolved. Therefore, (D) is false.

Step 3: Conclusion.
Oxygen gas is evolved in cases (B) and (C) only.

Final Answer:
\[ \boxed{(B) \text{ and } (C) \text{ only}} \]