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 :
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 :
Show Hint
(B) and (C) only
(A) and (D) only
(B) and (D) only
(A) and (C) only
The Correct Option is A
Solution and Explanation
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.
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