Question:
In the reaction $O_2(g)+4H^+(aq)+4e^- \rightarrow 2H_2O(l)$, the quantity of electricity needed to reduce one mole of gaseous oxygen is $(E^\circ_{cell}=1.23V)$
In the reaction $O_2(g)+4H^+(aq)+4e^- \rightarrow 2H_2O(l)$, the quantity of electricity needed to reduce one mole of gaseous oxygen is $(E^\circ_{cell}=1.23V)$
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Multiply electron moles by 96500 C to get required charge.
Updated On: Apr 24, 2026
- $386000$ Coulombs
- $96500$ Coulombs
- $24125$ Coulombs
- $19300$ Coulombs
- $579000$ Coulombs
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Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Concept:
Chemistry - Faraday’s Law of Electrolysis.
Step 1: Electrons required. Reaction: [ O_2+4H^++4e^- \rightarrow 2H_2O ] 1 mole of $O_2$ requires: [ 4 \text{ moles of electrons} ]
Step 2: Charge of 1 mole electrons. 1 mole electrons = 1 Faraday [ 1F=96500\text{ C} ]
Step 3: Total charge required. [ Q=4F=4\times96500 ] [ Q=386000\text{ C} ]
Step 4: Final answer. [ \boxed{386000\text{ Coulombs}} ]
Hence, correct option is (A).
Step 1: Electrons required. Reaction: [ O_2+4H^++4e^- \rightarrow 2H_2O ] 1 mole of $O_2$ requires: [ 4 \text{ moles of electrons} ]
Step 2: Charge of 1 mole electrons. 1 mole electrons = 1 Faraday [ 1F=96500\text{ C} ]
Step 3: Total charge required. [ Q=4F=4\times96500 ] [ Q=386000\text{ C} ]
Step 4: Final answer. [ \boxed{386000\text{ Coulombs}} ]
Hence, correct option is (A).
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