What quantity of electricity will liberate 0.125 mole of oxygen molecules during the electrolysis of dilute sodium chloride solution?
24,125 C
48,250 C
72,375 C
96,500 C
Explanation
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To determine the quantity of electricity required to liberate 0.125 mole of oxygen molecules () during the electrolysis of dilute sodium chloride solution, we follow these steps:
Step 1: Write the Half-Reaction for Oxygen Evolution
The oxidation half-reaction at the anode during the electrolysis of dilute sodium chloride solution is:
2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-
From this reaction, 4 moles of electrons (4 Faradays) are required to produce 1 mole of .
Step 2: Calculate the Charge for 0.125 Mole of
Since 1 mole of requires 4 Faradays of charge, 0.125 moles will require:
0.125 \times 4 = 0.5 \text{ Faraday}
Since 1 Faraday (F) = 96,500 C, the total charge required is:
0.5 \times 96,500 = 48,250 \text{ C}
Step 3: Identify the Correct Answer
Thus, the correct answer is:
B. 48,250 C
Method 2 remember N=Q/nf
N=0.125 n=4 f=96500 Q=?
Make Q the subject of formula
Therefore Q=Nnf
Q=0.125×4×96500
Q=48250✓✓
