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Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milliampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol-1)
- a)9.65 x 104 sec
- b)19.3 x 104 sec
- c)28.95 x 104 sec
- d)38.6 x 104 sec
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
Electrolysis of dilute aqueous NaCl solution was carried out by passin...
Given:
- Current passing through the electrolysis cell = 10 milliamperes = 10 × 10^-3 A
- Number of moles of H2 gas liberated at the cathode = 0.01 mol
- 1 Faraday = 96500 C mol^-1
To find:
- Time required to liberate 0.01 mol of H2 gas at the cathode
Solution:
The amount of substance liberated at an electrode during electrolysis is directly proportional to the amount of charge passed through the electrolyte. The relationship between charge (Q), current (I), and time (t) is given by the equation:
Q = I × t
We can rearrange this equation to solve for time:
t = Q / I
Step 1:
Calculate the charge (Q) required to liberate 0.01 mol of H2 gas.
Q = number of moles × Faraday's constant
Q = 0.01 mol × 96500 C mol^-1
Q = 965 C
Step 2:
Substitute the values of Q and I into the equation to find the time (t).
t = Q / I
t = 965 C / (10 × 10^-3 A)
t = 965 C / 0.01 A
t = 96500 s
Step 3:
Convert the time to seconds.
1 hour = 3600 seconds
t = 96500 s / 3600 s/hour
t ≈ 26.8 hours
Answer:
The time required to liberate 0.01 mol of H2 gas at the cathode is approximately 26.8 hours.
- Current passing through the electrolysis cell = 10 milliamperes = 10 × 10^-3 A
- Number of moles of H2 gas liberated at the cathode = 0.01 mol
- 1 Faraday = 96500 C mol^-1
To find:
- Time required to liberate 0.01 mol of H2 gas at the cathode
Solution:
The amount of substance liberated at an electrode during electrolysis is directly proportional to the amount of charge passed through the electrolyte. The relationship between charge (Q), current (I), and time (t) is given by the equation:
Q = I × t
We can rearrange this equation to solve for time:
t = Q / I
Step 1:
Calculate the charge (Q) required to liberate 0.01 mol of H2 gas.
Q = number of moles × Faraday's constant
Q = 0.01 mol × 96500 C mol^-1
Q = 965 C
Step 2:
Substitute the values of Q and I into the equation to find the time (t).
t = Q / I
t = 965 C / (10 × 10^-3 A)
t = 965 C / 0.01 A
t = 96500 s
Step 3:
Convert the time to seconds.
1 hour = 3600 seconds
t = 96500 s / 3600 s/hour
t ≈ 26.8 hours
Answer:
The time required to liberate 0.01 mol of H2 gas at the cathode is approximately 26.8 hours.
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Electrolysis of dilute aqueous NaCl solution was carried out by passin...
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Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milliampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol-1)a)9.65 x 104secb)19.3 x 104secc)28.95 x 104secd)38.6 x 104secCorrect answer is option 'B'. Can you explain this answer?
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Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milliampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol-1)a)9.65 x 104secb)19.3 x 104secc)28.95 x 104secd)38.6 x 104secCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milliampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol-1)a)9.65 x 104secb)19.3 x 104secc)28.95 x 104secd)38.6 x 104secCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milliampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol-1)a)9.65 x 104secb)19.3 x 104secc)28.95 x 104secd)38.6 x 104secCorrect answer is option 'B'. Can you explain this answer?.
Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milliampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol-1)a)9.65 x 104secb)19.3 x 104secc)28.95 x 104secd)38.6 x 104secCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milliampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol-1)a)9.65 x 104secb)19.3 x 104secc)28.95 x 104secd)38.6 x 104secCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milliampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol-1)a)9.65 x 104secb)19.3 x 104secc)28.95 x 104secd)38.6 x 104secCorrect answer is option 'B'. Can you explain this answer?.
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