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The amount of Aluminium deposited when 0.1 Faraday current is passed through aluminium chloride will be (R = 27)
- a)0.9 g
- b)0.3 g
- c)0.27 g
- d)2.7 g
Correct answer is option 'A'. Can you explain this answer?
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The amount of Aluminium deposited when 0.1 Faraday current is passed t...
3 Faraday current deposit 27 gram of of Al3 + ions.
therefore, 0.1 Faraday will deposit 27 x 0.1/ 3 = 0.9 gram of aluminium ions
therefore, 0.1 Faraday will deposit 27 x 0.1/ 3 = 0.9 gram of aluminium ions
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The amount of Aluminium deposited when 0.1 Faraday current is passed t...
Given information:
- Current passed through aluminium chloride = 0.1 Faraday
- R = 27
To find:
The amount of Aluminium deposited.
Solution:
The amount of substance deposited during electrolysis can be determined using Faraday's laws of electrolysis.
Faraday's first law:
According to Faraday's first law, the amount of substance deposited or liberated during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
Mathematically, the amount of substance deposited (m) is given by the formula:
m = Z * F * I * t
Where:
- Z is the electrochemical equivalent (constant for a given substance)
- F is the Faraday constant (96500 C/mol)
- I is the current in amperes
- t is the time in seconds
In this case, the current passed through aluminium chloride is 0.1 Faraday, which is equal to 0.1 * 96500 C.
So, the amount of Aluminium deposited (m) can be calculated as follows:
m = Z * (0.1 * 96500)
Faraday's second law:
According to Faraday's second law, the electrochemical equivalent (Z) is equal to the ratio of the atomic mass (M) to the Faraday constant (F) for a substance.
Mathematically, the electrochemical equivalent (Z) is given by the formula:
Z = M / F
For Aluminium, the atomic mass (M) is 27 g/mol (given).
Substituting the values in the equation, we get:
Z = 27 / 96500
Therefore, the amount of Aluminium deposited (m) can be calculated as follows:
m = (27 / 96500) * (0.1 * 96500)
m = 0.1 g
Hence, the correct answer is option 'A' - 0.1 g.
- Current passed through aluminium chloride = 0.1 Faraday
- R = 27
To find:
The amount of Aluminium deposited.
Solution:
The amount of substance deposited during electrolysis can be determined using Faraday's laws of electrolysis.
Faraday's first law:
According to Faraday's first law, the amount of substance deposited or liberated during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
Mathematically, the amount of substance deposited (m) is given by the formula:
m = Z * F * I * t
Where:
- Z is the electrochemical equivalent (constant for a given substance)
- F is the Faraday constant (96500 C/mol)
- I is the current in amperes
- t is the time in seconds
In this case, the current passed through aluminium chloride is 0.1 Faraday, which is equal to 0.1 * 96500 C.
So, the amount of Aluminium deposited (m) can be calculated as follows:
m = Z * (0.1 * 96500)
Faraday's second law:
According to Faraday's second law, the electrochemical equivalent (Z) is equal to the ratio of the atomic mass (M) to the Faraday constant (F) for a substance.
Mathematically, the electrochemical equivalent (Z) is given by the formula:
Z = M / F
For Aluminium, the atomic mass (M) is 27 g/mol (given).
Substituting the values in the equation, we get:
Z = 27 / 96500
Therefore, the amount of Aluminium deposited (m) can be calculated as follows:
m = (27 / 96500) * (0.1 * 96500)
m = 0.1 g
Hence, the correct answer is option 'A' - 0.1 g.
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The amount of Aluminium deposited when 0.1 Faraday current is passed through aluminium chloride will be (R = 27)a)0.9 gb)0.3 gc)0.27 gd)2.7 gCorrect answer is option 'A'. Can you explain this answer?
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