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During the electrolysis of a solution of AgNO₃, 9650 C of charge passes through the electroplating bath, the mass of silver deposited on the cathode will be
- a)1.08 g
- b)10.8 g
- c)21.6 g
- d)108 g
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
During the electrolysis of a solution of AgNO₃, 9650 C of charge...
1 mole of AgNO will produce one mole of monovalent silver ion on dissociation.
Therefore,
.number of electrons involved will also be one mole for one mole of AgNO .
Thus, 0.1 mole of silver will be produced by 0.1 mole of AgNO .
Since,
silver nitrate dissociates completely into Ag and NO ion.
Hence,
mass of silver = no. of moles x molar mass= 0.1 x 108 = 10.8 g
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Community Answer
During the electrolysis of a solution of AgNO₃, 9650 C of charge...
Given:
Charge passed (Q) = 9650 C
Solution of AgNO3 is used for electrolysis.
To find:
Mass of silver deposited on the cathode
Solution:
AgNO3 dissociates into Ag+ and NO3- ions in solution.
Ag+ ions will move towards the cathode and get reduced to Ag(s) on gaining electrons.
The quantity of electric charge is related to the amount of substance produced by the following equation:
n = Q/F
where n is the number of moles of substance produced, Q is the quantity of electric charge passed through the solution, and F is the Faraday constant (96500 C/mol).
The mass of substance produced can be calculated from the number of moles using the molar mass of the substance.
m = n*M
where m is the mass of substance produced, n is the number of moles of the substance produced, and M is the molar mass of the substance.
For silver, the molar mass is 107.87 g/mol.
Using these equations, we can calculate the mass of silver deposited on the cathode as follows:
n = Q/F = 9650 C/96500 C/mol = 0.1 mol
m = n*M = 0.1 mol * 107.87 g/mol = 10.8 g
Therefore, the mass of silver deposited on the cathode is 10.8 g, which is option (b).
Charge passed (Q) = 9650 C
Solution of AgNO3 is used for electrolysis.
To find:
Mass of silver deposited on the cathode
Solution:
AgNO3 dissociates into Ag+ and NO3- ions in solution.
Ag+ ions will move towards the cathode and get reduced to Ag(s) on gaining electrons.
The quantity of electric charge is related to the amount of substance produced by the following equation:
n = Q/F
where n is the number of moles of substance produced, Q is the quantity of electric charge passed through the solution, and F is the Faraday constant (96500 C/mol).
The mass of substance produced can be calculated from the number of moles using the molar mass of the substance.
m = n*M
where m is the mass of substance produced, n is the number of moles of the substance produced, and M is the molar mass of the substance.
For silver, the molar mass is 107.87 g/mol.
Using these equations, we can calculate the mass of silver deposited on the cathode as follows:
n = Q/F = 9650 C/96500 C/mol = 0.1 mol
m = n*M = 0.1 mol * 107.87 g/mol = 10.8 g
Therefore, the mass of silver deposited on the cathode is 10.8 g, which is option (b).
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During the electrolysis of a solution of AgNO₃, 9650 C of charge passes through the electroplating bath, the mass of silver deposited on the cathode will bea)1.08 gb)10.8 gc)21.6 gd)108 gCorrect answer is option 'B'. Can you explain this answer?
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