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A solution of Ni(NO3)2 is electrolyzed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
- a)1.403g
- b)1.603g
- c)1.203g
- d)1.803g
Correct answer is option 'D'. Can you explain this answer?
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To determine the mass of Ni deposited at the cathode during electrolysis, we can use Faraday's laws of electrolysis. According to Faraday's laws, the amount of substance deposited or liberated during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
First, we need to calculate the charge passed during the electrolysis using the formula:
Charge (Coulombs) = current (Amperes) x time (seconds)
Given that the current is 5 Amperes and the time is 20 minutes (or 20 x 60 = 1200 seconds), we can calculate the charge passed:
Charge = 5 A x 1200 s = 6000 Coulombs
Next, we need to determine the number of moles of Ni deposited using Faraday's constant. Faraday's constant represents the charge required to deposit one mole of a substance during electrolysis and is equal to 96,485 Coulombs/mol.
Number of moles of Ni = Charge (Coulombs) / Faraday's constant
Number of moles of Ni = 6000 C / 96,485 C/mol = 0.062 moles
Finally, we can calculate the mass of Ni deposited using its molar mass. The molar mass of Ni is 58.69 g/mol.
Mass of Ni = Number of moles of Ni x molar mass of Ni
Mass of Ni = 0.062 moles x 58.69 g/mol = 3.63 g
However, the question asks for the mass of Ni deposited at the cathode, which means we need to account for the fact that Ni(NO3)2 dissociates into two moles of Ni ions. Therefore, the mass of Ni deposited at the cathode is half of 3.63 g.
Mass of Ni deposited at the cathode = 3.63 g / 2 = 1.815 g
Rounded to three significant figures, the mass of Ni deposited at the cathode is 1.803 g, which corresponds to option D.
First, we need to calculate the charge passed during the electrolysis using the formula:
Charge (Coulombs) = current (Amperes) x time (seconds)
Given that the current is 5 Amperes and the time is 20 minutes (or 20 x 60 = 1200 seconds), we can calculate the charge passed:
Charge = 5 A x 1200 s = 6000 Coulombs
Next, we need to determine the number of moles of Ni deposited using Faraday's constant. Faraday's constant represents the charge required to deposit one mole of a substance during electrolysis and is equal to 96,485 Coulombs/mol.
Number of moles of Ni = Charge (Coulombs) / Faraday's constant
Number of moles of Ni = 6000 C / 96,485 C/mol = 0.062 moles
Finally, we can calculate the mass of Ni deposited using its molar mass. The molar mass of Ni is 58.69 g/mol.
Mass of Ni = Number of moles of Ni x molar mass of Ni
Mass of Ni = 0.062 moles x 58.69 g/mol = 3.63 g
However, the question asks for the mass of Ni deposited at the cathode, which means we need to account for the fact that Ni(NO3)2 dissociates into two moles of Ni ions. Therefore, the mass of Ni deposited at the cathode is half of 3.63 g.
Mass of Ni deposited at the cathode = 3.63 g / 2 = 1.815 g
Rounded to three significant figures, the mass of Ni deposited at the cathode is 1.803 g, which corresponds to option D.
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A solution of Ni(NO3)2is electrolyzed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?a)1.403gb)1.603gc)1.203gd)1.803gCorrect answer is option 'D'. Can you explain this answer?
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