Calculate the current that must be passed into a solution of copper(ii)salt for 70sec in order to deposit 29gram of copper metal which atomic mass of copper is for 64
96,485 coulombs will deposit 64/2 = 32 g Cu so it will take
96,485 x (29/32) = approx 87,000 coulombs needed BUT you must solve it more acc accurately than that.
coulombs = amperes x seconds
Substitute coulombs and time and solve for amperes.
To determine the current required to deposit 29 grams of copper metal from a copper(II) salt solution, we can use Faraday's law of electrolysis.
Step 1: Calculate the number of moles of copper metal.
The molar mass of copper (Cu) is 64 grams/mol.
Number of moles = mass / molar mass
Number of moles = 29 grams / 64 grams/mol
Number of moles = 0.4531 moles
Step 2: Determine the number of electrons required to deposit one mole of copper.
From the balanced equation of the electrolysis of copper(II) salt, we know that 2 moles of electrons are required to produce one mole of copper metal. Therefore, the number of electrons required to deposit 0.4531 moles of copper is:
Number of electrons = 0.4531 moles × 2
Number of electrons = 0.9062 moles
Step 3: Calculate the total charge required.
One mole of electrons carries a charge of 1 Faraday, which is equal to 96,485 Coulombs.
Total charge required = number of electrons × charge per electron
Total charge required = 0.9062 moles × 96,485 C/mol
Total charge required = 87,516 C
Step 4: Calculate the current.
The current is given by the formula I = Q / t, where I is the current, Q is the charge, and t is the time.
I = 87,516 C / 70 s
I = 1,250 Amps
Therefore, to deposit 29 grams of copper metal from a copper(II) salt solution, a current of 1,250 Amps must be passed for 70 seconds.
To calculate the current required to deposit a certain amount of copper metal, we can use Faraday's laws of electrolysis.
First, we need to determine the number of moles of copper metal that need to be deposited. This can be done using the atomic mass of copper:
Atomic mass of copper = 64 g/mol
Moles of copper = Mass of copper / Atomic mass of copper
Moles of copper = 29 g / 64 g/mol
Moles of copper ≈ 0.4531 mol
According to Faraday's first law of electrolysis, the amount of substance deposited on each electrode is directly proportional to the quantity of electricity passed through the electrolyte:
Amount of substance (in mol) = (Electric charge (in coulombs) / Faraday's constant)
Faraday's constant (F) is equal to 96,485 coulombs.
To calculate the electric charge (Q), we can use the equation:
Q = I * t
Where:
Q = Electric charge in coulombs
I = Current in amperes
t = Time in seconds
Now, let's calculate the current (I) required to deposit 0.4531 mol of copper in 70 seconds:
Q = I * t
Q = (I) * (70 sec)
Using Faraday's constant (F):
0.4531 mol = (I * 70 sec) / 96,485 C/mol
Simplifying the equation:
I = (0.4531 mol * 96,485 C/mol) / 70 sec
I ≈ 623.3 A
Therefore, the approximate current that must be passed into the solution for 70 seconds in order to deposit 29 grams of copper metal is 623.3 amperes.