For How Long Must A Current Of 1.5 Ampere Be Passed Through An Aqueous Solution Of A Copper Salt During Electrolysis In Order To Deposit 2.50g Of Copper (cu=63.5)
96,485 coulombs of electricity will deposit 63.5/2 or 31.25 g Cu.
Coulombs = amperes x seconds.
How many coulombs must you have? That's
96,485 x 2.5 g/31.25 g = ?
Then ? coulombs = 1.5 x seconds. You know coulombs and amperes. Solve for seconds. Post your work if you get stuck.
For how long a current of 1.5 ampere will be passed through an aqueous solution if copper salts during electrolysis in order to deposits 2.50g of copper ( cu=63.5)
576.5g
To determine the time required for a current of 1.5 Amperes to deposit 2.50 grams of copper (Cu) during electrolysis, we need to use the concept of Faraday's laws of electrolysis.
Faraday's laws state that the mass of the substance produced during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte. The relationship between the mass of a substance (m), the electric current (I), the time (t), and the electrochemical equivalent (Z) can be described by the equation:
m = I * t * Z
Where:
m is the mass of the substance deposited (in grams)
I is the electric current (in Amperes)
t is the time (in seconds)
Z is the electrochemical equivalent (in grams per Coulomb)
To calculate the time required, we need to find the electrochemical equivalent for copper.
The electrochemical equivalent of a substance is the mass of the substance deposited or liberated by the passage of one Coulomb of electric charge. For copper, the electrochemical equivalent is equal to its molar mass divided by the number of electrons involved in the reaction.
The molar mass of copper (Cu) is 63.5 g/mol, and during the electrolysis of copper, each copper ion (Cu²⁺) gains two electrons to form Cu. Hence, the electrochemical equivalent of copper is:
Z = (molar mass of Cu) / (number of electrons involved)
= 63.5 g/mol / 2
= 31.75 g/C
Now, we can rearrange the equation to solve for time (t):
t = m / (I * Z)
= 2.50 g / (1.5 A * 31.75 g/C)
≈ 0.052 seconds
Therefore, it would take approximately 0.052 seconds for a current of 1.5 Amperes to deposit 2.50 grams of copper during electrolysis.