Answer :
Final answer:
The correct answer is b) 50.6 minutes. The time to deposit 1.27 g of copper at the cathode when a current of 2 A is passed through a CuSO4 solution is 50.6 minutes. This calculates by using the formula derived from Faraday's laws of electrolysis.
Explanation:
The correct answer is b) 50.6 minutes. The question involves the calculation of the time required to deposit a certain mass of copper using a given current during an electroplating process. This process is commonly examined in electrolysis, which is a part of high school chemistry curricula.
First, using the Faraday's laws of electrolysis, we know that the amount of substance that is deposited (or dissolved) at each electrode is directly proportional to the amount of electricity that passes through the electrolyte. The mass \m\ of an ion involved in the electrode reaction is given by:
\m = (ItM)/(nF)\
For copper (Cu) deposition, copper ions (Cu2+) obtain two electrons to become copper atoms. Therefore, \(n=2\). The molar mass \(M\) of copper is approximately 63.546 g/mol.
So we have the following equation for copper:
\t = (m imes nF)/(IM)\
Plugging in our values:
\t = (1.27 g imes 2 imes 96500 C/mol)/(2 A imes 63.546 g/mol)\
Solving for \(t\):
\t = (1.27 imes 2 imes 96500)/(2 imes 63.546)\
\t = 3036.95 s\ which is approximately \50.6 minutes\ (because 1 minute = 60 seconds).
Therefore, the time to deposit 1.27 g of copper at the cathode while passing a current of 2 A through the solution of CuSO4 is \(\50.6 minutes\\).
