Problem 13 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Two Faraday of electricity is passed through a solution of \(\mathrm{CuSO}_{4}\). The mass of copper deposited at the cathode is (at. mass of \(\mathrm{Cu}=63.5 \mathrm{amu}\) ) [Main 2015] (a) \(2 \mathrm{~g}\) (b) \(127 \mathrm{~g}\) (c) \(0 \mathrm{~g}\) (d) \(63.5 \mathrm{~g}\)

Step-by-Step Solution

Answer

The mass of copper deposited is 63.5 g (option d).

Step 1: Understanding the Problem

We have a solution of \( \mathrm{CuSO}_4 \) and we need to find the mass of copper deposited when two Faradays of electricity are passed through it. We are given the atomic mass of copper as \( 63.5 \, \mathrm{amu} \).

Step 2: Identify Faraday's Law of Electrolysis

According to Faraday's second law of electrolysis, the mass \( m \) of a substance deposited at an electrode is given by \( m = \frac{E \times Q}{96500} \), where \( E \) is the equivalent weight of the substance and \( Q \) is the total charge passed.

Step 3: Calculate Charge Passed

One Faraday of electricity is equivalent to \( 96500 \) coulombs. Therefore, two Faradays is \( 2 \times 96500 = 193000 \) coulombs.

Step 4: Determine Equivalent Weight of Copper

The equivalent weight \( E \) of copper in the reaction \( \mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu} \) is \( \frac{M}{n} \), where \( M \) is the molar mass of copper (\( 63.5 \, \mathrm{g/mol} \)) and \( n \) is 2 because copper is divalent in this case. Therefore, \( E = \frac{63.5}{2} = 31.75 \, \mathrm{g/equiv} \).

Step 5: Apply Faraday's Law to Find Mass

Substitute the values into Faraday's Law equation: \( m = \frac{31.75 \times 193000}{96500} \). This simplifies to \( m = \frac{31.75 \times 2}{1} = 63.5 \, \mathrm{g} \).

Step 6: Conclude the Solution

The mass of copper deposited at the cathode is \( 63.5 \, \mathrm{g} \). This matches option (d).

Key Concepts

Copper Deposition

Copper deposition refers to the process in which copper ions in a solution collect on an electrode, forming a solid layer. This is usually observed in electrochemical cells where copper is chosen as the substance to be deposited from its ionic state in solution.

• In our specific context, we have a solution of copper sulfate (\( \mathrm{CuSO}_4 \)), where copper ions (\( \mathrm{Cu}^{2+} \)) are reduced at the cathode to form metallic copper.
  
• This process is governed by Faraday's Laws of Electrolysis, which relate the quantity of electric charge passed through the solution to the amount of substance deposited.
  
• The efficiency and amount of copper deposited depend on factors such as current and time, which determine how much charge passes through the solution.
  

Understanding copper deposition is significant in applications such as electroplating and the purification of copper through the method of electrorefining.

Electrochemistry

Electrochemistry is the branch of chemistry that deals with the relationship between electrical energy and chemical changes. In the context of the exercise, electrochemistry explains how copper ions are deposited as solid copper on the cathode.

• Chemical reactions in an electrochemical cell involve oxidation and reduction processes, also known as redox reactions.
  
• Here, the copper ions (\( \mathrm{Cu}^{2+} \)) gain electrons (are reduced) at the cathode, thus depositing copper metal.
  
• The principles of electrochemistry help calculate how different conditions such as potential difference, type of electrolyte, and concentration can affect the efficiency of the deposition process.
  

Electrochemistry is foundational in many areas, including battery technology, electrolysis processes, and corrosion prevention.

Charge and Mass Relationship

The relationship between charge and mass is central to understanding how electric current can lead to the deposition of a solid substance in electrochemistry. Faraday's Law of Electrolysis provides a quantitative description of this relationship:

• According to the law, the mass (\( m \)) of a substance deposited is directly proportional to the total electric charge (\( Q \)) that passes through the electrolyte.
  
• The constant of proportionality is the equivalent weight (\( E \)) of the substance, which is derived from the molar mass divided by the valency of the ions.
  
• In our problem, the formula \( m = \frac{E \times Q}{96500} \) is used, where \( 96500 \) is the Faraday constant, representing the charge of one mole of electrons.
  

Through accurate calculations using this formula, one can determine the precise mass of copper deposited given a certain charge, highlighting the precision of electrochemical processes.