Problem 20 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A solution of copper sulphate \(\left(\mathrm{CuSO}_{4}\right)\) is electrolysed for 10 minutes with a current of \(1.5\) amperes. The mass of copper deposited at the cathode (at. mass of \(\mathrm{Cu}=63 \mathrm{u}\) ) is: [Main Online April 25, 2013] (a) \(0.3892 \mathrm{~g}\) (b) \(0.2938 \mathrm{~g}\) (c) \(0.2398 \mathrm{~g}\) (d) \(0.3928 \mathrm{~g}\)

Step-by-Step Solution

Answer

The correct answer is (b) \(0.2938\, \mathrm{g}\).

Step 1: Write the Chemical Equation

The electrolysis of copper sulfate (CuSO₄) involves the following reaction at the cathode: \( \text{Cu}^{2+} + 2\text{e}^{-} \rightarrow \text{Cu} \). This indicates that two moles of electrons are required to deposit one mole of copper.

Step 2: Use Faraday's First Law of Electrolysis

Faraday's First Law states that the amount of substance deposited during electrolysis is directly proportional to the quantity of electricity passed through the solution. The formula is \( m = \frac{EIt}{nF} \), where \(m\) is the mass of the substance, \(E\) is the equivalent weight, \(I\) is the current, \(t\) is the time, \(n\) is the number of electrons exchanged (2 in this case), and \(F\) is Faraday's constant \(96500\, \mathrm{C/mol}\).

Step 3: Calculate the Equivalent Weight of Copper

Equivalent weight \(E\) of copper \(\text{Cu}\) is calculated as \( \text{atomic mass} / n \). Given atomic mass is \(63 \) and \(n = 2\), \(E = \frac{63}{2} = 31.5\, \mathrm{g/mol}\).

Step 4: Plug in Values to the Formula

Use \(I = 1.5\, \mathrm{A}\), \(t = 10 \times 60 = 600\, \mathrm{s}\), \(n = 2\), and \(F = 96500\, \mathrm{C/mol}\):\[m = \frac{31.5 \times 1.5 \times 600}{2 \times 96500}\]

Step 5: Solve the Equation

Calculate the mass \(m\):\[m = \frac{31.5 \times 1.5 \times 600}{193000} \approx 0.2938\, \mathrm{g}\]Thus, the mass of copper deposited is approximately \(0.2938\, \mathrm{g}\).

Step 6: Determine the Correct Answer

From the options given, option (b) \(0.2938\, \mathrm{g}\) matches the calculated mass. Hence, the correct answer is (b) \(0.2938\, \mathrm{g}\).

Key Concepts

Faraday's Law of Electrolysis

Faraday's Law of Electrolysis is a fundamental concept in electrochemistry that provides a clear understanding of how electrical currents interact with chemical compounds.
It states that the amount of substance deposited at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the solution.
This means that if you increase the current or the time of electrolysis, more substance will be deposited. In mathematical terms, this relationship is defined as: \[ m = \frac{EIt}{nF} \]Where:

• \( m \) is the mass of the substance deposited.
• \( E \) is the equivalent weight of the substance.
• \( I \) is the current in amperes.
• \( t \) is the time in seconds.
• \( n \) is the number of electrons exchanged.
• \( F \) is Faraday's constant, approximately \( 96500 \, \mathrm{C/mol} \).

To apply this law, you'd need to know the equivalent weight, which can be found by dividing the atomic mass by the number of electrons involved in the reaction.
This makes Faraday's Law a practical tool for calculating how much of a substance can be obtained from a given amount of current over a specific time.

Copper Electrolysis

Copper electrolysis is a common process used to purify copper from a solution of copper sulfate (CuSO₄).
During the process, copper ions (\( \text{Cu}^{2+} \)) in the solution are reduced at the cathode by electrons to form solid copper metal. The chemical reaction occurring at the cathode is:\[ \text{Cu}^{2+} + 2\text{e}^{-} \rightarrow \text{Cu} \]This indicates that two moles of electrons are necessary to reduce one mole of copper ions to copper metal. In practical applications, copper electrolysis is used in copper refining, where impure copper is purified.
It ensures that the copper deposited at the cathode is of high purity, removing other impurities present in the ore. For the calculations:

• The atomic mass of copper is \( 63 \, \mathrm{u} \).
• Due to the transfer of two electrons, the equivalent weight \( E \) is \( \frac{63}{2} = 31.5 \, \mathrm{g/mol} \).

When electricity is passed through the solution, copper ions gain electrons and deposit as metallic copper at the cathode.
This ability to extract and purify metals makes electrolysis a valuable process in various industries.

Electrochemical Reactions

Electrochemical reactions are reactions that involve the transfer of electrons between chemical species.
They occur in systems called electrochemical cells, where two reactions—oxidation and reduction—take place at two different electrodes. In electrochemical reactions:

• Oxidation occurs at the anode, where a species loses electrons.
• Reduction occurs at the cathode, where a species gains electrons.

Copper electrolysis is an example of such a reaction.
Here, copper ions in the solution gain electrons and are reduced to form solid copper at the cathode. These reactions are harnessed in various applications:

• Electroplating, where a thin metal layer coats an object.
• Electrorefining, such as in copper purification.
• Electrochemical sensors, which detect chemical changes.

Understanding these reactions is essential in fields like chemistry and materials science, where the focus is on the manipulation of chemical properties and reactions.
This deeper understanding makes it possible to conduct industrial processes more efficiently and develop innovative technologies.