Problem 49 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

In an electrolysis experiment, current, was passed for 5 hours through two cells connected in series. The first cell contains a solution of gold and the second contains copper sulphate solution. \(9.85 \mathrm{~g}\) of gold was deposited in the first cell. If the oxidation number of gold is \(+3\), find the amount of copper deposited on the cathode of the second cell. Also calculate the magnitude of the current in amperes. (1 faraday \(=96,500\) coulombs \()\)

Step-by-Step Solution

Answer

47.65 g of copper is deposited; the current is approximately 8.05 A.

Step 1: Understanding Faraday's Law of Electrolysis

Faraday's first law of electrolysis states that the mass of a substance deposited or liberated at an electrode is directly proportional to the amount of electricity passed through the electrolyte. This is mathematically expressed as \( m = \frac{Q}{F} \times \text{(Equivalent Mass)} \) where \( Q \) is the charge in coulombs, \( F \) is the Faraday constant, and the Equivalent Mass is \( \frac{M}{n} \) where \( M \) is the molar mass and \( n \) is the valency.

Step 2: Calculate Charge Using Gold Deposition

To find the charge \( Q \), use the mass of gold deposited. Given mass of gold \( m = 9.85 \mathrm{~g} \), the molar mass \( M = 197 \text{ g/mol} \), and valency \( n = 3 \). Calculate equivalent mass, \( \frac{197}{3} \approx 65.67 \). Then, \( 9.85 = \frac{Q}{96500} \times 65.67 \) giving \( Q \approx \frac{9.85 \times 96500}{65.67} \approx 144974 \) C.

Step 3: Calculate Mass of Copper Deposited

Use the charge calculated to find the mass of copper deposited. The equivalent mass of copper is \( \frac{63.5}{2} \approx 31.75 \) since copper's oxidation state is \( +2 \). Then, \( m_{Cu} = \frac{Q}{F} \times \text{Eq. Mass Cu} = \frac{144974}{96500} \times 31.75 \approx 47.65 \) g of copper.

Step 4: Calculate Current Per Hour

To find the current, use the formula \( Q = I \times t \) where \( t = 5 \times 3600 \) s. Rearrange to find \( I = \frac{Q}{t} = \frac{144974}{18000} \approx 8.05 \) A.

Key Concepts

Faraday's Law

Faraday's Law of electrolysis forms the foundation for understanding how substances are deposited or liberated in electrolytic processes. This principle helps us calculate how much material we'll find at an electrode after an electrical current passes through a solution.
Faraday's First Law states that the mass of a substance changed at an electrode during electrolysis is proportional to the total electric charge passing through the circuit.

• It is mathematically expressed as:
  \[ m = \frac{Q}{F} \times \text{(Equivalent Mass)} \]
  where \( m \) is the mass of the substance, \( Q \) is the total electric charge, \( F \) is Faraday's constant, and Equivalent Mass is defined as \( \frac{M}{n} \).
• The law helps predict the outcome of an electrolysis experiment based on the amount of energy used.

This relationship is crucial for designing and controlling different electrochemical processes, as it allows us to calculate exactly how much of a substance will be deposited or consumed at an electrode.

Equivalent Mass

Equivalent mass is a key concept in electrolysis that ties into Faraday's Law by determining how a substance will respond to an electrical current. It is a way of translating the chemical nature of a substance into a form that can be plugged into the equations of Faraday's Law.

• It is calculated using the formula:
  \[ \text{Equivalent Mass} = \frac{M}{n} \]
  where \( M \) is the molar mass and \( n \) is the valency (number of moles of electrons involved in the redox reaction).
• This value allows us to determine how much of a substance (in grams) will either be deposited or dissolved per Faraday of electricity.

Understanding equivalent mass ensures we can predict how much of a substance is involved in a chemical change during electrolysis. It is critical when working with various substances, each having different chemical characteristics.

Oxidation Number

The oxidation number, also known as oxidation state, is an indicator of the degree of oxidation (loss of electrons) of an atom in a chemical compound. It is essential in balancing chemical reactions, particularly redox reactions, where electrons are transferred between atoms.
Knowing the oxidation number:

• Helps in identifying the element that is oxidized (loses electrons) and the one that is reduced (gains electrons).
• Aids in realizing which reactions are feasible under given conditions, which is crucial for the process of electrolysis.

For example:

• In our electrolysis example, gold with an oxidation number of \(+3\) means each atom donates three electrons during its deposition.
• Copper has an oxidation number of \(+2\), meaning it accepts two electrons per atom in its electrolyte.

The oxidation number helps piece the puzzle together in an electrochemical reaction, allowing calculation of how much material will change form.

Electrochemical Cells

Electrochemical cells are devices capable of either generating electrical energy from chemical reactions or facilitating chemical reactions through the introduction of electrical energy.

• They consist of two main types: Voltaic (or Galvanic) cells, which produce electricity, and Electrolytic cells, where external electrical energy drives a non-spontaneous reaction.
• In the context of electrolysis, we're dealing with electrolytic cells. Here is where energy is applied to drive a chemical change.

This setup involves:

• A cathode, where reduction occurs (electron gain).
• An anode, where oxidation happens (electron loss).

In our experiment, gold deposition occurs at one cathode and copper deposition at another. These processes illustrate how electrochemical cells allow us to convert electrical energy into a tangible form of chemical energy, resulting in metal deposition. Understanding the function and structure of electrochemical cells is essential for harnessing these processes in areas such as material recovery and refining.