Problem 71 - A Complete Resource Book in Chemistry for JEE Main

Question

A certain current liberates \(0.504 \mathrm{~g}\) of hydrogen in 2 hour. The amount (or mass) of copper deposited at the cathode during the electrolysis of \(\mathrm{CuSO}_{4}\) (aq) solution by the same current flowing for the same time is (atomic masses \(\mathrm{H}=1.0, \mathrm{Cu}=63.5\) ) (a) \(16.0 \mathrm{~g}\) (b) \(32.0 \mathrm{~g}\) (c) \(8.6 \mathrm{~g}\) (d) \(1.60 \mathrm{~g}\)

Step-by-Step Solution

Answer

The mass of copper deposited is 16.0 g.

Step 1: Understand the Problem

We are given that a current liberates 0.504 grams of hydrogen in 2 hours. We need to determine the mass of copper that would be deposited by the same current and duration during the electrolysis of \(CuSO_4\) solution.

Step 2: Use Faraday's First Law of Electrolysis

Faraday's First Law states that the mass \(m\) of substance deposited or liberated is proportional to the quantity of electricity (charge \(Q\)) passed and can be expressed as \(m = \frac{Q imes M}{z imes F}\), where \(M\) is the molar mass, \(z\) is the valency number, and \(F\) is Faraday’s constant (approximately 96500 C/mol).

Step 3: Calculate Moles of Hydrogen Liberated

The molar mass of hydrogen is 2 g/mol (since \(H_2\) is liberated). Therefore, the moles of hydrogen liberated is \( \text{moles of } H_2 = \frac{0.504}{2} = 0.252 \text{ moles} \).

Step 4: Determine Charge Passed for Hydrogen

Since hydrogen \((H_2)\) involves a charge of 2 Faraday per mole (because \(H_2\) is 2H\^- forming one \(H_2).\), the total charge \(Q\) used is \(0.252 \, \text{mol} \times 2 \, F\), which is \(0.504 \, F\).

Step 5: Calculate Copper Deposition

Copper \( (Cu^{2+})\) has a valency of 2, so it also requires 2 Faraday per mole. The number of moles of copper deposited using \(0.504 \, F\) is \( \frac{0.504}{2} = 0.252 \text{ moles} \).

Step 6: Calculate Mass of Copper Deposited

Multiply the moles of copper by its molar mass (63.5 g/mol): \( 0.252 \, \text{mol} \times 63.5 \, \text{g/mol} = 15.99 \, \text{g} \).

Step 7: Final Answer Determination

Round the copper mass deposited to the nearest value from the provided options. Therefore, the amount of copper deposited is approximately 16.0 g.

Key Concepts

Faraday's First Law

Faraday's First Law of Electrolysis is a fundamental concept in understanding how substances are deposited or liberated during electrolysis. This law states that the mass of an element deposited at an electrode is directly proportional to the quantity of electricity (charge) passing through the electrolyte. In simpler terms, the more electricity you pass through an electrolytic solution, the more substance you'll deposit.
The formula for this law is \[ m = \frac{Q \times M}{z \times F} \] where \( m \) is the mass of the substance deposited, \( Q \) is the total electric charge passed through the solution, \( M \) is the molar mass of the substance, \( z \) is the valency number (number of electrons exchanged), and \( F \) is Faraday's constant (approximately 96500 C/mol).

• Proportionality: Indicates the direct relationship between substance mass and electrical charge.
• Valency & Molar Mass: Key players in determining the deposition quantity.

Copper Deposition

Copper deposition is the process of depositing copper ions onto an electrode during electrolysis. This occurs commonly in solutions such as \( CuSO_4 \) (copper(II) sulfate) where copper ions move toward the cathode.
In this process, the copper ions \((Cu^{2+})\) undergo a reduction, meaning they gain electrons to form copper metal \((Cu)\).
The reaction at the cathode can be represented as: \[ Cu^{2+} + 2e^- \rightarrow Cu \]
For each mole of copper deposited, two moles of electrons are required because copper has a valency of 2. Copper deposition is useful in various applications like electroplating, electroforming, and refining copper metals.
When considering copper deposition:

• Understand the role of valency: Copper is a divalent atom.
• The molar mass of copper helps calculate the total amount deposited.

Molar Mass Calculation

Molar mass is a critical component for solving electrolysis problems since it directly influences the amount of substance deposited.
Molar mass is the mass of one mole of a substance, generally expressed in grams per mole (g/mol).
In this scenario, copper has a molar mass of 63.5 g/mol. This means that one mole of copper atoms weighs 63.5 grams. Knowing this value allows us to convert the number of moles calculated during electrolysis to actual mass.

• Molar mass units: Always expressed as g/mol for easy calculations.
• Conversion use: Turns theoretical calculations into practical mass measurements.

In electrolysis, converting moles to grams correctly ensures accurate predictions and measurements.

Valency in Electrolysis

Valency refers to the ability of an element to combine with other elements, which is determined by the number of electrons it can lose, gain, or share.
In electrolysis, valency plays a crucial role because it dictates how many moles of electrons are required to deposit or dissolve one mole of ions.
This influences the overall calculation of the electrolysis process.
For example, copper \((Cu^{2+})\) has a valency of 2. This indicates that two electrons are needed for one copper ion to be deposited as solid copper on the electrode.

• Valency units: Usually expressed as an integer, indicating electron exchange.
• Significance: Essential for calculating the exact charge needed for reactions.

Understanding valency ensures proper alignment and execution of electrolytic reactions for the desired outcomes.