Problem 207 - A Complete Resource Book in Chemistry for JEE Main

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 \(\mathrm{Cu}=63.5\) amu) (a) \(0 \mathrm{~g}\) (b) \(63.5 \mathrm{~g}\) (c) \(2 \mathrm{~g}\) (d) \(127 \mathrm{~g}\)

Step-by-Step Solution

Answer

The mass of copper deposited is 63.5 g; thus, the answer is option (b).

Step 1: Understanding the Problem

To find the mass of copper deposited at the cathode, we need to use Faraday's laws of electrolysis. One faraday of electricity will deposit the equivalent mass of a substance, the atomic mass divided by the valency. Copper has a valency of 2.

Step 2: Calculate Equivalent Mass

The equivalent mass of copper (Cu) can be calculated as follows: the atomic mass of copper is given as 63.5 amu, and the valency of copper is 2. Thus, the equivalent mass is \( \frac{63.5}{2} = 31.75 \mathrm{~g/equiv} \).

Step 3: Determine Mass Deposited

Since two faradays of electricity are passed through the solution, the mass of copper deposited will be two times the equivalent mass of copper. Therefore, mass deposited is \( 2 \times 31.75 \mathrm{~g} = 63.5 \mathrm{~g} \).

Step 4: Solution & Conclusion

By following the steps to calculate the mass of copper deposited via electrolysis, we conclude that the correct answer is (b) \( 63.5 \mathrm{~g} \). This is obtained by multiplying two faradays by the equivalent mass per faraday of copper.

Key Concepts

Equivalent Mass

Understanding the concept of equivalent mass is crucial in electrolysis. Equivalent mass is essentially the mass of a substance that will react with or be released by one faraday of electricity. It acts as a bridge between the atomic mass and the behavior of elements during chemical reactions. To find it, we divide the atomic mass by the number of electrons involved in the reaction, which is also known as the valency.
In the case of copper, the atomic mass is 63.5 atomic mass units (amu), and its usual valency is 2. This way of calculating is pivotal because one faraday, by definition, will deposit the equivalent mass of an element. Therefore, for copper, the equivalent mass is \( \frac{63.5}{2} = 31.75 \mathrm{~g/equiv} \).
This property allows us to predict how much of a substance will change during an electrolysis process when a certain amount of charge is passed through a system.

Valency of Copper

Valency is a fundamental concept in chemistry that refers to the combining power of an element. It's the number of electrons an atom can gain, lose, or share when it reacts with other atoms. Knowing the valency of an element is essential for understanding how it will interact in chemical processes like electrolysis.
For copper, its prevalent valency is 2. This means in reactions, like the one in a copper sulfate solution during electrolysis, each copper ion will typically lose two electrons to form metallic copper at the cathode. This attribute is essential for calculating the equivalent mass, as valency is used in the denominator of the equivalent mass equation. By understanding its valency, we thus get a clearer insight into the deposition processes occurring at the electrodes.

Mass of Copper Deposition

To calculate the mass of copper deposition, Faraday’s laws of electrolysis are indispensable. These laws state that the mass of an element deposited is directly proportional to the quantity of electricity passed through the electrolyte. Therefore, for copper's deposition on the cathode, we first find the equivalent mass, which we previously calculated as 31.75 g/equiv.
When two faradays of electricity are applied, the mass deposited can be easily found by multiplying the number of faradays by the equivalent mass. Thus, the equation becomes \(2 \times 31.75 \mathrm{~g}=63.5 \mathrm{~g}\).
This calculation is straightforward with a clear understanding of equivalent mass and its role in electrolysis. In this specific problem, it resolved the initial query and confirmed that passing two faradays of electricity would result in the deposition of 63.5 grams of copper at the cathode.