Assertion (A): Passage of 48,250 coulombs of electricity through cupric sulphate and ferrous sulphate solutions result in the deposition \(0.5\) moles each of iron and copper at the respective cathodes. Reason (R): Number of equivalents of a metal deposited at cathode is equal to the number of faradays of electricity passed through the electrolyte. (a) Both "A" and "R" are true and "R" is the correct explanation of "A". (b) Both "A" and "R" are true and "R" is not the correct explanation of "A". (c) "A" is correct and "R" is wrong. (d) "A" is wrong and "R" is correct.

Short Answer

Expert verified
Explain your answer using Faraday's laws of electrolysis. Answer: No, the passage of 48,250 coulombs of electricity results in the deposition of 0.25 moles each of iron and copper at the respective cathodes, not 0.5 moles. This is determined using Faraday's laws of electrolysis, which states that the number of equivalents of a metal deposited at the cathode is equal to the number of faradays of electricity passed through the electrolyte.

Step by step solution

01

Analyze Assertion (A)

First, we need to determine if the passage of 48,250 coulombs of electricity through cupric sulfate and ferrous sulfate solutions result in the deposition of 0.5 moles each of iron and copper at the respective cathodes. Copper (Cu) has a valency of +2, and hence its equivalent weight is atomic mass / 2 = 63.5 / 2. Iron (Fe) also has a valency of +2, and its equivalent weight is atomic mass / 2 = 55.8 / 2. Using Faraday's Law, we know that the number of equivalents = charge passed (in coulombs) / (n * Faraday's constant), where n is the valency of the metal. Faraday's constant is approximately 96,500 C/mol.
02

Calculate the number of equivalents of Cu and Fe deposited on the respective cathodes

For copper: Number of equivalents of Cu = Charge passed / (2 * 96,500) = 48,250 / (2 * 96,500) = 0.25 mol For iron: Number of equivalents of Fe = Charge passed / (2 * 96,500) = 48,250 / (2 * 96,500) = 0.25 mol Since we have calculated the number of moles of each metal deposited at the cathodes and it doesn't match the amount stated in Assertion (A), we can conclude that Assertion (A) is incorrect.
03

Analyze Reason (R)

Now let's analyze the Reason (R). The statement says that "Number of equivalents of a metal deposited at cathode is equal to the number of faradays of electricity passed through the electrolyte." This statement describes Faraday's Law of electrolysis and it's correct. According to Faraday's law, the amount of substance transformed in an electrolytic cell is proportional to the amount of electrical charge passed through the cell.
04

Determine the relationship between "A" and "R"

Now we have the conclusions about Assertion (A) and Reason (R). 1. Assertion (A) is incorrect, as the passage of 48,250 coulombs of electricity will not result in the deposition of 0.5 moles each of iron and copper at the respective cathodes, but instead 0.25 moles each. 2. Reason (R) is correct, as it accurately describes Faraday's Law of electrolysis. Therefore, the correct answer is (d) "A" is wrong and "R" is correct.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Ratio of number of Faradays of electricity required to deposit magnesium, aluminium and sodium in equimolar ratio on electrolysis of their respective molten salts is (a) \(2: 3: 2\) (b) \(2: 1: 1\) (c) \(2: 3: 1\) (d) \(4: 6: 1\)

Pairs of electrodes and their corresponding standard electrode potentials are given. Arrange the cells constructed by these electrodes in increasing order of emf values. (1) Electrode \(\rightarrow \mathrm{A}, \mathrm{E}^{\circ}=-2.92 \mathrm{~V}\) and Electrode \(\rightarrow\) \(\mathrm{B}, \mathrm{E}^{\circ}=-2.71 \mathrm{~V}\) (2) Electrode \(\rightarrow \mathrm{C}, \mathrm{E}^{\circ}=-0.760 \mathrm{~V}\) and Electrode \(\rightarrow\) \(\mathrm{D}, \mathrm{E}^{\circ}=-0.44 \mathrm{~V}\) (3) Electrode \(\rightarrow \mathrm{B}, \mathrm{E}^{\circ}=-2.71 \mathrm{~V}\) and Electrode \(\rightarrow\) \(\mathrm{D}, \mathrm{E}^{\circ}=-0.44 \mathrm{~V}\) (4) Electrode \(\rightarrow \mathrm{A}, \mathrm{E}^{\circ}=-2.92 \mathrm{~V}\) and Electrode \(\rightarrow\) \(\mathrm{C}, \mathrm{E}^{\circ}=-0.76 \mathrm{~V}\) (a) 2143 (b) 1243 (c) 2413 (d) 2431

State whether the following statements are true or false. The negative sign of the reduction potential indicates that reduction occurs at that electrode when connected to the standard hydrogen electrode.

In a Daniel cell, if \(A\left(E_{0}=-0.76 v\right)\) and \(B(=-2.36 v)\) half- cells are taken then (a) \(\mathrm{B}\) acts as an anode (b) \(\mathrm{A}\) acts as an anode (c) \(\mathrm{B}\) acts as a cathode (d) cannot be predicted

The aqueous solution of non-electrolytes contains (a) atoms (b) ions (c) electrons (d) molecules

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free