Class 12 Exam  >  Class 12 Questions  >  How much electricity in terms of Faraday is r... Start Learning for Free
How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl2. (ii) 40.0 g of Al from molten Al2O3.?
Most Upvoted Answer
How much electricity in terms of Faraday is required to produce (i) 20...
(1) Ca2++2e−→Ca


1 mole (40 g) of Ca will require 2 F of electricity.
20 g (0.5 mole) of Ca will require 1 F of electricity.


(2) Al3++3e−→Al


1 mole (27 g) of Al will require 3 moles of electrons.
40 g of Al will require 273×40F​=4.4 F
Community Answer
How much electricity in terms of Faraday is required to produce (i) 20...
Electrolysis and Faraday's Law

Electrolysis is a process in which an electric current is used to cause a chemical reaction. Faraday's laws of electrolysis provide a quantitative relationship between the amount of substance produced or consumed during electrolysis and the amount of electricity passed through the electrolyte.

Faraday's First Law of Electrolysis

Faraday's first law states that the mass of a substance produced or consumed during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte. Mathematically, it can be expressed as:

m = Z * F * Q

where:
m is the mass of substance produced or consumed,
Z is the electrochemical equivalent (the amount of substance produced or consumed per unit of electricity),
F is Faraday's constant (the charge of one mole of electrons, approximately 96,485 C/mol),
Q is the quantity of electricity passed through the electrolyte in coulombs.

Calculating the Quantity of Electricity

To determine the amount of electricity required to produce a given amount of a substance, we must first calculate the quantity of electricity (Q) using Faraday's first law equation:

Q = m / (Z * F)

where:
m is the mass of the substance produced or consumed,
Z is the electrochemical equivalent of the substance.

Calculating the Electrochemical Equivalent

The electrochemical equivalent can be determined experimentally. It represents the amount of substance produced or consumed per unit of electricity passed through the electrolyte. For example, to calculate the electrochemical equivalent of calcium (Ca), we can perform the following steps:

1. Determine the molar mass of calcium (40.08 g/mol).
2. Calculate the number of moles of calcium produced (20.0 g / 40.08 g/mol = 0.499 mol).
3. Using the balanced equation for the electrolysis of molten CaCl2, determine the stoichiometric ratio between calcium and electricity. The equation is:
Ca2+ + 2e- -> Ca
From the equation, we can see that 2 moles of electrons are required to produce 1 mole of calcium.
4. Calculate the electrochemical equivalent of calcium:
Z = 0.499 mol / 2 = 0.2495 mol/C

Calculating the Quantity of Electricity Required

Using the electrochemical equivalent, we can now calculate the quantity of electricity required to produce the given amount of substance. For example:

Q = m / (Z * F)
Q = 20.0 g / (0.2495 mol/C * 96,485 C/mol)
Q ≈ 0.082 C

Therefore, approximately 0.082 coulombs of electricity are required to produce 20.0 grams of calcium from molten CaCl2.

Similarly, the quantity of electricity required to produce 40.0 grams of aluminum (Al) from molten Al2O3 can be calculated using the same steps as outlined above. The electrochemical equivalent of aluminum and the stoichiometric ratio between aluminum and electricity should be determined from the balanced equation for the electrolysis of molten Al2O3.
Explore Courses for Class 12 exam
How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl2. (ii) 40.0 g of Al from molten Al2O3.?
Question Description
How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl2. (ii) 40.0 g of Al from molten Al2O3.? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl2. (ii) 40.0 g of Al from molten Al2O3.? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl2. (ii) 40.0 g of Al from molten Al2O3.?.
Solutions for How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl2. (ii) 40.0 g of Al from molten Al2O3.? in English & in Hindi are available as part of our courses for Class 12. Download more important topics, notes, lectures and mock test series for Class 12 Exam by signing up for free.
Here you can find the meaning of How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl2. (ii) 40.0 g of Al from molten Al2O3.? defined & explained in the simplest way possible. Besides giving the explanation of How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl2. (ii) 40.0 g of Al from molten Al2O3.?, a detailed solution for How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl2. (ii) 40.0 g of Al from molten Al2O3.? has been provided alongside types of How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl2. (ii) 40.0 g of Al from molten Al2O3.? theory, EduRev gives you an ample number of questions to practice How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl2. (ii) 40.0 g of Al from molten Al2O3.? tests, examples and also practice Class 12 tests.
Explore Courses for Class 12 exam
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev