Faraday's Law: Calculating the Mass of Copper Deposited
One of the most essential topics in the Electrochemistry chapter is Faraday's First Law of Electrolysis. Many students struggle with the formula W = Z × I × t, getting lost in equivalent weights and seconds.
However, there is a much faster and more reliable method: Stoichiometry! Let's solve a classic JEE Main Previous Year Question (PYQ) using Abhishek Sir's direct mole concept approach.
Video Tutorial: The Direct Mole Approach
Watch Abhishek Sengar sir from CHEMCA break down exactly how to relate the charge in Faradays directly to the mass of the metal deposited without memorizing complex formulas.
Step-by-Step Problem Breakdown
Problem Statement:
2 Faradays of electricity are passed through a solution of CuSO4. The mass of copper deposited at the cathode is? (Molar mass of Cu = 63.5 g/mol).
1 Faraday (F) of charge is exactly equal to the charge of 1 Mole of Electrons.
If a reaction requires n moles of electrons to reduce 1 mole of an ion, it will require exactly n Faradays of electricity!
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Ionize the Electrolyte:
Copper Sulphate (CuSO4) breaks down in aqueous solution.
CuSO4 → Cu2+ + SO42- -
Write the Cathode Reduction Reaction:
At the cathode, reduction (gain of electrons) takes place. The Cu2+ ions will gain electrons to form solid Copper metal.
Cu2+ + 2e- → Cu(s) -
Read the Stoichiometry:
From the balanced half-reaction above, we can clearly read:
1 Mole of Cu2+ accepts 2 Moles of electrons to deposit 1 Mole of Cu metal. -
Convert to Faradays and Mass:
Since 2 moles of electrons = 2 Faradays (2F), and 1 mole of Copper weighs 63.5 grams, we can state:
Passing 2 Faradays of charge will exactly deposit 63.5 g of Copper.
Fig: Logical flow from Charge to Mass using Moles.
Because the question specifically asks what happens when exactly 2 Faradays are passed, the answer is directly 63.5 grams! No extra multiplication or division needed.
Practice Questions for JEE & NEET
Apply this exact same logic to these variations. Always start by writing the reduction half-reaction!
Question 1: If only 1 Faraday of electricity was passed through the same CuSO4 solution, what mass of Copper would be deposited?
Answer: 31.75 g
Reasoning:
- We established that 2 Faradays deposit 1 mole of Cu (63.5 g).
- Using simple unitary method: If the charge is halved to 1 Faraday, the mass deposited will also be halved.
- Mass = 63.5 / 2 = 31.75 g.
Question 2: How many grams of Aluminum (Al) will be deposited at the cathode if 3 Faradays of electricity are passed through a molten solution of AlCl3? (Molar mass of Al = 27 g/mol).
Answer: 27 g
Reasoning:
- Ionization: AlCl3 → Al3+ + 3Cl-.
- Cathode Reaction: Al3+ + 3e- → Al(s).
- This tells us that 3 moles of electrons (3 Faradays) are required to deposit exactly 1 mole of Aluminum.
- Since 1 mole of Aluminum weighs 27 grams, passing 3 Faradays will deposit exactly 27 g of Al.
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