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#copper deposition

8 public questions tagged with this topic.

CHEMISTRY

During the electrolysis of aqueous CuSO₄ using copper electrodes, what mass of copper is deposited at the cathode if a c

Given: During the electrolysis of aqueous CuSO₄ using copper electrodes, what mass of copper is deposited at the cathode if a current of 1.5 A flows for 10 minutes? (Molar mass of Cu = 63 g/mol, F = 96500 C/mol) These values define the system as per NCERT data. Formula: Charge, Q = I × t = 1.5 × 600 = 900 C. This is standard NCERT relation. Substitution & Calculation: For Cu²⁺ + 2e⁻ → Cu(s), 2F (2 × 96500 C) deposits 63 g of Cu. Mass = 63 × 900/2 × 96500 approx 0.294 g . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Electrochemistry, Topic: Faraday's first law, charge to reduce Al³⁺ to Al, 3F = 3 × 96500 C.

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Electrochemical Cells - Galvanic Cells and Electrode Potential

A current of 1 A deposits 0.635 g of Cu from CuSO₄ in 1930 s. What is the time required to deposit 0.54 g of Al from Al₂

Cu: Cu²⁺ + 2e⁻ → Cu , Moles = (0.635/63.5) = 0.01 mol , Charge = 0.01 × 2 × 96500 = 1930 C , matches 1 × 1930 . Al: Al³⁺ + 3e⁻ → Al , Moles = (0.54/27) = 0.02 mol , Charge = 0.02 × 3 × 96500 = 5790 C . t = (5790/1) = 5790 s .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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Electrochemical Cells - Galvanic Cells and Electrode Potential

A current of 1 A deposits 0.635 g of Cu from CuSO₄ in 1930 s. What is the time required to deposit 0.54 g of Al from Al₂

Cu: Cu²⁺ + 2e⁻ → Cu , Moles = (0.635/63.5) = 0.01 mol , Charge = 0.01 × 2 × 96500 = 1930 C , matches 1 × 1930 . Al: Al³⁺ + 3e⁻ → Al , Moles = (0.54/27) = 0.02 mol , Charge = 0.02 × 3 × 96500 = 5790 C . t = (5790/1) = 5790 s .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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Corrosion and Applications of Electrochemistry

A current of 0.2 A deposits 0.127 g of Cu from CuSO₄ in 9650 s. What is the current required to deposit 0.27 g of Al fro

Cu: Charge = 0.2 × 9650 = 1930 C , Moles = (0.127/63.5) = 0.002 mol , Charge = 0.002 × 2 × 96500 = 1930 C , matches. Al: Al³⁺ + 3e⁻ → Al , Moles = (0.27/27) = 0.01 mol , Charge = 0.01 × 3 × 96500 = 2895 C . I = (2895/9650) = 0.3 A .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry

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