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#Faraday's law

51 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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CHEMISTRY

What is the mass of nickel deposited at the cathode when a 5 A current flows through Ni(NO₃)₂ solution for 20 minute

Given: What is the mass of nickel deposited at the cathode when a 5 A current flows through Ni(NO₃)₂ solution for 20 minutes? (Molar mass of Ni = 58.7 g/mol, F = 96500 C/mol) These values define the system as per NCERT data. Formula: Q = 5 × 1200 = 6000 C. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For Ni²⁺ + 2e⁻ → Ni(s), 2F (193000 C) deposits 58.7 g. Mass = 58.7 × 6000/193000 approx 1.824 g . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Solutions, Topic: Concentration terms and colligative properties.

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Variation of Conductivity with Concentration and Measurement

What is the time (in seconds) required to deposit 0.355 g of cobalt from a CoSO₄ solution using a current of 0.2 A? (Ato

Co²⁺ + 2e⁻ → Co . 1 mol Co (59 g) requires 2F. Moles = (0.355/59) = 0.006017 mol , Charge = 0.006017 × 2 × 96500 = 1161.24 C . t = (Q/I) = (1161.24/0.2) = 5806.2 s .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement

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Electrolytic Cells and Electrolysis and Faraday's Laws

In electrolysis of molten MgCl₂, 0.24 g of Mg (atomic mass 24 g/mol) is deposited. What volume of Cl₂ gas (STP) is produ

Cathode: Mg²⁺ + 2e⁻ → Mg . Moles = (0.24/24) = 0.01 mol , Charge = 0.01 × 2 × 96500 = 1930 C . Anode: 2Cl⁻ → Cl₂ + 2e⁻ . Moles Cl₂ = 0.01 mol , Volume = 0.01 × 22.4 = 0.224 L .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws

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Electrolytic Cells and Electrolysis and Faraday's Laws

A dry cell delivers 0.1 A for 19300 s. What mass of MnO₂ (molar mass 87 g/mol) is reduced at the cathode? (F = 96500 C/m

Charge = 0.1 × 19300 = 1930 C . Cathode: MnO₂ + H⁺ + e⁻ → MnO(OH) , 1 mol MnO₂ requires 1F. Faradays = (1930/96500) = 0.02 F , Moles = 0.02 mol , Mass = 0.02 × 87 = 1.74 g .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws

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Electrolytic Cells and Electrolysis and Faraday's Laws

A current of 0.5 A deposits 0.3175 g of Cu from CuSO₄ in 3860 s. What mass of Mg would be deposited from MgCl₂ with the

Charge = 0.5 × 3860 = 1930 C . Cu: Cu²⁺ + 2e⁻ → Cu , Moles = (0.3175/63.5) = 0.005 mol , Charge = 0.005 × 2 × 96500 = 1930 C , matches. Mg: Mg²⁺ + 2e⁻ → Mg , Moles = (1930/96500 × 2) = 0.01 mol , Mass = 0.01 × 24 = 0.24 g .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws

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