#coulombs

Electrolytic Cells and Electrolysis and Faraday's Laws

During electrolysis of aqueous Na₂SO₄ with inert electrodes, 0.224 L of gas (STP) is collected at the cathode. How many

Cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻ . 1 mol H₂ (22.4 L) requires 2F. Moles = (0.224/22.4) = 0.01 mol , Charge = 0.01 × 2 × 96500 = 1930 C .

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

How many coulombs are required to deposit 0.81 g of magnesium from MgSO₄ solution? (Atomic mass of Mg = 24 g/mol, F = 96

Mg²⁺ + 2e⁻ → Mg . 1 mol Mg (24 g) requires 2F. Moles = (0.81/24) = 0.03375 mol , Charge = 0.03375 × 2 × 96500 = 6513.75 C .

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

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Conductance - Conductivity Molar Conductivity and Kohlrausch Law

How many coulombs are required to deposit 2.07 g of lithium from a LiCl solution? (Atomic mass of Li = 6.9 g/mol, F = 96

Li⁺ + e⁻ → Li . 1 mol Li (6.9 g) requires 1F. Moles = (2.07/6.9) = 0.3 mol , Charge = 0.3 × 96500 = 28950 C .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law

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Conductance - Conductivity Molar Conductivity and Kohlrausch Law

How many coulombs are required to deposit 2.07 g of lithium from a LiCl solution? (Atomic mass of Li = 6.9 g/mol, F = 96

Li⁺ + e⁻ → Li . 1 mol Li (6.9 g) requires 1F. Moles = (2.07/6.9) = 0.3 mol , Charge = 0.3 × 96500 = 28950 C .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law

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Batteries - Primary Secondary and Fuel Cells

In a dry cell, 0.435 g of MnO₂ (molar mass 87 g/mol) is reduced to MnO(OH). How many coulombs are consumed? (F = 96500 C

MnO₂ + H⁺ + e⁻ → MnO(OH) . Moles = (0.435/87) = 0.005 mol , Charge = 0.005 × 96500 = 482.5 C .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells

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Batteries - Primary Secondary and Fuel Cells

A fuel cell produces 0.448 L of H₂O vapor (STP) at the cathode from O₂. How many coulombs are involved? (F = 96500 C/mol

Cathode: O₂ + 4H⁺ + 4e⁻ → 2H₂O . 2 mol H₂O (44.8 L) requires 4F. Moles = (0.448/22.4) = 0.02 mol , Charge = (0.02/2) × 4 × 96500 = 0.04 × 96500 = 3860 C .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells

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Batteries - Primary Secondary and Fuel Cells

A fuel cell produces 0.448 L of H₂O vapor (STP) at the cathode from O₂. How many coulombs are involved? (F = 96500 C/mol

Cathode: O₂ + 4H⁺ + 4e⁻ → 2H₂O . 2 mol H₂O (44.8 L) requires 4F. Moles = (0.448/22.4) = 0.02 mol , Charge = (0.02/2) × 4 × 96500 = 0.04 × 96500 = 3860 C .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells

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Batteries - Primary Secondary and Fuel Cells

In a lead storage battery during discharge, 0.1515 g of PbSO₄ (molar mass 303 g/mol) is formed at the anode. How many co

Anode: Pb + SO₄²⁻ → PbSO₄ + 2e⁻ . Moles = (0.1515/303) = 0.0005 mol , Charge = 0.0005 × 2 × 96500 = 96.5 C .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells

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Batteries - Primary Secondary and Fuel Cells

In a lead storage battery during discharge, 0.1515 g of PbSO₄ (molar mass 303 g/mol) is formed at the anode. How many co

Anode: Pb + SO₄²⁻ → PbSO₄ + 2e⁻ . Moles = (0.1515/303) = 0.0005 mol , Charge = 0.0005 × 2 × 96500 = 96.5 C .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells

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Batteries - Primary Secondary and Fuel Cells

In a dry cell, the cathode reaction produces MnO(OH). If 0.87 g of MnO₂ (atomic mass Mn = 55, O = 16) is reduced, how ma

MnO₂ + H⁺ + e⁻ → MnO(OH) . Molar mass = 55 + 32 = 87 g/mol . Moles = (0.87/87) = 0.01 mol , Charge = 0.01 × 96500 = 965 C .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells

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Batteries - Primary Secondary and Fuel Cells

In a fuel cell, the anode reaction produces 0.224 L of H₂O vapor at STP from H₂. How many coulombs are involved? (F = 96

Anode: H₂ + 2OH⁻ → 2H₂O + 2e⁻ . 1 mol H₂O (22.4 L as vapor) requires 2F. Moles = (0.224/22.4) = 0.01 mol , Charge = 0.01 × 2 × 96500 = 1930 C .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells

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