7 public questions tagged with this topic.
Cathode: HgO + H₂O + 2e⁻ → Hg + 2OH⁻ . HgO is the cathode material.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws
Anode: Zn(Hg) + 2OH⁻ → ZnO(s) + H₂O + 2e⁻ . Product is ZnO.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
Cathode: PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O . In PbSO₄, Pb = +2.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells
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
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
Anode: Pb (0) + SO₄²⁻ → PbSO₄ (+2) + 2e⁻ , Change = +2. Cathode: PbO₂ (+4) + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ (+2) + 2H₂O , Change = -2. Total change = |+2| + |-2| = 4 .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells
Charge = 1.5 × 10 × 3600 = 54000 C . Anode: Pb + SO₄²⁻ → PbSO₄ + 2e⁻ . 1 mol PbSO₄ requires 2F. Faradays = (54000/96500) = 0.5596 F , Moles = (0.5596/2) = 0.2798 mol , Mass = 0.2798 × 303 = 84.78 g .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells
