In a fuel cell, what is the cathode reaction?
Cathode: O₂ + 4H⁺ + 4e⁻ → 2H₂O .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
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Cathode: O₂ + 4H⁺ + 4e⁻ → 2H₂O .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
Δ E = -(0.059/n) log ([H⁺]₂/[H⁺]₁) , 1.17 - 1.23 = -0.06 = -(0.059/n) log (1/0.1) . -0.06 = -(0.059/n) × 1 , n = (0.059/0.06) ≈ 1 , but cathode reaction O₂ + 4H⁺ + 4e⁻ , so n = 4 .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Nernst Equation and Gibbs Energy and Equilibrium Constant
Cathode: O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq) .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells
Cathode: O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq) .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells
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
Charge = 1 × 4825 = 4825 C . Cathode: PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O , 1 mol PbSO₄ requires 2F. Faradays = (4825/96500) = 0.05 F , Moles = (0.05/2) = 0.025 mol , Mass = 0.025 × 303 = 7.575 g .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells
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