Practice question
Question
In a fuel cell operating at 298 K, the cell potential decreases from 1.23 V to 1.17 V when the [H⁺] at the cathode increases from 0.1 M to 1 M. What is the number of electrons involved in the cathode reaction?
Explanation
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 .
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