Practice question
Question
A hydrogen electrode operates at 298 K with PH_₂ = 1.5 atm and [H⁺] = 0.0001 M . What is its potential? (Given: E°H⁺/H_₂ = 0.00 V )
Explanation
E = E° - (0.059/2) log (PH_₂/[H⁺]²) . E = 0 - 0.0295 log (1.5/10⁻⁸) = 0 - 0.0295 × 8.176 = -0.2412 V .