A hydrogen electrode in a solution with PH_₂ = 2 atm and [H⁺] = 0.002 M operates at 298 K. What is its potential? (Given
E = E° - (0.059/2) log (PH_₂/[H⁺]²) . E = 0 - 0.0295 log (2/0.000004) = 0 - 0.0295 × 5.699 = -0.1681 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
