#standard hydrogen electrode

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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 .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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Corrosion and Applications of Electrochemistry

E = E° - (0.059/n) log (1/[H⁺]) , n = 1 . E = 0 - 0.059 × 3 = -0.177 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry

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Corrosion and Applications of Electrochemistry

E = E° - (0.059/n) log (1/[H⁺]) , n = 1 . E = 0 - 0.059 × 3 = -0.177 V .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry

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