4 public questions tagged with this topic.
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
E = E° - (0.059/n) log (1/[H⁺]) , [H⁺] = 10⁻⁵ , n = 1 . E = 0 - 0.059 × 5 = -0.295 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
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
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
