8 public questions tagged with this topic.
E = E° - (0.059/2) log (PH_₂/[H⁺]²) , [H⁺] = 10⁻³ . E = 0 - 0.0295 log (0.5/10⁻⁶) = 0 - 0.0295 × 5.301 = -0.1564 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
E = E° - (0.059/n) log (1/[H⁺]) , [H⁺] = 10⁻⁷ , n = 1 . E = 0 - 0.059 × 7 = -0.413 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⁺] = 10⁻⁷ , n = 1 . E = 0 - 0.059 × 7 = -0.413 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
E°cell = 0.34 - (-0.44) = 0.78 V . Ecell = 0.78 - (0.059/2) log (0.001/0.1) = 0.78 + 0.059 = 0.839 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
E = E° - (0.059/2) log (PH_₂/[H⁺]²) , [H⁺] = 10⁻⁴ . E = 0 - 0.0295 log (0.1/10⁻⁸) = 0 - 0.0295 × 7 = -0.2065 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
E = E° - (0.059/2) log (PH_₂/[H⁺]²) , [H⁺] = 10⁻⁴ . E = 0 - 0.0295 log (0.1/10⁻⁸) = 0 - 0.0295 × 7 = -0.2065 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⁺] = 10⁻¹⁰ , n = 1 . E = 0 - 0.059 × 10 = -0.59 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
E = E° - (0.059/n) log (1/[H⁺]) , [H⁺] = 10⁻¹⁰ , n = 1 . E = 0 - 0.059 × 10 = -0.59 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
