17 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⁺] = 10⁻² , n = 1 . E = 0 - 0.059 × 2 = -0.118 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
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⁺] = 10⁻⁷ , 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 = 0.00 - (-0.44) = 0.44 V . Ecell = 0.44 - (0.059/2) log (0.1/0.001²) = 0.44 - 0.0885 = 0.3515 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential
E°cell = 0.00 - (-0.44) = 0.44 V . Ecell = 0.44 - (0.059/2) log (0.1/0.001²) = 0.44 - 0.0885 = 0.3515 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⁺]) , [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° - (0.059/2) 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
