197 public questions tagged with this topic.
Positive E° indicates Cu cannot displace H 2 from acids, as it is less reactive than hydrogen.
Ref: NCERT Class 12 Chemistry > Chapter 4: d and f Block Elements > Topic: Applications of d and f Block Elements
E°cell = 0.00 - (-0.76) = 0.76 V . Ecell = 0.76 - (0.059/2) log (0.001/1²) = 0.76 + 0.0885 = 0.8485 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
Ecell = E°cell - (0.059/n) log Q , 1.50 = 1.44 - (0.059/2) log Q . 0.06 = -0.0295 log Q , log Q = -(0.06/0.0295) = -2.034, Q = 10⁻².⁰³⁴ ≈ 0.0092 . Since Q = ([Anode]/[Cathode]) = (0.001/0.1) = 0.01 , matches approximately.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
Cathode: O₂ + 4H⁺ + 4e⁻ → 2H₂O .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
E°cell = (0.059/n) log Kc . 0.236 = (0.059/2) log Kc , log Kc = 8, Kc = 10⁸ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
Δr G° = -n F E°cell , -142610 = -n × 96500 × 0.74 . n = (142610/96500 × 0.74) = (142610/71410) ≈ 2 .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
E°cell = 0.80 - (-0.28) = 1.08 V . Ecell = 1.08 - (0.059/2) log ([Co²⁺]/[Ag⁺]²) = 1.08 - 0.0295 log (0.01/0.000001) = 1.08 - 0.177 = 0.903 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) , [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°cell = 0.34 - (-0.25) = 0.59 V . Ecell = 0.59 - (0.059/2) log (0.1/0.001) = 0.59 - 0.059 = 0.531 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
E°cell = (0.059/n) log Kc , 0.59 = (0.059/n) × 20 , n = (1.18/0.59) = 2 .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
Al³⁺ + 3e⁻ → Al . 1 mol Al (27 g) requires 3F. Moles = (0.27/27) = 0.01 mol , Charge = 0.01 × 3 × 96500 = 2895 C .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
E°cell = E°cathode - E°anode = -0.25 - (-0.28) = 0.03 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement
