NTA Abhyas NEET2020ChemistryElectrochemistryPractice
A fuel cell involves combustion of the butane at at 1 atm and 298 K C 4 H 10 ( g ) + 13 2 O 2 ( g ) → 4 C O 2 ( g ) + 5 H 2 O ( l ) Δ G ° = - 2746 kJ/mole The value of E c e l l o is nearly?
Options
- A0.8 V
- B1 V
- C1.2 V
- D1.4 V
Correct answer
B. 1 V
Step-by-step solution
C 4 - 10 H 10 ( g ) + 13 2 O 2 ( g ) → 4 C O 2 + 4 ( g ) + 5 H 2 O ( l ) Here total change in oxidation number of C-atoms = + 16 - ( - 10 ) = + 26 It means total number of e - involved = 26 As E c e l l o - - Δ G ° n F = - ( - 2746 ) × 1000 26 × 96500 = + 1.09 V ≈ 1.0 V