NTA Abhyas NEET2020ChemistryElectrochemistryPractice
A fuel cell involves combustion of the butane 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 k J / m o l What is E ° of a cell?
Options
- A+4.74 V
- B+0.547 V
- C+1.09 V
- D+4.37 V
Correct answer
C. +1.09 V
Step-by-step solution
In the reaction C 4 H 10 g - 10 + 10 + 13 2 O 2 g → 4 C O 2 g + 4 - 4 + 5 H 2 O l Change in oxidation number of carbon = + 16 - ( - 10 ) = + 26 So number of electrons involved in cell process will be 26. E ° = - Δ G ° n F = ( - 2746 ) × 1000 26 × 96500 = + 1.09 V