NTA Abhyas NEET2020ChemistryElectrochemistryPractice
The rusting of iron takes place as 2 H + + 2 e + 1 2 O 2 → H 2 O ( l ) ; E ° = + 1.23 V F e 2 + + 2 e → F e ( s ) ; E ° = - 0.44 V Thus, Δ G ° for the net process is
Options
- A-322 kJ/mol
- B-161 kJ/mol
- C-1522 kJ/mol
- D-76 kJ/mol
Correct answer
A. -322 kJ/mol
Step-by-step solution
E ° = E O P F e o + E R P H 2 O o = 0.44 + 1.23 = 1.67 V So Δ G ° = - nFE ° = - 2 × 1.67 × 96500 = - 322.31 kJ/mole.