NTA Abhyas JEE Main2020ChemistryElectrochemistryPractice
Consider the following cell reaction 2Fe s + O 2 g + 4 H + aq → 2F e 2 + aq + 2 H 2 O l E o = 1 . 67 V At F e 2 + = 1 0 - 3 M , P O 2 = 0.1 atm and pH = 3 , the cell potential at 25 o C is
Options
- A1.77 V
- B1.57 V
- C1.87 V
- D1.47 V
Correct answer
B. 1.57 V
Step-by-step solution
2Fe s + O 2 g + 4 H + aq → 2F e 2 + aq + 2 H 2 O l N = 4 (no. of moles of electrons involved) From Nernst equation, E c e l l = E c e l l o - 0.0591 n log Q = 1.67 - 0.0591 4 log log 1 0 - 3 2 0.1 × 1 0 - 3 4 H + = 1 0 - p H = 1 . 67 - 0.106 = 1 . 57 V