NEET2019ChemistryElectrochemistryActual
For the cell reaction 2 F e 3 + a q + 2 I - a q → 2 F e 2 + a q + I 2 a q E c e l l 0 = 0.24   V at 298   K . The standard Gibbs energy Δ r G - of the cell reaction is: [Given that Faraday constant F = 96500   C m o l - 1 ]
Options
- A- 46.32 k J m o l - 1
- B- 23.16 k J m o l - 1
- C46.32 k J m o l - 1
- D23.16 k J m o l - 1
Correct answer
A. - 46.32 k J m o l - 1
Step-by-step solution
For the cell reaction 2 F e 3 + a q + 2 I - a q → 2 F e 2 + a q + I 2 a q n = 2 F = 96500 E cell o = 0.24   V Δ G o = - n F E cell o = - 2 × 96500 × 0.24 = - 46320   J = - 46.32   k J