NTA Abhyas JEE Main2020ChemistryElectrochemistryPractice
E.M.F. diagram for some ions is given as F e O 4 2 - → E o = + 2.20 V F e 3 + → E o = + 0.77 V F e 2 + → E o = - 0.445 V F e o Determine the value of E F e O 4 2 - / F e 2 + o
Options
- A1.84 V
- B1.42 V
- C1.3 V
- D2.0 V
Correct answer
A. 1.84 V
Step-by-step solution
F e 6 + + 3 e → F e 3 + ; - Δ G 1 o = 3 × 2.20 × F …(i) F e 3 + + e → F e 2 + ; - Δ G 2 o = 1 × 0.77 × F …(ii) F e 2 + + 2 e → F e o ; - Δ G 3 o = 2 × ( - 0.445 ) × F …(iii) By equation (i) and (ii) F e 6 + + 4 e → F e 2 + ; - Δ G 4 o = - Δ G 1 o - Δ G 2 o Hence 4 × E 4 o × F = 3 × 2.20 × F + 1 × 0.77 × F So E 4 o = 1.84 V or F e 6 + + 4 e → F e 2 + ; E o = 1.84 V