JEE Main2014ChemistryElectrochemistryActual
Given below are the half - cell reactions : Mn 2 + + 2 e - → Mn ; E ∘ = - 1.18 V 2 Mn 3 + + e - → Mn 2 + ; E ∘ = + 1.51 V The E ∘ for 3 Mn 2 + → Mn + 2 Mn 3 + will be :
Options
- A- 2.69 V ; the reaction will not occur
- B- 2.69 V ; the reaction will occur
- C- 0.33 V ; the reaction will not occur
- D- 0.33 V ; the reaction will occur
Correct answer
A. - 2.69 V ; the reaction will not occur
Step-by-step solution
Δ G ∘ = - nF E ∘ Mn + 2 + 2 e - → Mn ........... (1) Δ G = - 2 × F × - 1.18 = + 2.36 F 2 Mn + 3 + 2 e - → 2 Mn + 2 ....... (2) Δ G = - 2 × F × +1.51 = - 3.02 F (1) - (2) 3 Mn + 2 → Mn + 2 Mn + 3 Δ G = + 2.36F +   3.02F = 5.38 F Δ G = - 2 × F × E E = 5.38F - 2 × F = - 2.69 V We know that when E cell < 0, cell is non-spontaneous.