NEET2022ChemistryElectrochemistryActual
Given below are half cell reactions : MnO 4 - + 8 H + + 5 e - → Mn 2 + + 4 H 2 O , E Mn 2 + / MnO 4 - o = - 1 . 510   V 1 2 O 2 + 2 H + + 2 e - → H 2 O E O 2 / H 2 O o = + 1 . 223   V Will the permanganate ion, MnO 4 - liberate O 2 from water in the presence of an acid?
Options
- ANo, because E cell o = - 0 . 287   V
- BYes, because E cell o = + 2 , 733   V
- CNo, because E cell o = - 2 . 733   V
- DYes, because E cell o = + 0 . 287   V
Correct answer
D. Yes, because E cell o = + 0 . 287   V
Step-by-step solution
The reaction between MnO 4 - and H 2 O is E o for the cell reaction:- E ° = E MnO 4 - / Mn + 2 o   +   E H 2 O / O 2 o = 1 . 510   V - 1 . 223   V E o = + 0 . 287   V As E ° of the cell reaction is positive, so this reaction is feasible. Therefore permanganate ion MnO 4 - liberate O 2 from water in presence of an acid. Hence, option D is correct.