Given below are half cell reactions: MnO4- + 8H+ + 5e- Mn2+ + 4H2O, E Mn2+/MnO4- = -1.510 V 12O2 + 2H+ + 2e- H2O E O2/H2O = +1.223 V Will the permanganate ion, MnO4- liberate O2 from water in the presence of an acid?
- Yes, because E cell = +0.287 V
- No, because E cell = -0.287 V
- Yes, because E cell = +2.733 V
- No, because E cell = -2.733 V
Answer
(1) Yes, because E cell = +0.287 V
Solution
E Mn2+/MnO4- = -1.510 V (oxidation potential) ⇒ reduction potential E MnO4-/Mn2+ = +1.510 V.
For MnO4- to liberate O2: MnO4- is reduced (cathode) and water is oxidised to O2 (anode).
4MnO4- + 12H+ 4Mn2+ + 6H2O + 5O2
E cell = E cathode - E anode
= 1.510 - 1.223
= +0.287 V
∵ E cell is positive ⇒ G < 0 and the reaction is feasible; so yes, MnO4- can liberate O2 from acidified water.
Answer: (1)