Given below are half cell reactions:
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O, E° = 1.510 V
(1/2)O2 + 2H+ + 2

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Question

Given below are half cell reactions:
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O, E° = 1.510 V
(1/2)O2 + 2H+ + 2e- → H2O, E° = 1.223 V
Will the permanganate ion, MnO4- liberate O2 from water in the presence of an acid?
(1) No, because E°cell = -2.733 V
(2) Yes, because E°cell = +0.287 V
(3) No, because E°cell = -0.287 V
(4) Yes, because E°cell = +2.733 V

Accepted Answer

Correct Answer: B - Yes, because E°cell = +0.287 VStep-by-step Solution:To check if MnO4- can oxidise H2O to O2, we set up the cell where MnO4- acts as the oxidising agent (cathode/reduction) and H2O is oxidised to O2 (anode/oxidation).Cathode (reduction): MnO4- + 8H+ + 5e- → Mn2+ + 4H2O; E°red = +1.510 V.Anode (oxidation): H2O → (1/2)O2 + 2H+ + 2e-; E°ox = -(+1.223) = -1.223 V.E°cell = E°cathode - E°anode = 1.510 - 1.223 = +0.287 V.Since E°cell > 0, the reaction is spontaneous, and MnO4- will liberate O2 from water in acidic medium.Why other options are incorrect: Option C (-0.287 V) would mean non-spontaneous. Options A and D have incorrect magnitudes (2.733 V = 1.510 + 1.223, incorrectly summed instead of subtracted).Key Concepts:E°cell = E°cathode - E°anode; if positive, reaction is spontaneous.KMnO4 is a powerful oxidising agent; can oxidise water to O2 in acidic medium.In acidic medium: MnO4- → Mn2+ (change of +7 to +2, gain of 5 electrons).

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