Given below are half cell reactions: MnO_4^- + 8H^+ + 5e^- → Mn^2+ + 4H_2O, E^_MnO_4^-/Mn^2+ = +1.510 V 1/2O_2 + 2H^+ + 2e^- → H_2O, E^_O_2/H_2O = +1.223 V Will the permanganate ion, MnO_4^- , liberate O_2 from water in the presence of an acid?

Given below are half cell reactions: MnO_4^- + 8H^+ + 5e^- → Mn^2+ + 4H_2O, E^_MnO_4^-/Mn^2+ = +1.510 V 1/2O_2 + 2H^+ + 2e^- → H_2O, E^_O_2/H_2O = +1. - Chemistry MCQ NEET 2022 2026 | ExamTest.live

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Given below are half cell reactions: MnO_{4}^{-} + 8 H^{+} + 5 e^{-} \to Mn^{2 +} + 4 H_{2} O, E_{MnO_{4}^{-} / Mn^{2 +}}^{\circ} = + 1.510 V \frac{1}{2} O_{2} + 2 H^{+} + 2 e^{-} \to H_{2} O, E_{O_{2} / H_{2} O}^{\circ} = + 1.223 V Will the permanganate ion, MnO_{4}^{-}, liberate O_{2} from water in the presence of an acid?

For the reaction to liberate O_{2}, MnO_{4}^{-} must be reduced (Cathode) and water must be oxidized to O_{2} (Anode).Reaction: 2 MnO_{4}^{-} + 6 H^{+} \to 2 Mn^{2 +} + 3 H_{2} O + \frac{5}{2} O_{2} (Overall stoichiometry varies but potential logic remains). E_{ce ll}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ} E_{ce ll}^{\circ} = E_{MnO_{4}^{-} / Mn^{2 +}}^{\circ} - E_{O_{2} / H_{2} O}^{\circ} E_{ce ll}^{\circ} = 1.510 - 1.223 = + 0.287 V Since E_{ce ll}^{\circ} is positive, the reaction is spontaneous. Yes.

Try it yourself before checking the explanation above.

Given below are half cell reactions: MnO_{4}^{-} + 8 H^{+} + 5 e^{-} \to Mn^{2 +} + 4 H_{2} O, E_{MnO_{4}^{-} / Mn^{2 +}}^{\circ} = + 1.510 V \frac{1}{2} O_{2} + 2 H^{+} + 2 e^{-} \to H_{2} O, E_{O_{2} / H_{2} O}^{\circ} = + 1.223 V Will the permanganate ion, MnO_{4}^{-}, liberate O_{2} from water in the presence of an acid?

Yes, because E_{ce ll}^{\circ} = + 0.287 V

No, because E_{ce ll}^{\circ} = - 0.287 V

Yes, because E_{ce ll}^{\circ} = + 2.733 V

No, because E_{ce ll}^{\circ} = - 2.733 V

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