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Given below are half cell reactions : MnO_4^-+8H^++5e^-→ Mn^2++4H_2O, E_ Mn^2+/ MnO_4^-^o=-1.510V 1 2 O_2+2H^++2e^-→ H_2O E_O_2/H_2O^o=+1.223V Will…

NEET UG 2022 — Chemistry Physical Chemistry

2022mcqmedium

Given below are half cell reactions :

MnO4−+8H++5e−→Mn2++4H2O{\mathrm{MnO}}_{4}^{-}+8{H}^{+}+5{e}^{-}\rightarrow {\mathrm{Mn}}^{2+}+4{H}_{2}OMnO4−​+8H++5e−→Mn2++4H2​O,

EMn2+/MnO4−o=−1.510V{E}_{{\mathrm{Mn}}^{2+}/{\mathrm{MnO}}_{4}^{-}}^{o}=-1.510VEMn2+/MnO4−​o​=−1.510V

12O2+2H++2e−→H2O\frac{1}{2}{O}_{2}+2{H}^{+}+2{e}^{-}\rightarrow {H}_{2}O21​O2​+2H++2e−→H2​O

EO2/H2Oo=+1.223V{E}_{{O}_{2}/{H}_{2}O}^{o}=+1.223VEO2​/H2​Oo​=+1.223V

Will the permanganate ion, MnO4−{\mathrm{MnO}}_{4}^{-}MnO4−​ liberate O2{O}_{2}O2​ from water in the presence of an acid?

Official previous-year question

Held on 30 Apr 2022 · Verified 9 Jul 2026.

Options

  1. A

    No, because Ecello=−0.287V{E}_{\mathrm{cell}}^{o}=-0.287VEcello​=−0.287V

  2. B

    Yes, because Ecello=+2,733V{E}_{\mathrm{cell}}^{o}=+2,733VEcello​=+2,733V

  3. C

    No, because Ecello=−2.733V{E}_{\mathrm{cell}}^{o}=-2.733VEcello​=−2.733V

  4. D

    Yes, because Ecello=+0.287V{E}_{\mathrm{cell}}^{o}=+0.287VEcello​=+0.287V

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