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Electrochemistry question

2022 · Q111
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Electrochemistry question

2022 · Q111

NEETChemistryElectrochemistryMCQ+4 / −1

Given below are half cell reactions:

MnO4−+8H++5e−→Mn2++4H2OMnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}OMnO4−​+8H++5e−→Mn2++4H2​O,

EMn2+/MnO4−o=−1.510 VE_{M{n^{2 + }}/MnO_4^ - }^o = - 1.510\,VEMn2+/MnO4−​o​=−1.510V

12O2+2H++2e−→H2O{1 \over 2}{O_2} + 2{H^ + } + 2{e^ - } \to {H_2}O21​O2​+2H++2e−→H2​O

EO2/H2Oo=+1.223 VE_{{O_2}/{H_2}O}^o = + 1.223\,VEO2​/H2​Oo​=+1.223V

Will the permanganate ion, MnO4−MnO_4^ -MnO4−​ liberate O2 from water in the presence of an acid?

  1. A
    Yes, because EcelloE_{cell}^oEcello​ = + 0.287 V
  2. B
    No, because EcelloE_{cell}^oEcello​ = −-−0.287 V
  3. C
    Yes, because EcelloE_{cell}^oEcello​ = + 2.733 V
  4. D
    No, because EcelloE_{cell}^oEcello​ = −-− 2.733 V
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