Given below are half cell reactions: {MnO}_{4}^{-}+8 {H}^{+}+5 {e}^{-} \to {Mn}^{2+}+4 {H}_{2} {O} {E}_{ {Mn}^{2+} \/ {MnO}_{4}^{-}}^{\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? [NEET-2022]

Correct Answer is : Yes, because Ecell ∘=+0.287V

Yes, because {E}_{\text{cell }}^{\circ}=+0.287 {V}

No, because {E}_{\text{cell }}^{\circ}=-0.287 {V}

Yes, because {E}_{ {cell}}^{\circ}=+2.733 {V}

No, because {E}_{\text{cell }}^{\circ}=-2.733 {V}

NEET 2022 Physics and Chemistry

Question : 81 of 100

Marks: +1, -0

Given below are half cell reactions:
MnO4−+8H++5e−→Mn2++4H2O
EMn2+∕MnO4−∘=−1.510V
‌

O2+2H++2e−→H2O
EO2∕H2O∘=+1.223V
Will the permanganate ion, MnO4−liberate O2 from water in the presence of an acid?

[NEET-2022]

Go to Question:

Prev Question

Next Question