Given below are half cell reactions:MnO4−+8H++5e−→Mn2++4H2O\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}MnO4−+8H++5e−→Mn2++4H2O, EMn2+/MnO4−∘=−1.510 VE^\circ_{\text{Mn}^{2+}/\text{MnO}_4^-} = -1.510\,\text{V}EMn2+/MnO4−∘=−1.510V12O2+2H++2e−→H2O\dfrac{1}{2}\text{O}_2 + 2\text{H}^+ + 2e^- \rightarrow \text{H}_2\text{O}21O2+2H++2e−→H2O, EO2/H2O∘=+1.223 VE^\circ_{\text{O}_2/\text{H}_2\text{O}} = +1.223\,\text{V}EO2/H2O∘=+1.223VWill the permanganate ion, MnO4−\text{MnO}_4^-MnO4− liberate O2\text{O}_2O2 from water in the presence of an acid?A.Yes, because Ecell=+0.287 VE_\text{cell} = +0.287\,\text{V}Ecell=+0.287VB.No, because Ecell=−0.287 VE_\text{cell} = -0.287\,\text{V}Ecell=−0.287VC.Yes, because Ecell=+2.733 VE_\text{cell} = +2.733\,\text{V}Ecell=+2.733VD.No, because Ecell=−2.733 VE_\text{cell} = -2.733\,\text{V}Ecell=−2.733VShow answerCorrect option: A[NEET 2022]
