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Electrochemistry: NEET 2022

Hard· Electrochemical Cell (Electrode Potential and EMF of a Cell)· +4 / −1
Given below are half cell reactions:
MnO4−+8H++5e−→Mn2++4H2O,EMn2+/MnO4−∘=−1.510 V\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O},\quad E^\circ_{\mathrm{Mn^{2+}/MnO_4^-}}=-1.510\ \text{V}
12O2+2H++2e−→H2O,EO2/H2O∘=+1.223 V\dfrac{1}{2}\mathrm{O_2 + 2H^+ + 2e^- \rightarrow H_2O},\quad E^\circ_{\mathrm{O_2/H_2O}}=+1.223\ \text{V}
Will the permanganate ion, MnO4−\mathrm{MnO_4^-}, liberate O2\mathrm{O_2} from water in the presence of an acid?
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