O2+2H++2e− . . . (ii) EO2∕H2O∘=1.223V Using 2× (i) +5× (ii), net cell reactions is 2MnO4−+6H+⟶2Mn2++
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O2+3H2O Ecell ∘=EC∘−EA∘=EMnO4−∕Mn2+∘−EO2∕H2O∘=1.51−1.223=0.287V Since Ecell 0>0, therefore net cell reaction is spontaneous and so MnO4−liberate O2 from H2O in presence of an acid.