MnO−4+8H++5e−→Mn2++4H2O;EO=1.51V; ΔG01=−5×1.51×FMnO2+ – InterviewSolution

1.	MnO−4+8H++5e−→Mn2++4H2O;EO=1.51V; ΔG01=−5×1.51×FMnO2+4H++2e−→Mn2++2H2O;EO=1.23V; ΔG02=−5×1.23×FE∘MnO−4\|MnO2 is
Answer» $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O; E^{O} = 1.51 V; Δ G_{1}^{0} = - 5 \times 1.51 \times F$ $M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O; E^{O} = 1.23 V; Δ G_{2}^{0} = - 5 \times 1.23 \times F$ $E_{M n O_{4}^{-} \| M n O_{2}}^{\circ}$ is

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