MnO4−+8H++5e−→Mn2++4H2O Eo =1.51 VThe quantity of electricity – InterviewSolution

1.	MnO4−+8H++5e−→Mn2++4H2O Eo =1.51 VThe quantity of electricity required in Faraday to reduce five moles of MnO−4 is ____.
Answer» $M n {O_{4}}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O E^{o} = 1.51 V$ The quantity of electricity required in Faraday to reduce five moles of $M n O_{4}^{-}$ is ____.

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