19.Electrode potential for the following half cell reactions are Zn=Zn^+2+2e;E^°=+0.75VFe=Fe^+2+2e;E^°=+0.44V The EMF for the cell reaction Fe^+2+Zn=Zn^+2+Fe will be

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Z n \to Z n^{2 +} + 2 e^{-} (E_{(Z n^{2 +} / Z n)}^{0} = - 0.76 V)

F e \to F e^{2 +} + 2 e^{-} E^{0} = 0.41 V

F e^{2 +} + Z n \to Z n^{2 +} + F e

Z n^{2 +} + 2 e^{-} \to Z n; E = - 0.76 V

F e^{2 +} + 2 e^{-} \to F e; E = - 0.44 V

F e^{2 +} = Z n^{-} \to Z n^{2 +} + F e

E^{o}

Z n \to Z n^{2 +} + 2 e^{-}, E^{\circ} = + 0.76 V F e \to F e^{2 +} + 2 e^{-}, E^{\circ} = + 0.41 V

F e^{2 +} + Z n \to Z n^{2 +} + F e

E^{o}

Z n \to Z n^{2 +} + 2 e^{-}; E^{o} = - 0.76 V

F e^{2 +} + 2 e^{-} \to F e; E^{o} = + 0.41 V

F e^{2 +} + Z n \to Z n^{2 +} + F e

E^{⊝}

Z n \to Z n^{2 +} + 2 e^{-}; E^{⊝} = + 0.76 V

F e \to F e^{2 +} + 2 e^{-}; E^{⊝} = 0.41 V

F e^{2 +} Z n \to Z n^{2 +} + F e

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