The standard electrode potential of the half cells is given below- Zn2-2e- Zn- E-0-76 V Fe2-2e- Fe- E-0-44 VThe emf of the cell Fe2-Zn- Zn2-Fe

The correct option is C 0.32V E _ cell ^ 0 =( E _ Fe ^ 2+ |Fe ^ 0 + E _ Zn| Zn ^ 2+ ^ 0 ) #160; #160; #160; #160; #160; =( 0.44 ...

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Redox Reactions and Couple

The standard ...

Question

The standard electrode potential of the half cells is given below:
$Z n^{2 +} + 2 e^{-} \to Z n; E = - 0.76 V$
$F e^{2 +} + 2 e^{-} \to F e; E = - 0.44 V$
The emf of the cell $F e^{2 +} = Z n^{-} \to Z n^{2 +} + F e$

1.54 V

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- 1.54 V

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0.32 V

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+ 0.19 V

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Solution

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Similar questions

Z n^{2 +} + 2 e^{-} \to Z n; E_{1} = - 7.62 V

F e^{2 +} + 2 e^{-} \to F e; E_{2} = - 7.81 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

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^{\circ}

Z n \to Z n^{2 +} + 2 e^{-}; E^{\circ} = + 0.76 v o l t

F e \to F e^{2 +} + 2 e^{-}; E^{\circ} = + 0.41 v o l t

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

E^{o}

Z n ⟶ {Z n}^{2 +} + 2 e^{-}

E^{o} = + 0.76 V

F e ⟶ {F e}^{2 +} + 2 e^{-}

E^{o} = + 0.41 V

{F e}^{2 +} + Z n ⟶ {Z n}^{2 +} + F e

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