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Redox Reactions and Couple

Given the sta...

Question

Given the standard half-cell potentials $(E^{o})$ of the following as

$Z n = Z n^{2 +} + 2 e^{-}$ ; $E^{o} = + 0.76$ V
$F e = F e^{2 +} + 2 e^{-}$ ; $E^{o} = + 0.41$ V
Then the standard e.m.f. of the cell with the reaction $F e^{2 +} + Z n \to Z n^{2 +} + F e$ is?

- 0.35

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+ 0.35

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+ 1.17

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- 1.17

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Solution

The correct option is B $+ 0.35$ V
$Z n^{+ 2} + 2 e^{-} \to Z n$ ; $E^{o} = - 0.76 V$ (SRP) (Anode)

$F e^{+ 2} + 2 e^{-} \to F e$ ; $E^{o} = - 0.41 V$ (SRP) (Cathode)

$E_{c e l l}^{o} = E_{F e^{+ 2} / F e}^{+} - E_{Z n^{+ 2} / Z n}^{o}$

$= - 0.41 + 0.76 = 0.35 V$ .

Hence, the correct option is $B$

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Similar questions

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

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

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

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