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For the react...

Question

For the reaction, $F e^{3 +} + 3 e^{-} ⇌ F e, E^{\circ}$ is $- 0.036 v o l t$ and the standard electrode potential for $F e^{3 +} + e^{-} ⇌ F e^{2 +}$ is $0.771 v o l t$ . Calculate the $E^{\circ}$ for $F e^{2 +} + 2 e^{-} ⇌ F e$ .

0.4395 v o l t

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1.4395 v o l t

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- 0.4395 v o l t

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- 2.4395 v o l t

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Solution

The correct option is D $- 0.4395 v o l t$
$F e^{3 +} + 3 e^{-} ⇌ F e, E^{\circ} = - 0.036 V; △ G^{\circ} = - n F E_{c e l l}^{\circ} = 10422 V \dots (1)$
$F e^{3 +} + e^{-} ⇌ F e^{2 +}, E^{\circ} = 0.771 v o l t; △ G^{\circ} = - 74401.5 V ⟶ (2)$
Subtracting $(2)$ from $(1)$ , we get,
$F e^{2 +} + 2 e^{-} ⇌ F e; △ G^{\circ} = 84823.5$
$∴ E_{c e l l}^{\circ} = \frac{- △ G^{\circ}}{n F} = \frac{- 84823.5}{2 \times 96500} = - 0.4395 V$

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Similar questions

F e^{3 +} + 3 e^{-} \to F e; E^{\circ} = - 0.036 v o l t

F e^{2 +} + 2 e^{-} \to F e; E^{\circ} = - 0.440 v o l t

E^{\circ}

F e^{3 +} + e^{-} \to F e^{2 +}

{F e}^{3 +} + {3 e}^{-} ⟶ F e; E^{o} = - 0.036 v o l t

{F e}^{2 +} + {2 e}^{-} ⟶ F e; E^{o} = - 0.440 v o l t

E^{o}

{F e}^{3 +} + e^{-} ⟶ {F e}^{2 +}

F e^{2} + 2 e^{-} \to F e, E^{\circ} = - 0.440 V

F e^{3 +} + 3 e^{-} \to F e, E^{\circ} = - 0.036 V

(E^{\circ})

F e^{3 +} + e^{-} \to F e^{2 +}

Given standard electrode potentials:

$F e^{3 +} + 3 e^{-} \to F e; E^{0} = - 0.036 V$

$F e^{2 +} + 2 e^{-} \to F e; E^{0} = - 0.440 V$

The standard electrode potential $E^{o}$ for $F e^{3 +} + e^{-} \to F e^{2 +}$ is:

F e^{3 +} + 3 e^{-} \to F e

E^{\circ} = - 0.036 V

F e^{2 +} + 2 e^{-} \to F e

E^{\circ} = - 0.44 V

F e^{3 +} + e^{-} \to F e^{2 +}

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