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Gibb's Energy and Nernst Equation

Given standar...

Question

Given standard electrode potentials :
${F e}^{3 +} + {3 e}^{-} ⟶ F e; E^{o} = - 0.036 v o l t$
${F e}^{2 +} + {2 e}^{-} ⟶ F e; E^{o} = - 0.440 v o l t$
The standard electrode potential $E^{o}$ for
${F e}^{3 +} + e^{-} ⟶ {F e}^{2 +}$ is:-

0.476

volt

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0.404

volt

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0.440

volt

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0.772

volt

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Solution

The correct option is D $0.772$ volt
${F e}^{3 +} + {3 e}^{-} ⟶ F e Δ G_{1} = - n F E_{c e l l}^{o} = - 3 F E_{c e l l}^{o}$
${F e}^{3 +} + {2 e}^{-} ⟶ F e Δ G_{2} = - 2 F E_{c e l l}^{o}$
${F e}^{3 +} + e^{-} ⟶ {F e}^{2 +} Δ G_{3} = - F E_{c e l l}^{o}$
required $∵ {Δ G}_{3} = {Δ G}_{1} - {Δ G}_{2}$
$- F E_{c e l l} = - 3 F E_{c e l l}^{o} + 2 F E_{c e l l}$
${- E}_{c e l l} = - 3 \times (- 0.036) + 2 (- 0.440)$
${- E}_{c e l l} = - 0.772 E_{c e l l} = + 0.772 V$

Suggest Corrections

Similar questions

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

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

(E^{o})

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

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

⟶

F e; E^{o} = - 0.440 V

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

⟶

F e; E^{o} = - 0.036 V

(E^{o}

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

\to

F e^{2 +}

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 +}

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

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

- 0.440 V

- 0.036 V

(E^{o})

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

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