Given that E F e 3 + ∣ F e ∘ and E F e 2 + ∣ F e ∘ are − 0.036 V and − 0.439 V respectively. The value of E F e 3 + ∣ F e 2 + ∘ would be:

F e 2 + + 2 e − → F e F e 3 + + 3 e − → F e F e 3 + + e − → F e 2 + Δ G ∘ = − n F E ∘ E 3 ∘ = 1 3 E 2 ∘ − 2 E 1 ∘ E ∘ = 3 ( − 0.036 ) − 2 ( − 0.439 )

Chemistry - Electrochemistry Question with Solution | TestHub

ChemistryElectrochemistryRelationship between Eº, DGº & KeqMedium2 minQB

ChemistryMediumsingle choice

Given that $E_{F e^{3 +} ∣ F e}^{\circ}$ and $E_{F e^{2 +} ∣ F e}^{\circ}$ are $- 0.036 V$ and $- 0.439 V$ respectively. The value of $E_{F e^{3 +} ∣ F e^{2 +}}^{\circ}$ would be:

Options:

Answer:

Solution:

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

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

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

$Δ G^{\circ} = - n F E^{\circ}$

$E_{3}^{\circ} = \frac{3 E _{2}^{\circ} - 2 E _{1}^{\circ}}{1}$

$E^{\circ} = 3 (- 0.036) - 2 (- 0.439)$

Stream:JEESubject:ChemistryTopic:ElectrochemistrySubtopic:Relationship between Eº, DGº & Keq

⏱ 2mℹ️ Source: QB

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