Given electrode potentials: Fe³⁺ + e⁻ ⟶ Fe²⁺; E ∘ = 0.771 V I₂ + 2e⁻ ⟶ 2I⁻; E ∘ = 0.536 V E ce ll ∘ for the cell reaction 2 F e 3 + + 2 I − ⟶ 2 F e 2 + + I 2 is:

E ce ll ∘ = E ∘ ( F e ) 3 + − E ∘ ( I 2 ) = 0.771 − 0.536 = 0.235 V

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ChemistryElectrochemistryRelationship between Eº, DGº & KeqMedium2 minQB

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Given electrode potentials:

Fe³⁺ + e⁻ ⟶ Fe²⁺; $E^{\circ} = 0.771 V$

I₂ + 2e⁻ ⟶ 2I⁻; $E^{\circ} = 0.536 V$

$E_{ce ll}^{\circ}$ for the cell reaction $2 F e^{3 +} + 2 I^{-} ⟶ 2 F e^{2 +} + I_{2}$ is:

Answer:

Solution:

$E_{ce ll}^{^{^{\circ}}} = E^{\circ} (F e)^{3 +} - E^{\circ} (I_{2}) = 0.771 - 0.536 = 0.235 V$

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

⏱ 2mℹ️ Source: QB

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