The value of E ∘ Z n / Zn 2 + = 0.76 V and that of E ∘ Fe 2 + / Fe = − 0.41 V . The E ∘ cell of the cell with net cell reaction Zn + Fe 2 + ⟶ Zn 2 + + Fe is

E ce ll ∘ = E re d u c t i o n ∘ + E o x i d a t i o n ∘ = − 0.41 V + 0.76 V = 0.35 V

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Chemistry - Electrochemistry Question with Solution | TestHub

ChemistryElectrochemistryRelationship between Eº, DGº & KeqMedium2 minQB

ChemistryMediumsingle choice

The value of $E^{\circ}_{Z n / Zn^{2 +}} = 0.76 V$ and that of $E^{\circ}_{Fe^{2 +} / Fe} = - 0.41 V$ . The $E^{\circ}_{cell}$ of the cell with net cell reaction $Zn + Fe^{2 +} ⟶ Zn^{2 +} + Fe$ is

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Solution:

$E_{ce ll}^{\circ} = E_{re d u c t i o n}^{\circ} + E_{o x i d a t i o n}^{\circ} = - 0.41 V + 0.76 V = 0.35 V$

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

⏱ 2mℹ️ Source: QB

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