The E cell ∘ of the reaction MnO 4 − + Fe 2 + + H + ⟶ Mn 2 + + Fe 3 + + H 2 O is 0.59 V at 2 5 ∘ C . The equilibrium constant for the reaction is

For the given half-cell reaction: − n F E ce ll ∘ = − RT ln K e q − 5 × 96500 × 0.59 = − 2.303 × 8.314 × 298 lo g K e q K e q ≈ 1 0 50

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ChemistryElectrochemistryRelationship between Eº, DGº & KeqMedium2 minQB

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The $E_{cell}^{\circ}$ of the reaction $MnO_{4}^{-} + Fe^{2 +} + H^{+} ⟶ Mn^{2 +} + Fe^{3 +} + H_{2} O$ is 0.59 V at $2 5^{\circ} C$ . The equilibrium constant for the reaction is

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Answer:

Solution:

For the given half-cell reaction:

$- n F E_{ce ll}^{\circ} = - RT ln K_{e q}$

$- 5 \times 96500 \times 0.59 = - 2.303 \times 8.314 \times 298 lo g K_{e q}$

$K_{e q} \approx 1 0^{50}$

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

⏱ 2mℹ️ Source: QB

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