The standard e.m.f. of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V at 2 5 ∘ C . The equilibrium constant of the reaction would be (Given F = 96500 C mol − 1 ; R = 8.314 J − 1 mol − 1 )

E ∘ = n 0.0591 lo g 10 K Here, n = 2 , E ∘ = 0.295 ∴ lo g 10 K = 0.0591 2 × 0.295 = 10 or K = 1 0 10

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ChemistryElectrochemistryGalvanic cells/Nernst Equation/Concentration CellsEasy2 minPYQ_2022

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The standard e.m.f. of a galvanic cell involving cell reaction with $n = 2$ is found to be $0.295 V$ at $2 5^{\circ} C$ . The equilibrium constant of the reaction would be (Given $F = 96500 C mol^{- 1}$ ; $R = 8.314 J^{- 1} mol^{- 1}$ )

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Answer:

Solution:

$E^{\circ} = \frac{0.0591}{n} lo g_{10} K$ Here, $n = 2, E^{\circ} = 0.295$ $∴ lo g_{10} K = \frac{2 \times 0.295}{0.0591} = 10 or K = 1 0^{10}$

Stream:VITEEESubject:ChemistryTopic:ElectrochemistrySubtopic:Galvanic cells/Nernst Equation/Concentration Cells

⏱ 2mℹ️ Source: PYQ_2022

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