For the electrochemical cell, M ∣ M + ∥ X − ∣ X , E M + ∣ M ∘ = 0.44 V and E X ∣ X − ∘ = 0.33 V . From this data we can deduce that:

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ChemistryElectrochemistryGalvanic cells/Nernst Equation/Concentration CellsMedium2 minQB

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For the electrochemical cell, $M ∣ M^{+} ∥ X^{-} ∣ X, E_{M^{+} ∣ M}^{\circ} = 0.44 V$ and $E_{X ∣ X^{-}}^{\circ} = 0.33 V$ . From this data we can deduce that:

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Solution:

The species with the higher standard reduction potential is reduced. Therefore M⁺ is reduced and X⁻ is oxidized.

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

⏱ 2mℹ️ Source: QB

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