All the energy released from the reaction X → Y , Δ r G o = - 193 kJ mo l - 1 is used for oxidizing M + as M + → M 3 + + 2 e - , E o = - 0.25 V . Under standard conditions, the number of moles of M + oxidized when one mole of X is converted to Y is [F = 96500 C m o l - 1 ]

Δ G o = - n F E o - n = 193 × 1 0 3 96500 × 0.25 = 8 So No. of moles of M + oxidized by 1 moles of X = 8 2 = 4

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ChemistryElectrochemistryMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasynumerical

All the energy released from the reaction $X \to Y, Δ_{r} G^{o} = - 193 kJ mo l^{- 1}$ is used for oxidizing $M^{+}$ as $M^{+} \to M^{3 +} + 2 e^{-}, E^{o} = - 0.25 V .$
Under standard conditions, the number of moles of $M^{+}$ oxidized when one mole of X is converted to Y is [F = 96500 C $m o l^{- 1}$ ]

Answer:

4.00

Solution:

$Δ G^{o} = - {n F E}^{o}$

$- n = \frac{193 \times 1 0^{3}}{96500 \times 0.25} = 8$

So No. of moles of $M^{+}$ oxidized by 1 moles of $X = \frac{8}{2} = 4$

Stream:NTA_ABHYASSubject:ChemistryTopic:ElectrochemistrySubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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