All the energy released from the reaction X → Y, Δ_rGº = –193 kJ mol^–1 is used for oxidizing M^+ as M^+ → M^3+ + 2e^– , Eº = – 0.25 V. Under standard conditions, the number of moles of M^+ oxidized…

Question

All the energy released from the reaction X → Y, Δ_rGº = –193 kJ mol^–1 is used for oxidizing M^+ as M^+ → M^3+ + 2e^– , Eº = – 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 mol^–1]

Accepted Answer

Correct answer: 4. (4) Sol. m^+ → m^3+ + 2e^– ΔG^0 = –nFE^0 For 1 mole of m^+ ΔG^0 = –2 × 96500 × (–0.25) J = + 48250 J/mole = 48.25 KJ/mole Energy released by conversion of 1 mole of x → y ΔG = –193 KJ Hence mole of m^+ convert 1 9 34 8.2 5= 4 m^+ → m^3+ + 2e^– ΔG^0 = –nFE^0 m^+ , ΔG^0 = –2 × 96500 × (–0.25) J = + 48250 J/mole = 48.25 KJ/mole x → y ΔG = –193 KJ 1 9 34 8.2 5= 4

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

Solution

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