All the energy released from the reaction X → Y, ΔrGº = -193 kJ mol- is used for oxidizing M+ as M+ → M3+ + 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]

Question

All the energy released from the reaction X → Y, ΔrGº = -193 kJ mol- is used for oxidizing M+ as M+ → M3+ + 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] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

M+ → M3+ + 2e- nM+ → nM3+ + 2ne- ΔGº = -2n FEº And: |ΔrGº| = |ΔGº| ∴ 193 × 1000 = 2 × n × 96500 × 0.25 n = 4

Q. All the energy released from the reaction X→Y,ΔrGº=−193kJmol− is used for oxidizing M+ as M+→M3++2e−,Eº=−0.25V. Under standard conditions, the number of moles of M+ oxidized when one mole of X is converted to Y is [F=96500Cmol−1]

M+→M3++2e− nM+→nM3++2ne−ΔGº=−2nFEº And : ∣ΔrGº∣=∣ΔGº∣ ∴193×1000=2×n×96500×0.25 n=4

$M^{+} \to M^{3 +} + 2 e^{-}$
$n M^{+} \to n M^{3 +} + 2 n e^{-} Δ G º = - 2 n FE º$
And :
$∣Δ r G º∣ = ∣Δ G º∣$
$∴ 193 \times 1000 = 2 \times n \times 96500 \times 0.25$
$n = 4$