All the energy released from the reaction X arrow Y , Δ r G °=-193 kJ mol -is used for oxidizing M +as M + arrow M 3++2 e -, E °=-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 arrow Y , Δ r G °=-193 kJ mol -is used for oxidizing M +as M + arrow M 3++2 e -, E °=-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 + arrow M 3++2 e - nM + arrow nM 3++2 ne - Δ G °=-2 n FE ° And: |Δ r G °|=|Δ G °| ∴ 193 × 1000=2 × n × 96500 × 0.25 n =4

Q. All the energy released from the reaction X→Y,Δr​G∘=−193kJmol−is used for oxidizing M+as M+→M3++2e−,E∘=−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−ΔG∘=−2nFE∘ And : ∣Δr​G∘∣=∣ΔG∘∣ ∴193×1000=2×n×96500×0.25 n=4

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