All the released energy from the reaction X - Y - - r G 0-193 kJ mol -1 is used for oxidizing M - M 3-2 e - E 0-0-25 VUnder standard conditions- the number of moles of M -oxidized when one mole of X is converted to Y is - F -96500 C mol -1-

M^+→ M^3++2e^ Δ G^0= nFE^0 For 1 mole of M^+ Δ G^0= 2× 96500× (0.25) J =+48250 J/mole =48.25 kJ/mole Energy released by conversion of 1 mole of X→ Y Δ G= 19 ...

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Standard XII

Chemistry

Gibb's Energy and Nernst Equation

All the relea...

Question

All the released energy from the reaction $X \to Y, Δ_{r} G^{0} = - 193 {kJ mol}^{- 1}$ is used for oxidizing $M^{+} \to M^{3 +} + 2 e^{-}, E^{0} = - 0.25 V$
Under standard conditions, the number of moles of $M^{+}$ oxidized when one mole of $X$ is converted to $Y$ is [ $F = 96500 {C mol}^{- 1}$ ]

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Solution

$M^{+} \to M^{3 +} + 2 e^{-}$
$Δ G^{0} = - n F E^{0}$ For 1 mole of $M^{+}$
$Δ G^{0} = - 2 \times 96500 \times (0.25) J$
$= + 48250 J/mole$
$= 48.25 kJ/mole$
Energy released by conversion of 1 mole of
$X \to Y Δ G = - 193 kJ$
Hence, moles of $M^{+}$ converted
$\frac{193}{48.25} = 4$

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All the released energy from the reaction X→Y,ΔrG0=−193 kJ mol−1 is used for oxidizing M+→M3++2e−, E0=−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]

M+→M3++2e− ΔG0=−nFE0 For 1 mole of M+ ΔG0=−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, moles of M+ converted 19348.25=4

All the released energy from the reaction $X \to Y, Δ_{r} G^{0} = - 193 {kJ mol}^{- 1}$ is used for oxidizing $M^{+} \to M^{3 +} + 2 e^{-}, E^{0} = - 0.25 V$
Under standard conditions, the number of moles of $M^{+}$ oxidized when one mole of $X$ is converted to $Y$ is [ $F = 96500 {C mol}^{- 1}$ ]

$M^{+} \to M^{3 +} + 2 e^{-}$
$Δ G^{0} = - n F E^{0}$ For 1 mole of $M^{+}$
$Δ G^{0} = - 2 \times 96500 \times (0.25) J$
$= + 48250 J/mole$
$= 48.25 kJ/mole$
Energy released by conversion of 1 mole of
$X \to Y Δ G = - 193 kJ$
Hence, moles of $M^{+}$ converted
$\frac{193}{48.25} = 4$