All the energy released from the reaction X- Y- - G- -193 kJ mol -1 is used for oxidizing M- as M- - M3- - 2e- E- - -0-25 V- Under standard conditions the number of moles of M- isGiven - -F-96500 C mol-1-

M^+ → M^3+ +2e^ E^∘ = 0.25V Δ G^∘ = nFE^∘ ⇒Δ G^∘ = 2× 96500 × ( 0.25)J ⇒Δ G^∘ =48250 J ⇒Δ G^∘ = 48.25 kJ Number of moles of M^+ oxidized =19348.25=4 ...

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Standard VII

Chemistry

Gibb's Energy and Nernst Equation

All the energ...

Question

All the energy released from the reaction $X \to Y; Δ G^{\circ} = - 193 {kJ mol}^{- 1}$ is used for oxidizing $M^{+}$ as $M^{+} \to M^{3 +} + 2 e^{-}; E^{\circ} = - 0.25 V$ . Under standard conditions the number of moles of $M^{+}$ is
(Given : $[F = 96500 C m o l^{- 1}]$ )

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Solution

$M^{+} \to M^{3 +} + 2 e^{-} E^{\circ} = - 0.25 V$
$Δ G^{\circ} = - n F E^{\circ}$
$\Rightarrow Δ G^{\circ} = - 2 \times 96500 \times (- 0.25) J$
$\Rightarrow Δ G^{\circ} = 48250 J$
$\Rightarrow Δ G^{\circ} = 48.25 k J$
Number of moles of $M^{+}$ oxidized
$= \frac{193}{48.25} = 4$

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All the energy released from the reaction X→Y;ΔG∘=−193 kJ mol−1 is used for oxidizing M+ as M+→M3++2e−; E∘=−0.25V. Under standard conditions the number of moles of M+ is (Given : [F=96500Cmol−1])

M+→M3++2e− E∘=−0.25V ΔG∘=−nFE∘ ⇒ΔG∘=−2×96500×(−0.25)J ⇒ΔG∘=48250J ⇒ΔG∘=48.25kJ Number of moles of M+ oxidized =19348.25=4

All the energy released from the reaction $X \to Y; Δ G^{\circ} = - 193 {kJ mol}^{- 1}$ is used for oxidizing $M^{+}$ as $M^{+} \to M^{3 +} + 2 e^{-}; E^{\circ} = - 0.25 V$ . Under standard conditions the number of moles of $M^{+}$ is
(Given : $[F = 96500 C m o l^{- 1}]$ )

$M^{+} \to M^{3 +} + 2 e^{-} E^{\circ} = - 0.25 V$
$Δ G^{\circ} = - n F E^{\circ}$
$\Rightarrow Δ G^{\circ} = - 2 \times 96500 \times (- 0.25) J$
$\Rightarrow Δ G^{\circ} = 48250 J$
$\Rightarrow Δ G^{\circ} = 48.25 k J$
Number of moles of $M^{+}$ oxidized
$= \frac{193}{48.25} = 4$