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(D) The energy released from the reaction $X \rightarrow Y$ is $\Delta_{r}G^0 = -193 \ kJ \ mol^{-1} = -193000 \ J \ mol^{-1}$.
For the oxidation rea

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$4$

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(D) The energy released from the reaction $X ightarrow Y$ is $\Delta_{r}G^0 = -193 \ kJ \ mol^{-1} = -193000 \ J \ mol^{-1}$.
For the oxidation reaction $M^{+} ightarrow M^{3+} + 2e^-$,the number of electrons involved is $n = 2$ and the standard electrode potential is $E^0 = -0.25 \ V$.
The Gibbs free energy change for the oxidation of $1 \ mol$ of $M^{+}$ is $\Delta G^0 = -nFE^0 = -2 	imes 96500 	imes (-0.25) = 48250 \ J \ mol^{-1} = 48.25 \ kJ \ mol^{-1}$.
The number of moles of $M^{+}$ oxidized by $1 \ mol$ of $X$ is the ratio of the energy released to the energy required per mole of $M^{+}$:
$	ext{Number of moles}$ $= \frac{|\Delta_{r}G^0_{X 	o Y}|}{\Delta G^0_{M^{+} 	o M^{3+}}} = \frac{193 	ext{ kJ mol}^{-1}}{48.25 	ext{ kJ mol}^{-1}} = 4 	ext{ mol}$

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