All the energy released from the reaction X→Y,ΔrG0=−193kJ mol-1 is used for oxidizing M+ as M+→M3++2e−,E0=−0.25V 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

ΔG0=−nFE0cellΔG0=−2×96,500×(−0.25)=+0.5×96,500=48,250=48,250kJ M+→M+3+2e-The number of moles of M+ oxidized from (X→Y).19348.25=4

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

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Detailed Solution

$\begin{array}{l} Δ G^{0} = - n F E^{0} cell \\ Δ G^{0} = - 2 \times 96, 500 \times (- 0.25) \\ = + 0.5 \times 96, 500 = 48, 250 \\ = 48, 250 kJ \end{array}$

$M^{+} \to M^{+ 3} + 2 e^{-}$

The number of moles of M⁺ oxidized from (X→Y).

$\frac{193}{48.25} = 4$

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