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

Question

All the energy released from the reaction

X ⟶ Y, {\Delta}_{r}{G}^{o} = - 193kJmo{l}^{- 1}

is used for oxidizing

{M}^{+}

as

{M}^{+} ⟶ {M}^{3 +} + 2{e}^{-}, {E}^{o} = - 0.25V

Under standard conditions, the number of moles of

{M}^{+}

oxidized when one mole of X is converted to Y is

[ F = 96500Cmo{l}^{- 1} ]

Quantrex Academy · Accepted Answer

Correct answer: 4

M ^ + → M ^ 3 + + 2 e ^ - ∆ G ^ o = - nF E ^ o for 1 mole of M ^ + ∆ G ^ o = - 2 × 96500 × (- 0.25)J = + 48250 J/mole = 48.25KJ /mole Energy released by conversion of 1 mole of X → Y ∆ G = - 193KJ Hence mole of M ^ + convert (193)/(48.25) = 4

All the energy released from the reaction X ⟶ Y, Δ _ r G ^ o = -...

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