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

Question

All the energy released from the reaction

X ⟶ Y, Δ_{r} G^{o} = - 193 k J m o l^{- 1}

is used for oxidizing

M^{+}

as

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

Aicharya · Accepted Answer

M + &#8594; M 3 + + 2 e - &#8710; G o = - n F E o for 1 mole of M + &#8710; G o = - 2 &#215; 96500 &#215; - 0.25 J = + 48250 J/mole = 48.25 &#160; K J /mole Energy released by conversion of 1 mole of X &#8594; Y &#8710; G = - 193 &#160; K J Hence mole of M + convert 193 48.25 = 4 &#160;

All the energy released from the reaction X ⟶ Y ,   Δ r G o = - 193   k J   m o l…

Solution

All the energy released from the reaction X &#10230; Y , &#160; &#916; r G o = - 193 &#160; k J &#160; m o l…

Solution

All the energy released from the reaction X ⟶ Y , Δ r G o = - 193 k J m o l…