QJEEAdv15Ch8ecc: JEE Advanced 2015 Chemistry - Class 12 JEEAdv15Ch8ecc Solution: All the energy released from the reaction = -193 kJ mol -1 is…

Given :

X $\to$ Y; $\Delta$ _rG $^\circ$ = $-$ 193 kJ mol ^{$-$ 1}

M⁺ $\to$ M³⁺ + 2e ^$-$ ; E $^\circ$ = $-$ 0.25 V

F = 96500 C mol ^{$-$ 1}

Let 193 kJ is used for oxidising x moles of M⁺.

For 1 mole of M⁺,

$\Delta$ G $^\circ$ = $-$ nFE $^\circ$

= $-$ 2 $\times$ 96500 $\times$ ( $-$ 0.25)

= 48250 J mol ^{$-$ 1} = 48.25 kJ mol ^{$-$ 1}

Thus, no. of moles of M⁺ oxidized when one mole of X is converted to Y = ${{193} \over {48.25}} = 4$ .