JEE Advanced · Chemistry · 17. Electrochemistry

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
$[F = 96500 C m o l^{- 1}]$

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Answer & Solution

Correct Answer

(B) 4

Step-by-step Solution

Detailed explanation

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

∆ G^{o} = - n F E^{o}

for 1 mole of

M^{+}

∆ G^{o} = - 2 \times 96500 \times (- 0.25) J

= + 48250

J/mole

= 48.25 K J

/mole
Energy released by conversion of 1 mole of

X \to Y

∆ G = - 193 K J

Hence mole of

M^{+}

convert

\frac{193}{48.25} = 4

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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All the energy released from the reaction X⟶Y, ΔrGo=-193 kJ mol-1 is used for oxidizing M+ as M+⟶M3++2e- , Eo= -0.25 VUnder standard conditions, the number of moles of M+ oxidized when one mole of X is converted to Y is[F=96500 C mol-1]

Step-by-step Solution