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

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Redox Reactions and Electrochemistry

All the energy released from the reaction

X \to Y, Δ_{r} G^{0} = - 193 kJ mol^{- 1}

is used for oxidizing

M^{+}

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

[F = 96500 C mol^{- 1}]

Answer: 4

Solution

For the oxidation

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

Δ G^{0} = - n F E^{0} = - 2 \times 96500 \times (- 0.25) = + 48250

J/mol = 48.25 kJ/mol

This is the energy required to oxidize 1 mole of

M^{+}

.

Energy available from X→Y = 193 kJ

Number of moles of

M^{+}

oxidized =

\frac{193}{48.25} = 4

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Integer answer

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