Electrochemistry - Result Question 105
####53. All the energy released from the reaction $X \rightarrow Y, \Delta _r G^{\circ}=-193 kJmol^{-1}$ is used for oxidising $M^{+}$as $M^{+} \rightarrow M^{3+}+2 e^{-}, E^{\circ}=-0.25 V$.
Under standard conditions, the number of moles of $M^{+}$ oxidised when one mole of $X$ is converted to $Y$ is $\left[F=96500 C mol^{-}\right]$
(2015 Adv.)
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Answer:
Correct Answer: 53. $(4 mol)$
Solution:
- Energy obtained as one mole $X$ is converted into $Y$ is $193 kJ$.
Energy consumed in converting one mole of $M^{+}$to $M^{3+}$
$$ \begin{aligned} & =-n E^{\circ} F=2 \times 96500 \times 0.25 J=\frac{96500}{2} J \ \Rightarrow \quad 193 \times 10^{3} & =n\left(\frac{96500}{2}\right) \Rightarrow n=4 mol \end{aligned} $$
