Moles of Metal Oxidized by Reaction Free Energy

To find the number of moles of $\text{M}^+$ oxidized, we calculate the standard Gibbs free energy change ($\Delta G^0$) for both processes.

Step 1: Energy liberated by the process $\text{X} \rightarrow \text{Y}$ When $1\text{ mole}$ of $\text{X}$ is transformed into $\text{Y}$, the free energy liberated is: $\text{Energy available} = |\Delta_r G^0| = 193\text{ kJ} = 193000\text{ J}$

Step 2: Free energy required to oxidize $1\text{ mole}$ of $\text{M}^+$ The oxidation half-reaction is given by: $\text{M}^+ \rightarrow \text{M}^{3+} + 2e^- \quad \left(E^0 = -0.25\text{ V}\right)$

For this oxidation process, $n = 2$ moles of electrons are released per mole of $\text{M}^+$. The standard Gibbs free energy change required to oxidize $1\text{ mole}$ of $\text{M}^+$ is: $\Delta G^0_{\text{ox}} = -n F E^0$

Substitute the given values ($n = 2$, $F = 96500\text{ C mol}^{-1}$, $E^0 = -0.25\text{ V}$): $\Delta G^0_{\text{ox}} = -2 \times 96500\text{ C mol}^{-1} \times (-0.25\text{ V})$ $\Delta G^0_{\text{ox}} = 48250\text{ J mol}^{-1} = 48.25\text{ kJ mol}^{-1}$

Step 3: Calculate the number of moles of $\text{M}^+$ oxidized Let $n_{\text{M}^+}$ be the total number of moles of $\text{M}^+$ oxidized by utilizing all the energy liberated: $n_{\text{M}^+} = \frac{\text{Total energy available}}{\text{Energy required per mole of }\text{M}^+}$

$n_{\text{M}^+} = \frac{193000\text{ J}}{48250\text{ J mol}^{-1}} = 4\text{ moles}$

Thus, the quantity of moles of $\text{M}^+$ oxidized is 4.