Animated Solution for Chemistry - Electrochemistry: All the energy released from the reaction $\text{X}\to \text{Y}$, ${\Delta }_r{G}^{\circ }=-193{\text{ kJ mol}}^{-1}$ is used for the oxidizing ${\text{M}}^{+}$ and ${\text{M}}^{+}\to {\text{M}}^{3+}+2e^{-}$, $E^{\circ }=-0.25\text{ V}$. Under standard conditions, the number of moles of ${\text{M}}^{+}$ oxidized when one mole of X is converted to Y is $[F=96500{\text{ C mol}}^{-1}]$

Imagine a microscopic power plant. In one sector, a chemical reaction occurs spontaneously, releasing a massive burst of energy. In another sector, a different reaction is waiting—it needs energy to proceed. What if we could wire these two sectors together? This is the essence of coupled reactions, a fundamental concept in both industrial chemistry and biological systems.

In our problem, we have a spontaneous reaction where substance X converts to Y. This process is generous; it releases a standard Gibbs free energy, ${\Delta }_r{G}^{\circ }$, of $-193{\text{ kJ mol}}^{-1}$. The negative sign is simply nature's way of saying, "I am giving this energy away."

On the receiving end, we have the oxidation of a metal ion, ${\text{M}}^{+}$, to a higher oxidation state, ${\text{M}}^{3+}$. This process is not spontaneous under standard conditions. It has a standard electrode potential, $E^{\circ }$, of $-0.25\text{ V}$. The negative potential indicates that this reaction requires an input of energy to occur. Our goal is to find out exactly how many moles of ${\text{M}}^{+}$ can be oxidized using the energy provided by exactly one mole of X.

To solve this, we first need to understand the energy appetite of our metal ion. How much energy does it take to oxidize just one mole of ${\text{M}}^{+}$ to ${\text{M}}^{3+}$?

We turn to the master equation that bridges electrochemistry and thermodynamics:

Let's break down the components of this equation for our specific oxidation reaction:

1. The Number of Electrons ($n$): By observing the change in oxidation state from $+1$ to $+3$, it is clear that each ion of M loses two electrons. Therefore, for one mole of the reaction, $n=2$ moles of electrons are transferred. 2. Faraday's Constant ($F$): This is the charge of one mole of electrons, given as $96500{\text{ C mol}}^{-1}$. 3. Standard Electrode Potential ($E^{\circ }$): The problem states this is $-0.25\text{ V}$.

Now, we substitute these values into our master equation. Watch out for the minus signs!

The two negative signs cancel each other out, confirming that the overall free energy change will be positive—meaning energy is absorbed.

Since the energy released by the first reaction is given in kilojoules, it is crucial to maintain consistent units. Let's convert our result to kilojoules by dividing by 1000:

This tells us that every single mole of ${\text{M}}^{+}$ requires exactly $48.25\text{ kJ}$ of energy to be successfully oxidized to ${\text{M}}^{3+}$.

Now we bring both halves of the problem together. We have a total energy budget provided by the conversion of one mole of X to Y.

We assume that all of this energy is perfectly transferred to the oxidation process, with zero losses. If $x$ represents the number of moles of ${\text{M}}^{+}$ that are oxidized, then the total energy consumed by this process will be $x$ times the energy required per mole.

By the principle of conservation of energy, the total energy available must equal the total energy consumed:

We are now one simple algebraic step away from our answer. To isolate $x$, we divide the total available energy by the energy required per mole:

If you look closely at the numbers, you might notice a neat relationship. Let's multiply the denominator by 4 to see what happens:

The division is perfect.

This means that the energy released from the conversion of exactly one mole of X is perfectly sufficient to drive the oxidation of exactly 4 moles of ${\text{M}}^{+}$. It is a beautifully clean result that highlights the elegance of thermodynamic calculations when energy is perfectly conserved and coupled.