Animated Solution for Chemistry - Electrochemistry: The correct option(s) about entropy ($S$) is(are) [$R=\text{gas constant}$, $F=\text{Faraday constant}$, $T=\text{Temperature}$]

This problem is a beautiful amalgamation of thermodynamics, electrochemistry, and coordination chemistry. It tests your ability to connect the abstract concept of entropy to tangible physical and chemical processes. Let's embark on a detailed journey through each option to uncover the underlying principles.

The Temperature Coefficient

We start with the fundamental thermodynamic relationship that links the Gibbs free energy change ($\Delta G$) to enthalpy ($\Delta H$) and entropy ($\Delta S$):

If we differentiate this equation with respect to temperature at constant pressure, we obtain a direct relationship for the entropy change:

In electrochemistry, the Gibbs free energy change is related to the cell potential by the equation $\Delta G=-nF{E}_{\text{cell}}$. Substituting this into our derivative gives us a powerful tool to measure entropy changes experimentally:

Now, let's look at the specific reaction provided:

In this reaction, the metal $\text{M}$ is oxidized from an oxidation state of $0$ to $+2$, meaning it loses $2$ electrons. Therefore, the number of moles of electrons transferred, $n$, is $2$. The problem states that the temperature coefficient $\frac{\text{d}{E}_{\text{cell}}}{\text{d}T}=\frac{R}{F}$. Plugging these values into our formula yields:

Option A claims that the entropy change is $R$, which contradicts our calculated value of $2R$. Thus, Option A is incorrect.

The Concentration Cell

Option B presents us with a specific type of electrochemical cell known as a concentration cell:

In a concentration cell, both the anode and the cathode consist of the same chemical species (in this case, hydrogen electrodes). The only difference is the concentration of the electrolyte. Because the standard reduction potentials of both half-cells are identical, the standard cell potential, $E_{\text{cell}}^{\circ }$, is exactly zero.

We can determine the actual cell potential using the Nernst equation:

Substituting the given concentrations:

Since $E_{\text{cell}}$ is positive, the cell reaction is spontaneous, which means $\Delta G<0$. Now, let's consider the enthalpy change. In a concentration cell, the chemical reactions at the anode and cathode are exact opposites. The net process is simply the transfer of ions from a higher concentration to a lower concentration. No new types of bonds are formed or broken, so the enthalpy change, $\Delta H$, is zero.

Returning to our master equation, $\Delta G=\Delta H-T\Delta S$, and setting $\Delta H=0$, we get:

For $\Delta G$ to be negative (spontaneous), the term $-T\Delta S$ must be negative. Since temperature $T$ is always positive in Kelvin, $\Delta S$ must be positive. The reaction is driven entirely by the increase in entropy as the system moves towards a more uniform concentration. Therefore, Option B is correct.

The Thermodynamics of Racemization

Racemization is the process where a pure optically active enantiomer converts into a racemic mixture (an equimolar mixture of both enantiomers).

Enantiomers have identical physical properties and identical bond energies. Therefore, converting one enantiomer into another involves no net change in enthalpy ($\Delta H\approx 0$). However, a racemic mixture is a more disordered state than a pure enantiomer because it consists of two different types of molecules mixed together.

Since racemization occurs spontaneously over time for many compounds, $\Delta G$ must be negative. With $\Delta H=0$, the spontaneity is entirely driven by the increase in entropy ($\Delta S>0$). Thus, Option C is correct.

The Chelate Effect

Finally, let's examine the coordination chemistry reaction in Option D:

Here, a nickel complex with six monodentate water ligands reacts with three bidentate ethylenediamine (en) ligands. Let's count the number of independent particles on both sides of the equation.

On the reactant side, we have $1$ complex ion + $3$ ethylenediamine molecules = $4$ particles. On the product side, we have $1$ complex ion + $6$ water molecules = $7$ particles.

The reaction results in a net increase in the number of free-moving particles in the solution. More particles mean more degrees of freedom and greater randomness. This significant increase in entropy ($\Delta S>0$) when multidentate ligands replace monodentate ligands is famously known as the Chelate Effect. This makes the formation of chelate complexes highly favorable. Therefore, Option D is correct.

By systematically applying thermodynamic principles to electrochemistry, stereochemistry, and coordination chemistry, we have determined that options (B), (C), and (D) are correct, while option (A) fails due to an incorrect stoichiometric factor.