Problem 37 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Consider an electrochemical cell: \(A(\mathrm{~s})\left|A^{\mathrm{n}+}(\mathrm{aq}, 2 \mathrm{M})\right| B^{2 \mathrm{n}+}(\mathrm{aq}, 1 \mathrm{M}) \mid B(\mathrm{~s})\) The value of \(\Delta H^{\circ}\) for the cell reaction is twice that of \(\Delta G^{\circ}\) at \(300 \mathrm{~K}\). If the emf of the cell is zero, the \(\Delta S^{\circ}\) (in \(\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\) ) of the cell reaction per mole of \(B\) formed at \(300 \mathrm{~K}\) is (Given: \(\ln (2)=0.7, R\) (universal gas constant) \(=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} . H, S\) and \(G\) are enthalpy, entropy and Gibbs energy, respectively.)

Step-by-Step Solution

Answer

The entropy change \(\Delta S^{\circ}\) is zero.

Step 1: Understanding the Problem

We are given an electrochemical cell and asked to find the entropy change \(\Delta S^{\circ}\) for the cell reaction when certain conditions are given. The conditions include that the cell emf is zero and \(\Delta H^{\circ}\) is twice \(\Delta G^{\circ}\) at \(300 \, \text{K}\).

Step 2: Define Key Equations

We need the Gibbs free energy equation \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\). Since emf (electromotive force) is zero, we have \(\Delta G^{\circ} = 0\). Also given is \(\Delta H^{\circ} = 2 \Delta G^{\circ}\).

Step 3: Use Given Relationships

Substitute \(\Delta G^{\circ} = 0\) into the equation for \(\Delta H^{\circ}\). This implies \(\Delta H^{\circ} = 2 \times 0 = 0\). Since \(\Delta G^{\circ} = 0\), \(\Delta H^{\circ} = T\Delta S^{\circ}\).

Step 4: Solve for \(\Delta S^{\circ}\)

Set \(\Delta H^{\circ} = T \Delta S^{\circ}\) and note that \(T = 300 \, \text{K}\). Substituting \(\Delta H^{\circ} = 0\), we have \(0 = 300 \times \Delta S^{\circ}\). Solving this gives \(\Delta S^{\circ} = 0\).

Key Concepts

Gibbs Free Energy

Gibbs Free Energy (\( \Delta G \)) is a concept in thermodynamics that helps predict whether a chemical reaction will occur spontaneously. It combines three essential variables: enthalpy (\( \Delta H \)), entropy (\( \Delta S \)), and temperature (\( T \)). The equation is expressed as:
\[\Delta G = \Delta H - T\Delta S\]This equation tells us:

• Whether a reaction is energetically favorable. If \( \Delta G \) is negative, the reaction can occur spontaneously.
• If \( \Delta G \) is zero, the system is in equilibrium, meaning there is no net change as both forward and reverse reactions occur at the same rate.
• If \( \Delta G \) is positive, the reaction will require energy input.

Understanding Gibbs Free Energy is crucial for analyzing electrochemical cells. In an electrochemical reaction, change in Gibbs energy relates to the electric work done by the cell. Gibbs Free Energy also helps relate thermal and electrical energy aspects of the cell's behavior.
In our exercise, since the emf of the cell is zero, it means \(\Delta G^{\circ} = 0\). Therefore, the system is in equilibrium under the given conditions.

Entropy Change

Entropy (\( \Delta S \)), a measure of disorder or randomness in a system, is a crucial factor in predicting reaction spontaneity. A higher entropy indicates greater disorder and usually a more probable state. The change in entropy can affect the likelihood of a reaction progressing.
In thermodynamics and electrochemistry, entropy tells us:

• The level of disorder in a system's molecules.
• A positive \( \Delta S \) is indicative of increased entropy.
• A negative \( \Delta S \) suggests a more ordered system.

For the given electrochemical cell, the problem provides that the emf is zero and we calculate \( \Delta S^{\circ} \). Using the Gibbs equation:
\[ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \]
With \( \Delta G^{\circ} = 0 \), the entropy term becomes crucial. Given that \( \Delta H^{\circ} \) is also zero (as concluded in the exercise), we end up with:
\[ 0 = 300 \times \Delta S^{\circ} \]Thus, \( \Delta S^{\circ} = 0 \), indicating no change in disorder under these conditions.

Enthalpy

Enthalpy (\( \Delta H \)) is a concept that quantifies the total heat content of a system. It reflects the energy stored in chemical bonds plus the energy based on pressure and volume. Enthalpy changes are vital for understanding energy exchanges in chemical reactions.
Here's what you need to know about enthalpy:

• If \( \Delta H \) is positive, the reaction absorbs heat (endothermic).
• If \( \Delta H \) is negative, the reaction releases heat (exothermic).
• It is usually measured in Joules or Kilojoules.

In the exercise, we found that \( \Delta H^{\circ} \) is given by twice the value of \( \Delta G^{\circ} \), and since \( \Delta G^{\circ} \) is zero (due to zero emf), \( \Delta H^{\circ} \) also equals zero. This indicates no net heat changetakes place during the reaction under standard conditions. Understanding enthalpy helps in assessing the thermal dynamics of electrochemical cells and can aid in predicting reaction feasibility.