Problem 9 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For the following cell, \(\mathrm{Zn}(\mathrm{s})\left|\mathrm{ZnSO}_{4}(\mathrm{aq}) \| \mathrm{CuSO}_{4}(\mathrm{aq})\right| \mathrm{Cu}(\mathrm{s})\) when the concentration of \(\mathrm{Zn}^{2+}\) is 10 times the concentration of \(\mathrm{Cu}^{2+}\), the expression for \(\Delta G\) (in \(\mathrm{J} \mathrm{mol}^{-1}\) ) is \([F\) is Faraday constant; \(R\) is gas constant; \(T\) is temperature; \(E^{0}\) (cell) \(\left.=1.1 \mathrm{~V}\right]\) [Adv. 2017] (a) \(1.1 \mathrm{~F}\) (b) \(2.303 R T-2.2 F\) (c) \(2.303 R T+1.1 F\) (d) \(-2.2 F\)

Step-by-Step Solution

Answer

The correct expression is option (b): \(2.303RT - 2.2F\).

Step 1: Identify the Nernst Equation

The Nernst Equation is used to calculate the potential of an electrochemical cell under non-standard conditions. It is given by: \[ E = E^0 - \frac{RT}{nF} \ln Q \]where \( E \) is the cell potential under non-standard conditions, \( E^0 \) is the standard cell potential (1.1 V in this case), \( R \) is the gas constant, \( T \) is the temperature in Kelvin, \( n \) is the number of moles of electrons exchanged, \( F \) is the Faraday constant, and \( Q \) is the reaction quotient.

Step 2: Calculate the Reaction Quotient Q

For the given cell, the reaction is: \[ \text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s) \]The reaction quotient \( Q \) is given by:\[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \]Since \([\text{Zn}^{2+}]\) is 10 times \([\text{Cu}^{2+}]\), \( Q = 10 \).

Step 3: Determine the Expression for ΔG

The change in Gibbs free energy \( \Delta G \) is related to the cell potential by the equation:\[ \Delta G = -nFE \]Substituting for \( E \) from the Nernst equation:\[ \Delta G = -nF \left( E^0 - \frac{RT}{nF} \ln Q \right) \]\[ \Delta G = -nFE^0 + RT \ln Q\]

Step 4: Substitute Known Values and Simplify

Given that \( n = 2 \) because 2 moles of electrons are exchanged, \( E^0 = 1.1 \, \text{V} \), and \( Q = 10 \):\[ \Delta G = -2F(1.1) + 2.303RT \ln(10) \]Simplifying gives:\[ \Delta G = -2.2F + 2.303RT \]

Step 5: Identify the Correct Answer Choice

Look at the given options to find which expresses the derived equation for \( \Delta G \). The equation \( \Delta G = 2.303RT - 2.2F \) matches option (b), which is the correct expression for \( \Delta G \) under the given conditions.

Key Concepts

Nernst Equation

The Nernst Equation is a powerful tool used in electrochemistry to determine the potential of an electrochemical cell under non-standard conditions. It provides insights into how cell potential changes with varying concentrations of ions. The equation is expressed as:- \[ E = E^0 - \frac{RT}{nF} \ln Q \]- Here:

• \( E \) is the cell potential under actual conditions.
• \( E^0 \) refers to the standard cell potential.
• \( R \) is the universal gas constant, \( 8.314 \) J/(mol⋅K).
• \( T \) is the temperature in Kelvin.
• \( n \) denotes the number of moles of electrons involved in the reaction.
• \( F \) is Faraday's constant (approximately 96485 C/mol).
• \( Q \) is the reaction quotient, representing the ratio of product and reactant concentrations.

Breaking it down, the equation shows us how a difference in ion concentration can influence the voltage of the cell. For example, in our reaction, where the concentration of \( \mathrm{Zn}^{2+} \) is 10 times that of \( \mathrm{Cu}^{2+} \), the Nernst equation helps calculate the resulting cell potential, factoring in this concentration gradient.

Reaction Quotient

The reaction quotient, \( Q \), is a crucial aspect in understanding the Nernst Equation's utility. It is a measure of the ratio of concentrations of products to reactants at any point in a reaction that is not necessarily at equilibrium. The expression for \( Q \) changes depending on the balanced chemical equation. For our zinc-copper cell system, the reaction is:- \[ \text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s) \]- From this, \( Q \) is determined by:- \[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \]- Because the zinc ion concentration is provided as 10 times the copper ion concentration, this gives us \( Q = 10 \).
Reacting systems constantly aim to reach equilibrium. As such, the reaction quotient provides a snapshot of the reaction's status relative to equilibrium:

• If \( Q < K \) (equilibrium constant), the reaction proceeds forward.
• If \( Q = K \), the system is at equilibrium.
• If \( Q > K \), the reaction shifts backward.

In conjunction with the Nernst Equation, \( Q \) becomes a vital component in predicting cell potentials under varying conditions.

Gibbs Free Energy

Gibbs Free Energy (\( \Delta G \)) is a term that combines enthalpy and entropy to determine the spontaneity of a reaction. It tells us whether a process can occur under constant pressure and temperature. When it comes to electrochemical cells, \( \Delta G \) can be directly linked to the cell potential by:- \[ \Delta G = -nFE \]- Where:

• \( \Delta G \) is the change in Gibbs Free Energy.
• \( n \) refers to the number of moles of electrons exchanged in the reaction.
• \( F \) stands for Faraday's constant.
• \( E \) is the cell potential.

By substituting the Nernst equation into this expression, we can understand how changes in ion concentration and temperature affect \( \Delta G \):- \[ \Delta G = -nF \left( E^0 - \frac{RT}{nF} \ln Q \right) \]- For our problem:- \[ \Delta G = -2F(1.1) + 2.303RT \ln(10) \]- This shows how increasing the concentration of \( \mathrm{Zn}^{2+} \) impacts the Gibbs Free Energy, ultimately affecting the reaction's spontaneity. When \( \Delta G \) is negative, the reaction is spontaneous, moving towards equilibrium naturally; a positive \( \Delta G \) indicates non-spontaneity under the given conditions.