Problem 4 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The standard Gibbs energy for the given cell reaction in \(\mathrm{kJ} \mathrm{mol}^{-1}\) at \(298 \mathrm{~K}\) is: [Main April 9, 2019 (I)] \(\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})\) \(\mathrm{E}^{\circ}=2 \mathrm{~V}\) at \(298 \mathrm{~K}\) (Faraday's constant, \(\left.\mathrm{F}=96000 \mathrm{C} \mathrm{mol}^{-1}\right)\) (a) \(-384\) (b) 384 (c) 192 (d) \(-192\)

Step-by-Step Solution

Answer

The standard Gibbs energy is \(-384 \text{ kJ/mol}\) at 298 K, corresponding to option (a).

Step 1: Understand the Problem

We need to find the standard Gibbs free energy change (\( \Delta G^{\circ} \)) for the given electrochemical reaction using the standard cell potential (\( \mathrm{E}^{\circ} \)) and Faraday's constant.

Step 2: Recall the Formula

The relationship between the standard Gibbs free energy change (\( \Delta G^{\circ} \)) and the standard cell potential (\( \mathrm{E}^{\circ} \)) is given by the formula:\[\Delta G^{\circ} = -nFE^{\circ}\]where \( n \) is the number of moles of electrons transferred in the balanced cell equation, \( F \) is Faraday's constant, and \( E^{\circ} \) is the standard cell potential.

Step 3: Identify Values

From the given reaction:\( \text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu} \)there are 2 electrons transferred. Therefore, \( n = 2 \).\( F = 96000 \text{ C/mol} \)\( E^{\circ} = 2 \text{ V} \) as given.

Step 4: Substitute Values into the Formula

Substitute the known values into the formula to calculate \( \Delta G^{\circ} \):\[\Delta G^{\circ} = - (2)(96000)(2) = -384000 \text{ J/mol}\]Convert joules to kilojoules:\[-384000 \text{ J/mol} = -384 \text{ kJ/mol}\]

Step 5: Confirm Correct Answer

The calculated standard Gibbs free energy change is \(-384 \text{ kJ/mol}\), which corresponds to option (a) \(-384\).

Key Concepts

Standard Gibbs Free Energy

In electrochemistry, the concept of standard Gibbs free energy (\( \Delta G^{\circ} \)) is vital.It helps determine how a reaction will proceed and how much energy is available to do work.Gibbs free energy combines the system's entropic and enthalpic properties and provides insights into the reaction's spontaneity.For a chemical reaction in a state defined by standard conditions, the equation to calculate the Gibbs free energy is \[ \Delta G^{\circ} = -nFE^{\circ} \]

• \( n \) is the number of moles of electrons transferred.
• \( F \) is Faraday's constant.
• \( E^{\circ} \) is the standard cell potential.

A negative \( \Delta G^{\circ} \) value indicates a spontaneous reaction, meaning it will occur without additional energy input under standard conditions.In the exercise, a \( \Delta G^{\circ} \) of \(-384 \,\text{kJ/mol}\) means that the zinc and copper ion reaction proceeds spontaneously, favoring the formation of products.

Standard Cell Potential

The standard cell potential (\( E^{\circ} \)) of a reaction is a measure of the driving force behind an electrochemical reaction.It is the voltage difference between two half-cells under standard conditions, which is 1 molar concentration, 1 atm pressure, and 298 K.This potential is determined by the nature of the reactants and their tendency to gain or lose electrons.
For the reaction given in the exercise:Ul>

The zinc metal acts as an anode, undergoing oxidation by losing electrons.

Copper ion reduction occurs at the cathode, gaining electrons.

The standard cell potential \( E^{\circ} \) is calculated as\( 2\, \text{V} \), showing the inherent power of the cell to drive the reaction forward under these conditions.A higher \( E^{\circ} \) value generally translates to a more vigorous reaction and a more significant energy release when electrons flow through the cell.

Faraday's Constant

Faraday’s constant is a fundamental constant in electrochemistry, symbolized as \( F \).It represents the charge of one mole of electrons and is valued at 96000 coulombs per mole (C/mol).
This constant is vital for connecting the macroscopic quantities of chemical reactions to microscopic electrical phenomena.In any electrochemical process:

• \( F \) is often used to bridge the quantity of substance and electrical charge exchanged in reactions.
• It allows the conversion between moles of electrons and coulombs during calculations of energy changes.

By incorporating Faraday's constant together with cell potential, one can determine the Gibbs free energy change, as shown in the formula \( \Delta G^{\circ} = -nFE^{\circ} \).Knowing \( F \) simplifies predicting how much energy will be used or produced when a specific chemical reaction is run as an electrochemical cell.