Problem 33 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The Gibbs energy change (in \(\mathrm{J}\) ) for the given reaction at \(\left[\mathrm{Cu}^{2+}\right]=\) \(\left[\mathrm{Sn}^{2+}\right]=1 \mathrm{M}\) and \(298 \mathrm{~K}\) is : \(\mathrm{Cu}(\mathrm{s})+\mathrm{Sn}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Sn}(\mathrm{s})\) \(\left(\mathrm{E}_{\mathrm{Sa}^{2+} \mid \mathrm{Su}}^{0}=-0.16 \mathrm{~V}, \mathrm{E}_{\mathrm{Cu}^{2+} \mid \mathrm{Cu}}^{0}=0.34 \mathrm{~V}\right.\) Take \(\mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1}\) )

Step-by-Step Solution

Answer

The Gibbs energy change is 96500 J/mol.

Step 1: Determine Half-Reactions

Identify the half-reactions involved in the given redox reaction. The oxidation half-reaction is \( \mathrm{Cu} \rightarrow \mathrm{Cu}^{2+} + 2\mathrm{e}^- \) and the reduction half-reaction is \( \mathrm{Sn}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Sn} \).

Step 2: Calculate Standard Cell Potential

The standard cell potential \( E^0_{\text{cell}} \) is given by the difference between the standard reduction potentials of the cathode and anode: \( E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}} = -0.16 \mathrm{~V} - 0.34 \mathrm{~V} = -0.50 \mathrm{~V} \).

Step 3: Use Nernst Equation

Since concentrations are standard \((1 \text{M})\), the reaction quotient \( Q = 1 \), and thus the Nernst equation simplifies to \( \Delta G = -nFE^0_{\text{cell}} \). The reaction involves 2 moles of electrons \((n=2)\).

Step 4: Calculate Gibbs Energy Change

Substitute the known values into the equation: \( \Delta G = -(2)(96500)(-0.50) \). Therefore, \( \Delta G = 96500 \times 1 \mathrm{~J/mol} = 96500 \mathrm{~J/mol} \).

Step 5: Conclude the Calculation

The Gibbs energy change for the reaction, under the specified standard conditions, is \( \Delta G = 96500 \mathrm{~J/mol} \).

Key Concepts

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are chemical processes where the oxidation state of atoms changes. These transformations occur through the movement of electrons between substances. There are always two parts to a redox reaction: one substance loses electrons in the oxidation half-reaction, while another gains those electrons in the reduction half-reaction.
For example, in the exercise given, copper (\( \text{Cu}\)) is oxidized to copper ions (\( \text{Cu}^{2+} \)). As it loses electrons, it shifts from a metal to a positively charged ion. Meanwhile, tin ions (\( \text{Sn}^{2+} \)) gain those electrons to convert into metallic tin, completing the reduction half-reaction.

Key features to keep in mind with redox reactions include:

• Identification of oxidizing and reducing agents: the substance that gets oxidized (loses electrons) is the reducing agent, whereas the one that gets reduced (gains electrons) is the oxidizing agent.
• Balancing of electron transfer, ensuring that the total loss of electrons equals the total gain.
• Each redox system involves at least two electrodes, represented in voltaic cells where such reactions often take place.

Grasping these basics is essential for understanding how energy changes are harnessed in electrochemical cells.

Standard Cell Potential

Standard Cell Potential (\(E^0\)) is a key concept in electrochemistry that measures how easily an electrochemical cell can produce electricity under standard conditions. It represents the voltage difference between two half-cells in a battery when the concentrations of the electrolytes are at standard conditions (1 M, 25°C).
In our exercise, the standard potentials are given as -0.16 V for tin and 0.34 V for copper. To find the overall standard cell potential, you subtract the anode potential from the cathode potential, as follows:
\[ E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}} \]
\[ E^0_{\text{cell}} = -0.16 \, \text{V} - 0.34 \, \text{V} = -0.50 \, \text{V} \]

Important aspects of standard cell potential include:

• The standard cell potential can be a positive or negative value, indicating whether the reaction is spontaneous (positive) or non-spontaneous (negative) under standard conditions.
• It's calculated from standard reduction potentials, which are generated based on comparisons to the standard hydrogen electrode (SHE).
• This potential reflects the cell's ability to do work, correlating with the number of electrons transferred and their energy difference, affecting the Gibbs free energy calculation.

Understanding this concept is vital as it connects the chemical properties of reactions with their electrical characteristics.

Nernst Equation

The Nernst equation helps relate the Gibbs energy change in a reaction to the cell potential under non-standard conditions. It is particularly useful for determining cell potentials at any concentrations rather than at fixed standard conditions.
The basic form of the Nernst equation is:
\[E = E^0 - \frac{RT}{nF} \ln Q\]
Where:

• \(E\) is the cell potential under non-standard conditions.
• \(E^0\) is the standard cell potential.
• \(R\) is the universal gas constant (8.314 J/mol·K).
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of moles of electrons transferred in the reaction.
• \(F\) is Faraday's constant (96500 C/mol).
• \(Q\) is the reaction quotient, which is the ratio of concentrations of products to reactants.

In the exercise, since the reactants are at their standard 1 M concentrations, the Nernst equation simplifies because \(Q = 1\), and thus \(\ln Q = 0\), reducing the equation to:
\[\Delta G = -nFE^0_{\text{cell}}\]
This equation shows how \( \Delta G \), the Gibbs energy, is directly related to the number of electrons involved and the cell’s electrical potential. The Nernst equation elevates our understanding of how actual reaction conditions shift cell potentials from their calculated standard values.