Problem 34 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For the disproportionation reaction $$ 2 \mathrm{Cu}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \text { at } 298 \mathrm{~K}, \ln \mathrm{K} $$ (where \(\mathrm{K}\) is the equilibrium constant) is \(\times 10^{-1}\). Given [Main Sep. 02, 2020 (II)] $$ \left(\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}}^{0}=0.16 \mathrm{~V} ; \mathrm{E}_{\mathrm{Cu}^{+} / \mathrm{Cu}}^{0}=0.52 \mathrm{~V} ; \frac{\mathrm{RT}}{\mathrm{F}}=0.025\right) $$

Step-by-Step Solution

Answer

\( \ln K = -28.8 \)

Step 1: Write the Nernst Equation

The Nernst equation relates the standard cell potential (E°) of an electrochemical cell to its equilibrium constant (K). It is expressed as: \[E^0 = \frac{RT}{nF} \ln K,\] where \( E^0 \) is the standard cell potential, \( R \) is the gas constant, \( T \) is the temperature in Kelvin, \( F \) is Faraday's constant, and \( n \) is the number of moles of electrons exchanged in the reaction.

Step 2: Determine the Number of Electrons Transferred

For the given reaction: \[ 2 \mathrm{Cu}^+ \rightleftharpoons \mathrm{Cu} + \mathrm{Cu}^{2+},\]envision the half-reactions and find the number of electrons transferred:- \( \mathrm{Cu}^{+} \to \mathrm{Cu} \): 1 electron- \( \mathrm{Cu}^{+} \to \mathrm{Cu}^{2+} \): loss of 1 electron.The total electrons transferred in the overall reaction is \( n = 2 \) since each Cu+ provides one electron and two such ions react.

Step 3: Calculate the Standard Cell Potential

Calculate the standard cell potential \( E^0 \) using the given electrode potentials:- \( E_{\mathrm{Cu}^{2+}/\mathrm{Cu}^+}^0 = 0.16 \text{ V} \)- \( E_{\mathrm{Cu}^+/\mathrm{Cu}}^0 = 0.52 \text{ V} \)The net cell potential \( E^0 \) for the reaction is found by combining these potentials: \[ E^0 = E_{\mathrm{Cu}^{2+}/\mathrm{Cu}^+}^0 - E_{\mathrm{Cu}^+/\mathrm{Cu}}^0 = 0.16 \text{ V} - 0.52 \text{ V} = -0.36 \text{ V} \]

Step 4: Plug Values into the Nernst Equation

Substitute the values into the Nernst equation to solve for \( \ln K \): \[ -0.36 \text{ V} = \frac{0.025}{2} \ln K \]\[ -0.36 = 0.0125 \ln K \]

Step 5: Solve for \(\ln K\)

Isolate \( \ln K \) in the equation from Step 4: \[ \ln K = \frac{-0.36}{0.0125} \]\[ \ln K = -28.8 \]

Key Concepts

Electrochemical Cells

Electrochemical cells are devices that convert chemical energy into electrical energy through chemical reactions. They consist of two electrodes, each serving as a site for oxidation or reduction. These reactions occur in separate half-cells, and the flow of electrons from one electrode to the other through an external circuit generates an electric current.

The core principle involves redox reactions, where oxidation takes place at the anode and reduction occurs at the cathode. In our copper disproportionation reaction, the movement of electrons between copper ions is what powers the cell.

- **Anode**: The electrode where oxidation occurs. Electrons are released from this site.
- **Cathode**: The electrode where reduction occurs. Electrons are gained at this site.

Electrochemical cells are the basis for batteries and play a critical role in various industrial processes, from electroplating to chemical manufacturing. Understanding how they work is essential in chemistry and physics.

Standard Cell Potential

The standard cell potential, represented as \(E^0\), is a measure of the electromotive force (EMF) of an electrochemical cell when all species are in their standard states (usually at 1 M concentration, 1 atm pressure, and 25°C or 298 K). It is an indication of the tendency of a cell reaction to occur spontaneously.

The potentials of half-reactions help determine the overall standard cell potential. For a cell composed of two half-reactions, the difference between the cathode and anode potentials gives us the cell's standard potential.

In the context of our exercise, the standard half-cell potentials provided are:

• \(E^0_{\mathrm{Cu}^{2+}/\mathrm{Cu}^+} = 0.16 \text{ V}\)
• \(E^0_{\mathrm{Cu}^+/\mathrm{Cu}} = 0.52 \text{ V}\)

By calculating the difference between these potentials according to the formula for the cell under consideration, we find \(E^0\) to be \(-0.36 \text{ V}\), indicating non-spontaneous conditions under standard circumstances without external influence.

Equilibrium Constant

The equilibrium constant, \(K\), quantifies the position of equilibrium in a chemical reaction. It expresses the ratio of the concentration of products to reactants at equilibrium. A large \(K\) indicates that the reaction favors products, while a small \(K\) suggests the reactants are favored.

The Nernst equation ties \(K\) to the standard cell potential, \(E^0\). In this exercise, we use the relationship:\[ E^0 = \frac{RT}{nF} \ln K \]where \(R\) is the gas constant, \(T\) is temperature, \(n\) is the number of moles of electrons exchanged, and \(F\) is Faraday's constant. Plugging in the appropriate values and solving the Nernst equation allows us to determine \(\ln K\) and hence \(K\).

With a calculated \(\ln K\) of \(-28.8\), this reaction strongly favors the formation of reactants over products. It emphasizes that, at equilibrium, very little \(\mathrm{Cu}_s\) and \(\mathrm{Cu}^{2+}\) are formed relative to the initial \(\mathrm{Cu}^+\) ions.