Problem 180 - A Complete Resource Book in Chemistry for JEE Main

Question

For a cell reaction involving two electrons, the standard emf of the cell is found to be \(0.295 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). The equilibrium constant of the reaction at \(25^{\circ} \mathrm{C}\) will be \([\mathbf{2 0 0 3}]\) (a) \(1 \times 10^{-10}\) (b) \(29.5 \times 10^{-2}\) (c) 10 (d) \(1 \times 10^{10}\)

Step-by-Step Solution

Answer

The equilibrium constant \(K\) is \(1 \times 10^{10}\), so the answer is (d).

Step 1: Identify Known Values

For this electrochemical cell reaction, you know the number of electrons, which is 2, and the standard emf, which is given as \(E^\circ = 0.295 \mathrm{~V}\). The temperature is given as \(25^{\circ} \mathrm{C}\).

Step 2: Use Nernst Equation for Equilibrium

At equilibrium, the relationship between emf, temperature, number of moles exchanged, Faraday's constant, and the equilibrium constant \(K\) is given by the Nernst equation: \(E^\circ = \frac{RT}{nF} \ln K\).

Step 3: Insert Known Constants and Values

From the Nernst equation, plug in the constants: \(R = 8.314 \text{ J/mol K}\), \(T = 298 \text{ K}\) (which is \(25^{\circ} \mathrm{C} + 273\)), \(n = 2\), and \(F = 96500 \text{ C/mol}\). Rearrange the equation to find \(K\): \[0.295 = \frac{8.314 \times 298}{2 \times 96500} \ln K\]

Step 4: Solve for \(\ln K\)

Calculate the expression \(\frac{8.314 \times 298}{2 \times 96500}\) to simplify the equation and solve for \(\ln K\): \[0.295 = 0.0127 \ln K\]Thus, \(\ln K = \frac{0.295}{0.0127}\).

Step 5: Calculate and Find \(K\)

Complete the division to find \(\ln K\): \[\ln K \approx 23.23\]Then, find \(K\) by exponentiating the result: \[K = e^{23.23} \approx 1 \times 10^{10}\].

Step 6: Verify Against Options

Compare the calculated \(K\) with the options. The closest match is option (d) \(1 \times 10^{10}\).

Key Concepts

Nernst equation

The Nernst equation is an essential tool in electrochemistry. It connects the cell potential of an electrochemical reaction to its standard electromotive force (emf), temperature, and reaction quotient or equilibrium constant. This equation enables us to predict how changes in a system, such as concentration or temperature, influence the cell voltage.
The Nernst equation is expressed as:\[E = E^\circ - \frac{RT}{nF} \ln Q\]

• \(E^\circ\) is the standard emf of the cell in volts.
• \(R\) is the universal gas constant, approximately 8.314 J/mol K.
• \(T\) is the temperature in Kelvin. For most calculations, we use 298 K, which is equivalent to 25°C.
• \(n\) is the number of moles of electrons transferred in the reaction.
• \(F\) is the Faraday constant, about 96500 C/mol, representing the charge of one mole of electrons.
• \(Q\) denotes the reaction quotient, which adjusts to give the equilibrium constant \(K\) at equilibrium.

When the reaction reaches equilibrium, the cell potential \(E\) is zero, and the equation simplifies to relate standard emf to the equilibrium constant \(K\). This helps show how spontaneous a reaction is under standard conditions.

equilibrium constant

The equilibrium constant \(K\) quantifies the position of equilibrium for a chemical reaction. It indicates the proportion of products to reactants at equilibrium. A high \(K\) value signifies that, at equilibrium, products are greatly favored over reactants. Electrochemically, it illustrates the potential of a reaction to proceed.
In the context of the Nernst equation, it relates to the standard emf by:\[E^\circ = \frac{RT}{nF} \ln K\]By knowing \(E^\circ\), \(R\), \(T\), \(n\), and \(F\), we can calculate \(K\). For instance, with a standard emf of 0.295 V, using these constants in the equation, we deduce the equilibrium constant.
This is vital in predicting both the direction and extent of reactions:

• Large \(K\) values (much greater than 1) indicate extensive product formation.
• Small \(K\) values (much less than 1) indicate a greater amount of reactant remains.

Through calculations, like in the example from the exercise, we obtain \(K = 1 \times 10^{10}\), showing a strong tendency toward product formation at equilibrium.

standard emf

The standard electromotive force (emf), denoted \(E^\circ\), is a measure of the driving force behind an electrochemical reaction. It represents the potential difference between two half-cells in a galvanic cell under standard conditions.
The standard emf is determined using standard reduction potentials, which are typically found in tables. It is calculated by subtracting the reduction potential of the anode from that of the cathode:\[E^\circ = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\]These values provide insight into reaction spontaneity:

• A positive \(E^\circ\) signifies a spontaneous reaction.
• A negative \(E^\circ\) indicates non-spontaneity under standard conditions.

Understanding \(E^\circ\) is crucial because it relates directly to the equilibrium constant \(K\) as seen in the Nernst equation. A higher standard emf typically correlates with a higher equilibrium constant, suggesting a greater propensity for the reaction to produce products.
This way, the standard emf serves not only as a metric of reaction potential but also a bridge to connecting electrochemical reactions with chemical equilibria.