Problem 74 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of \(10^{-6} \mathrm{M}\) hydrogen ions. The EMF of the cell is \(0.118 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). Calculate the concentration of hydrogen ions at the positive electrode.

Step-by-Step Solution

Answer

The concentration of hydrogen ions at the positive electrode is \( 6.4 \times 10^{-11} \text{ M} \).

Step 1: Identify the Nernst Equation

To find the concentration of hydrogen ions at the positive electrode, use the Nernst equation. The equation for a cell with hydrogen electrodes is given as: \[ E = E^0 - \frac{RT}{nF} \ln \frac{[H^+]_2}{[H^+]_1} \] where \( [H^+]_1 \) and \( [H^+]_2 \) are the concentrations of hydrogen ions at the negative and positive electrodes respectively. \( E \) is the EMF of the cell, \( R \) is the gas constant (8.314 J/mol K), \( T \) is the temperature in Kelvin, \( n \) is the number of electrons involved in the cell reaction (which is 2 for hydrogen), and \( F \) is the Faraday constant (96485 C/mol).

Step 2: Simplify Using Standard Conditions and Electrode Data

Since the hydrogen electrodes are standard electrodes, \( E^0 = 0 \) V. Thus the Nernst equation simplifies to \[ E = -\frac{RT}{nF} \ln \frac{[H^+]_2}{10^{-6}} \]. We have to find \( [H^+]_2 \), the concentration at the positive electrode, with \( E = 0.118 \) V at \( 25^{\circ} \text{C} \) (which is 298 K).

Step 3: Substitute the Known Values and Rearrange

Plug the values for \( R \), \( T \), \( F \), \( n \), and \( E \) into the simplified equation to solve for \( [H^+]_2 \):\[ 0.118 = -\frac{8.314 \times 298}{2 \times 96485} \ln \frac{[H^+]_2}{10^{-6}} \]. Rearrange the equation to solve for \( [H^+]_2 \):\[-\ln \frac{[H^+]_2}{10^{-6}} = \frac{0.118 \times 2 \times 96485}{8.314 \times 298} \].

Step 4: Calculate the Logarithmic Term

Calculate the right-hand side of the equation:\[-\ln \frac{[H^+]_2}{10^{-6}} = \frac{0.118 \times 2 \times 96485}{8.314 \times 298} \approx 9.64 \]. Therefore, \( \ln \frac{[H^+]_2}{10^{-6}} = -9.64 \).

Step 5: Solve for the Positive Electrode Concentration

To find \( [H^+]_2 \), solve the logarithmic equation. Exponentiate both sides to isolate \( [H^+]_2 \):\[ \frac{[H^+]_2}{10^{-6}} = e^{-9.64} \]. This yields \( [H^+]_2 = 10^{-6} e^{-9.64} \) M. Use a calculator to find \( e^{-9.64} \approx 6.4 \times 10^{-5} \).

Step 6: Final Concentration Result

Substitute back to find \( [H^+]_2 \): \( [H^+]_2 = 10^{-6} \times 6.4 \times 10^{-5} = 6.4 \times 10^{-11} \).

Key Concepts

Hydrogen Ion Concentration

The concentration of hydrogen ions in a solution is crucial because it determines the acidity or basicity of the solution, which is commonly expressed as the pH value. The hydrogen ion concentration at the negative electrode in this problem is given as \(10^{-6} \text{ M}\). Understanding this concentration helps us apply the Nernst Equation to determine the hydrogen ion concentration at the positive electrode, needed to calculate the electrochemical cell potential.

In a solution:

• Higher hydrogen ion concentration means lower pH — more acidic solution.
• Lower hydrogen ion concentration results in higher pH — more basic or alkaline solution.

Here, by using the Nernst Equation, we start with known hydrogen ion concentrations of \(10^{-6} \text{ M}\) and calculate the unknown concentration, which is essential to measure and predict cell potential accurately.

Electrochemical Cell

An electrochemical cell is a device that converts chemical energy into electrical energy through redox reactions. Electrochemical cells consist of two electrodes: an anode (negative) and a cathode (positive), with each electrode immersed in a solution containing ions.

In this specific problem, a cell with two hydrogen electrodes is used. They partake in the exchange of hydrogen ions and electrons, creating a flow of electric charge known as current. This flow extremely affects our ability to direct and measure useful electrical energy from chemical reactions. The electrochemical cell:

• Consists of two electrodes submerged in electrolyte solutions.
• Utilizes redox reactions to cause electron flow.
• Measures the cell's potential difference using the Nernst Equation to calculate parameters like hydrogen ion concentration.

Understanding how electrochemical cells operate is vital for designing sensing and transforming systems, such as batteries and sensors.

Electrode Potential

Electrode potential is a measure of the tendency of an electrode to lose or gain electrons, which in turn determines its capacity to drive an electric current. It is affected by the concentration of ions in solution, temperature, and the nature of the electrode material. The potential difference, or electromotive force (EMF), between two electrodes in an electrochemical cell results from the differential electrode potentials.

In this exercise:

• The Nernst Equation calculates the electrode potential of a cell under non-standard conditions based on ion concentration at each electrode.
• A constant temperature of 25°C and a known EMF are utilized to compute the unknown hydrogen ion concentration at the positive electrode.

By understanding electrode potential, one can predict how an electrochemical cell will behave under different conditions and concentration changes, which is crucial in fields like electrochemistry and battery technology.