Problem 128 - A Complete Resource Book in Chemistry for JEE Main

Question

If the pressure of hydrogen gas is increased from \(1 \mathrm{arm}\) to 100 atm, keeping the hydrogen ion concentration constant at \(1 \mathrm{M}\), the voltage of the hydrogen half cell at \(25^{\circ} \mathrm{C}\) will be (a) \(-0.059 \mathrm{~V}\) (b) \(+0.059 \mathrm{~V}\) (c) \(5.09 \mathrm{~V}\) (d) \(0.259 \mathrm{~V}\)

Step-by-Step Solution

Answer

The voltage change is approximately +0.118 V, with the closest multiple-choice option being d) 0.259 V, though not exact.

Step 1: Identify the Nernst Equation

The Nernst Equation relates the cell potential to the concentration of ions and the partial pressures of the gas involved. For the hydrogen electrode, the Nernst equation is given by: \[ E = E^0 - \frac{0.0591}{n} \log \left( \frac{1}{P(\text{H}_2)} \right) \]where \(E^0\) is the standard electrode potential (0 V for hydrogen), \(n\) is the number of electrons involved in the reaction (1 for hydrogen), and \(P(\text{H}_2)\) is the partial pressure of hydrogen gas.

Step 2: Substitute Known Values

Substitute the given values into the Nernst equation. We know the pressure initially is 1 atm and changes to 100 atm:\[ E = 0 - \frac{0.0591}{1} \log \left( \frac{1}{100} \right) \]

Step 3: Simplify the Expression

Simplify the logarithmic expression to find the change in electrode potential:\[ E = -0.0591 \log (0.01) \]\[ E = -0.0591 \times (-2) \]where the logarithm \( \log (0.01) \) equals -2 since 0.01 is \(10^{-2}\).

Step 4: Calculate the Voltage

Calculate the resulting voltage value:\[ E = 0.1182 \mathrm{~V} \]The change is positive because the pressure increase causes the potential to increase. However, since the options provided must be evaluated, the closest numerical choice to our result given the options is not exact, but the logic behind the closest approximation will mark the value for further selection.

Key Concepts

Hydrogen Half-Cell

A hydrogen half-cell is an electrochemical cell component where hydrogen gas is bubbled over a platinum electrode in an acidic solution. This is often set up under standard conditions, which means a pressure of 1 atm for hydrogen gas and a 1 M concentration of hydrogen ions. The hydrogen half-cell follows the reaction:

• Hydrogen gas dissociates to form protons and electrons: the half-reaction is displayed as \( ext{H}_2(g) \rightarrow 2 ext{H}^+(aq) + 2e^-\)
• Electrons are released into the circuit
• Protons enter the solution

The hydrogen half-cell plays a critical role as it sets the reference point in the measurement of other electrode potentials.
This helps us understand how other cell potentials are compared. Thus, this half-cell is crucial in electrochemical calculations and energy conversion systems.

Electrode Potential

Electrode potential, often denoted as \(E\), represents the voltage developed by a half-cell in an electrochemical cell.
This is caused by the tendency of a species to gain or lose electrons.

• If electrons are easily lost (oxidation), the potential is higher
• If electrons are easily gained (reduction), the potential is lower

The standard electrode potential, denoted as \(E^0\), is measured under standard conditions (1 atm pressure, 1 M concentration, and 25°C) using the standard hydrogen electrode as a reference.
The Nernst equation gives us a way to calculate the electrode potential under non-standard conditions by adjusting for differences in ion concentration and partial gas pressures. This shows how adaptable electrode potential is with real-world scenarios, changing depending on different environments set up by experimental conditions.

Cell Potential

The cell potential, also known as electromotive force (emf) of a cell, is the difference in potential between two half-cells under certain conditions.
It measures the drive for a chemical reaction in the cell.

• The greater the cell potential, the stronger the drive for the reaction
• Calculated as the sum of the potentials of the two half-cells in a galvanic cell

Using the Nernst equation allows us to find the cell potential at any given conditions, not just at standard conditions.
For instance, changes in pressure or concentration affect the calculated potential, as seen when increasing hydrogen gas pressure in the exercise.
This knowledge is pivotal when designing batteries or any energy conversion device, as it determines the voltage output based on various experimental setups and requirements.