Problem 201 - A Complete Resource Book in Chemistry for JEE Main

Question

The reduction potential of hydrogen half-cell will be negative if: (a) \(\mathrm{p}\left(\mathrm{H}_{2}\right)=2 \mathrm{~atm}\) and \(\left[\mathrm{H}^{+}\right]=1.0 \mathrm{M}\) (b) \(\mathrm{p}\left(\mathrm{H}_{2}\right)=1 \mathrm{~atm}\) and \(\left[\mathrm{H}^{+}\right]=1.0 \mathrm{M}\) (c) \(\mathrm{p}\left(\mathrm{H}_{2}\right)=1 \mathrm{~atm}\) and \(\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}\) (d) \(\mathrm{p}\left(\mathrm{H}_{2}\right)=2\) atm and \(\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}\)

Step-by-Step Solution

Answer

The reduction potential is negative for options (c) and (d).

Step 1: Understanding the Nernst Equation

The Nernst Equation is used to calculate the electrode potential when conditions are not standard. For a hydrogen half-cell reaction: \( 2H^+ + 2e^- \rightarrow H_2 \), the Nernst Equation is: \[ E = E^0 - \frac{RT}{nF} \ln \frac{[H_2]}{[H^+]^2} \] At 25°C, this simplifies to: \[ E = E^0 + 0.0592 \log \frac{p_{H_2}}{[H^+]^2} \] where \(E^0 = 0\) V under standard conditions: \(1 \text{ atm}\) pressure of \(H_2\) and \(1.0 \text{ M}\) \(H^+\). A negative \(E\) indicates a condition where the standard hydrogen electrode has less tendency to gain electrons.

Step 2: Evaluate Option (a)

Substitute the values: \( p_{H_2} = 2 \text{ atm} \) and \( [H^+] = 1.0 \text{ M} \) into the equation. \[ E = 0 + 0.0592 \log \frac{2}{(1)^2} \] \[ E = 0.0592 \log 2 \approx 0.0592 \times 0.3010 = 0.0178 \text{ V} \] The potential is positive.

Step 3: Evaluate Option (b)

Substitute the values: \( p_{H_2} = 1 \text{ atm} \) and \( [H^+] = 1.0 \text{ M} \) into the equation. \[ E = 0 + 0.0592 \log \frac{1}{(1)^2} \] \[ E = 0.0592 \log 1 = 0 \text{ V} \] The potential remains zero, the same as standard conditions.

Step 4: Evaluate Option (c)

Substitute the values: \( p_{H_2} = 1 \text{ atm} \) and \( [H^+] = 2.0 \text{ M} \) into the equation. \[ E = 0 + 0.0592 \log \frac{1}{(2)^2} \] \[ E = 0.0592 \log \frac{1}{4} \] \[ E = 0.0592 \times (-0.6021) = -0.0356 \text{ V} \] The potential is negative.

Step 5: Evaluate Option (d)

Substitute the values: \( p_{H_2} = 2 \text{ atm} \) and \( [H^+] = 2.0 \text{ M} \) into the equation. \[ E = 0 + 0.0592 \log \frac{2}{(2)^2} \] \[ E = 0.0592 \log \frac{2}{4} \] \[ E = 0.0592 \log 0.5 \approx 0.0592 \times (-0.3010) = -0.0178 \text{ V} \] The potential is negative.

Key Concepts

Electrode Potential

Electrode potential is the measure of a cell's ability to drive an electric current through a circuit. It's related to the energy required to transfer electrons during a chemical reaction within the electrochemical cell. Electrode potential can be positive or negative, depending on the reaction's specific conditions. For hydrogen half-cells, standard electrode potential (denoted as \(E^0\)) is set as zero by convention. This is because the half-cell's potential is measured relative to the standard hydrogen electrode. Under standard conditions, this electrode potential should theoretically be zero if dealing with 1 atm pressure of hydrogen and a 1.0 M concentration of hydrogen ions. Factors that affect electrode potential include:

• The concentration of ions involved in the reaction
• The pressure of gaseous reactants (like \(H_2\))
• Temperature (though at 25°C/298 K, this effect is standardized in the Nernst Equation)

Understanding these variables is essential for predicting and manipulating cell potentials in electrochemical reactions.

Hydrogen Half-cell

The hydrogen half-cell is a reference in electrochemistry used for establishing an electrode system's standard potentials. It comprises a platinum electrode immersed in a solution with hydrogen ions and gas. When this setup is standardized, it is referred to as the standard hydrogen electrode (SHE). This is the baseline at which electrode potentials are measured. In the hydrogen half-cell reaction, hydrogen ions ([\(H^+\)]) combine with electrons to form hydrogen gas ([\(H_2\)]), which can either gain or lose electrons. The reaction is as follows: \[ 2H^+ + 2e^- \rightarrow H_2 \] The purpose of the hydrogen half-cell is to provide a solid foundation for comparison in electrochemical studies. By assigning it a potential of \(0\text{ V}\), it becomes a universal reference point for other electrodes' potentials. Understanding the hydrogen half-cell is crucial for comprehending how different factors can influence electrode potential.

Reduction Potential

Reduction potential, which is often measured in volts, represents the tendency of an electrochemical system's species to gain electrons, also known as being reduced. It helps determine the direction and extent of chemical reactions. A high positive reduction potential indicates a strong tendency to gain electrons, suggesting the substance is a good oxidizing agent. Conversely, a negative reduction potential suggests a substance is more likely to lose electrons, indicating it’s a better reducing agent. In electrochemical cells, knowing the reduction potential helps us predict the behavior of reactions under different conditions, using tools like the Nernst Equation. This widely-used formula allows us to calculate cell potentials under non-standard conditions by considering the concentration and pressure of reactants. By tailoring the conditions of the cell, such as concentration of \([H^+]\) or pressure of \(H_2\), we can manipulate the cell’s reduction potential, altering its electrical output for various applications.