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If the pressure of hydrogen gas is increased from \(1 \mathrm{arm}\) to \(100 \mathrm{~atm}\), 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) \(0.259 \mathrm{~V}\)

Short Answer

Expert verified
The voltage of the hydrogen half cell is \(-0.059 \, V\).

Step by step solution

01

Understand the Nernst Equation

The Nernst equation relates the voltage of an electrochemical cell to the concentrations of the reactants and products. It is given by \( E = E^0 - \frac{RT}{nF} \ln(Q) \), where \( E^0 \) is the standard electrode potential, \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, \( n \) is the number of electrons transferred in the reaction, \( F \) is the Faraday constant, and \( Q \) is the reaction quotient.
02

Simplify the Nernst Equation at Standard Conditions

At standard temperature (298 K), the Nernst equation simplifies to \( E = E^0 - \frac{0.059}{n} \log(Q) \). Here \( E^0 \) is zero for the hydrogen half-cell reaction \( 2H^+ + 2e^- \rightarrow H_2 \).
03

Determine the Reaction Quotient \(Q\)

For the reaction \( 2H^+ + 2e^- \rightarrow H_2 \), the reaction quotient \( Q \) is \( \frac{1}{[H^+]^2} \times P_{H_2} \), where \( P_{H_2} \) is the pressure of the hydrogen gas.
04

Substitute the Values into the Nernst Equation

Since \([H^+]=1 \, M\) and the pressure changes from \( 1 \, atm \) to \( 100 \, atm \), we have \( Q = \frac{100}{1^2} = 100 \). Substituting into the Nernst equation gives \( E = 0 - \frac{0.059}{2} \log(100) \).
05

Calculate the Voltage Change

Calculate \( E = -\frac{0.059}{2} \log(100) = -\frac{0.059}{2} \times 2 \). This further simplifies to \( E = -0.059 \, V \).
06

Choose the Correct Answer

The calculated voltage change is \(-0.059 \, V\). Thus, the correct answer is option \((a) -0.059 \, V\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electrochemical Cell Voltage
Electrochemical cell voltage, often represented as \( E \), refers to the difference in electric potential between two electrodes of an electrochemical cell. This voltage is responsible for driving the movement of electrons through the external circuit, which in turn facilitates the redox reactions occurring within the cell. The Nernst equation provides a way to calculate the cell voltage under non-standard conditions by incorporating the concentrations or pressures of the reactants and products involved in the reaction.

In a galvanic cell, the cell voltage is generally positive, indicating that the reaction is spontaneous. By altering the reaction conditions, such as the concentration of ions or the pressure of gases involved, one can change the cell voltage. This is precisely what happens in the exercise you are studying: increasing the pressure of hydrogen gas changes the cell voltage.

When considering alterations in reaction conditions, it is essential to use the Nernst equation to calculate the new voltage. This equation allows for adjustments in cell voltage calculations by taking into account factors like reactant concentrations, temperature, and the system's overall state, thus reflecting the real operating conditions of the cell.
Standard Electrode Potential
The standard electrode potential, denoted as \( E^0 \), is the voltage of an electrochemical cell when all components are in their standard states. For a hydrogen half-cell, the standard electrode potential is defined as \( 0 \, V \). This serves as the reference point for measuring the electrode potentials of other elements.

Standard conditions usually involve ions having a concentration of \( 1 \, M \), a pressure of \( 1 \, atm \) for gases, and a temperature of \( 298 \, K \). The standard electrode potential is integral to calculating the overall cell voltage using the Nernst equation, as it provides the baseline from which deviations are analyzed.

The concept of standard electrode potential allows scientists and engineers to predict the direction of electron flow and the feasibility of chemical reactions under set conditions. For instance, in the hydrogen half-cell reaction \( 2H^+ + 2e^- \rightarrow H_2 \), the standard electrode potential plays a crucial role in helping determine how changes such as increased pressure will affect the cell's behavior.
Reaction Quotient
The reaction quotient, symbolized as \( Q \), is a measure of the relative quantities of products and reactants present in a reaction at any given time. It helps determine the direction in which the reaction will proceed to attain equilibrium. Unlike the equilibrium constant \( K \), which applies only at equilibrium, \( Q \) can be calculated at any stage during the reaction.

In the context of the hydrogen half-cell reaction, \( Q \) is given by the expression \( \frac{P_{H_2}}{[H^+]^2} \), where \( P_{H_2} \) is the partial pressure of hydrogen gas, while \([H^+]\) is the concentration of hydrogen ions. A change in either the pressure or concentration affects \( Q \), and consequently the cell potential as calculated by the Nernst equation.

For the problem discussed, increasing the pressure of hydrogen gas from \( 1 \, atm \) to \( 100 \, atm \) alters the reaction quotient \( Q \) from \( 1 \) to \( 100 \). This change shifts the cell away from its standard state and affects the calculated voltage as predicted by the Nernst equation. Understanding \( Q \) is crucial as it provides insight into how far a reaction is from reaching equilibrium at any given time, thus influencing the overall cell voltage.

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Most popular questions from this chapter

Given that \(E^{\circ} \quad\left(\mathrm{Zn}^{2+} / \mathrm{Zn}\right)=-0.763 \mathrm{~V}\) and \(\mathrm{E}^{\circ}\left(\mathrm{Cd}^{2+} / \mathrm{Cd}\right)=-0.403 \mathrm{~V}\), the emf of the following cell \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{a}=0.04) \| \mathrm{Cd}^{2+}(\mathrm{a}=0.2)\right| \mathrm{Cd}\) is given by (a) \(E=+0.36+[0.059 / 2][\log (0.2 / 0.004)]\) (b) \(E=-0.36+[0.059 / 2][\log (0.2 / 0.004)]\) (c) \(E=+0.36+[0.059 / 2][\log (0.004 / 0.2)]\) (d) \(E=-0.36+[0.059 / 2][\log (0.004 / 0.2)]\)

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 \(\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}\)

In the electrolytic cell, flow of electrons is from (a) cathode to anode in solution (b) cathode to anode through external supply (c) cathode to anode through internal supply (d) anode to cathode through internal supply

Two electrochemical cells \(\mathrm{Zn}\left|\mathrm{Zn}^{2+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}\) and \(\mathrm{Fe}\left|\mathrm{Fe}^{2+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}\) are con- nected in series. What will be the net emf of the cell at \(25^{\circ} \mathrm{C} ?\) Given: \(\mathrm{Zn}^{2+} \mid \mathrm{Zn}=-0.73 \mathrm{~V}\), \(\mathrm{Cu}^{2+} \mid \mathrm{Cu}=+0.34 \mathrm{~V}\) \(\mathrm{Fe}^{2+} \mid \mathrm{Fe}=-0.41 \mathrm{~V}\) (a) \(+1.85\) (b) \(-1.85 \mathrm{~V}\) (c) \(+0.83 \mathrm{~V}\) (d) \(-0.83 \mathrm{~V}\)

The \(E\left(\mathrm{M}^{3+} / \mathrm{M}^{2+}\right)\) values for \(\mathrm{Cr}, \mathrm{Mn}, \mathrm{Fe}\) and \(\mathrm{Co}\) are \(-0.41,+1.57,+0.77\) and \(+1.97 \mathrm{~V}\) respectively. For which one of these metals the change in oxidation state form \(+2\) to \(+3\) is easiest? (a) \(\mathrm{Cr}\) (b) \(\mathrm{Mn}\) (c) \(\mathrm{Fe}\) (d) \(\mathrm{Co}\)

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