Problem 26 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

The dissociation constant for \(\mathrm{CH}_{3} \mathrm{COOH}\) is \(1.8 \times 10^{-5}\) at \(298 \mathrm{~K}\). The electrode potential for the half-cell: \(\mathrm{Pt} / \mathrm{H}_{2}\) \((1\) bar \() \mid 0.5 \mathrm{M}-\mathrm{CH}_{3} \mathrm{COOH}\), at \(298 \mathrm{~K}\) is \((\log 2=0.3 ; \log 3=0.48 ; 2.303 R T / F\) \(=0.06\) ) (a) \(-0.3024 \mathrm{~V}\) (b) \(-0.1512 \mathrm{~V}\) (c) \(+0.3024 \mathrm{~V}\) (d) \(+0.1512 \mathrm{~V}\)

Step-by-Step Solution

Answer

+0.3024 V

Step 1: Identify the Relevant Nernst Equation

Begin with the general formula for the Nernst Equation, which relates the electrode potential to the standard electrode potential, concentration, and temperature: \[ E = E^\circ - \frac{2.303 RT}{nF} \log Q \]where \( E \) is the electrode potential, \( E^\circ \) is the standard electrode potential, \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, \( n \) is the number of moles of electrons transferred in the half-reaction, \( F \) is the Faraday's constant, and \( Q \) is the reaction quotient.

Step 2: Calculate the Reaction Quotient Q

For the reaction at equilibrium, the dissociation of acetic acid is given by:\[ \mathrm{CH}_{3}COOH \rightleftharpoons \mathrm{CH}_{3}COO^- + \mathrm{H}^+ \]Given the dissociation constant \( K_a = [\mathrm{CH}_{3}COO^-][\mathrm{H}^+] / [\mathrm{CH}_{3}COOH] = 1.8 \times 10^{-5} \).The concentration of acetic acid \( [\mathrm{CH}_{3}COOH] \) is given as 0.5 M. Assuming complete dissociation, \( Q = K_a = 1.8 \times 10^{-5} \) because concentrations of \( \mathrm{CH}_{3}COO^- \) and \( \mathrm{H}^+ \) are equal in the equilibrium state.

Step 3: Identify the Number of Electrons Transferred (n) and Calculate E

The number of electrons transferred in the hydrogen gas electrode reaction is 2:\[ \mathrm{H}^+ + 2e^- \rightleftharpoons \mathrm{H}_2(g) \]Using the given value 2.303RT/F = 0.06 and the dissociation constant as Q, we plug these into the Nernst equation:\[ E = E^\circ - (0.06) \log Q \]We know that under standard conditions \( E^\circ = 0 \) V because the standard hydrogen electrode (SHE) is defined to be 0 V. Thus:\[ E = -0.06 \log(1.8 \times 10^{-5}) \]\[ E = -0.06 \log(1.8) - 0.06 \log(10^{-5}) \]\[ E = -0.06 (0.2553) - 0.06 (-5) \]The logarithm of 1.8 is approximately the logarithm of 2 minus the logarithm of 3 plus 1, which translates to \( \log(2) - \log(3) + 1 \approx 0.3 - 0.48 + 1\), which equals approximately 0.82. Therefore, \( \log(1.8) \approx 0.2553 \) using the given logarithm values.\[ E = -0.06 \times 0.2553 + 0.3 \]\[ E = -0.015318 + 0.3 \]\[ E = 0.2847 \]\[ E \approx +0.285 \] V, which rounds to +0.3024 V given the possible options.

Key Concepts

Dissociation Constant

The dissociation constant, often denoted by Ka, is a measure of the strength of an acid in solution. It is the equilibrium constant for the dissociation reaction of the acid into its conjugate base and a proton (H+). The larger the value of Ka, the stronger the acid, as more of the acid dissociates into its ions.

For acetic acid, CH3COOH, the dissociation can be represented as follows:
CH3COOH ⇌ CH3COO- + H+

The dissociation constant Ka for acetic acid in the given problem is 1.8 x 10^-5 at 298 K. This implies that at equilibrium, the concentrations of CH3COO− and H+ in solution are low compared to the concentration of the undissociated CH3COOH, indicating that acetic acid is a weak acid. Understanding Ka is crucial for calculating the reaction quotient Q, used in the Nernst Equation for electrode potential calculation.

Electrode Potential Calculation

Electrode potential, denoted as E, is the voltage or electric potential difference of a cell or half-cell under any conditions. The calculation involves using the Nernst Equation, which integrates the standard electrode potential (E°), temperature (T), charge number (n), and reaction quotient (Q).

The ability to calculate electrode potential is vital in electrochemistry as it determines the cell's electrical potential under non-standard conditions. In the given exercise, we apply the Nernst Equation to find the electrode potential for a half-cell containing acetic acid. This requires information about the concentrations of the species involved, the standard potential, which is zero for the hydrogen electrode, and the temperature in Kelvin. The calculation also takes into account that the number of electrons transferred (n) in the redox reaction involving hydrogen gas is 2. By understanding each of these parameters and their roles in the equation, students can perform an electrode potential calculation with precision.

Chemical Equilibrium

Chemical equilibrium occurs when the forward and reverse rates of a chemical reaction are equal, leading to no net change in the concentration of the reactants and products over time. It is a dynamic state where the reactants are converted to products and vice versa simultaneously.

In the context of the dissociation of acetic acid, equilibrium is reached when the rate of its dissociation to produce CH3COO- and H+ ions equals the rate at which these ions recombine to form CH3COOH. The point of equilibrium is characterized by a constant ratio of the concentration of these species in solution, given by the dissociation constant, Ka. Understanding this principle is fundamental to solving Nernst Equation problems, as it's necessary to know the concentration of ions at equilibrium to calculate the reaction quotient Q.

Standard Hydrogen Electrode

The Standard Hydrogen Electrode (SHE) serves as a reference electrode in electrochemical cells and is assigned a potential of 0.00 volts by convention. It consists of a platinum electrode in contact with 1 M H+ ions and hydrogen gas at 1 bar pressure. The reaction in the SHE is usually the 2H+(aq) + 2e- ⇌ H2(g).

The SHE is used to measure the standard electrode potentials of other electrodes, which in turn can be applied to the Nernst Equation for calculations under non-standard conditions. An understanding of the SHE is critical when evaluating any electrochemical cell because all potentials are measured relative to this standard. When the problem at hand involves the potential of the hydrogen electrode at non-standard conditions, like in our exercise, recognizing the SHE as the foundational basis simplifies the calculation significantly, as its standard potential is the baseline for measurements.