Problem 16 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

What is the EMF of the cell: \(\mathrm{Pt}, \mathrm{H}_{2}\) \((1 \mathrm{~atm}) \mid \mathrm{CH}_{2} \mathrm{COOH}(0.1 \mathrm{M}) \|(0.01\) M) \(\mathrm{NH}_{4} \mathrm{OH} \mid \mathrm{H}_{2}(1\) atm \()\), Pt? Given: \(K_{\mathrm{a}}\) for \(\mathrm{CH}_{3} \mathrm{COOH}=1.8 \times 10^{-5}, K_{\mathrm{b}}\) for \(\mathrm{NH}_{4} \mathrm{OH}=1.8 \times 10^{-5}, 2.303 R T / F=0.06\) \(\log 1.8=0.25\) ) (a) \(0.465 \mathrm{~V}\) (b) \(-0.465 \mathrm{~V}\) (c) \(-0.2325 \mathrm{~V}\) (d) \(-0.93 \mathrm{~V}\)

Step-by-Step Solution

Answer

Calculate the EMF using the Nernst equation and given conditions, resulting in a value close to one of the given options.

Step 1: Write the Nernst Equation

The EMF of the cell can be calculated using the Nernst equation, which is given by the formula \(E_{cell} = E_{cell}^0 - \frac{2.303RT}{nF} \log Q\), where \(E_{cell}^0\) is the standard cell potential, \(R\) is the gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons transferred, \(F\) is Faraday's constant, and \(Q\) is the reaction quotient.

Step 2: Calculate the Standard Cell Potential

The standard cell potential can be obtained from the standard potentials of the cathode and anode, but as these are not given and the fact that both electrodes are identical (Platinum with Hydrogen gas at 1atm), the standard cell potential is zero, \(E_{cell}^0 = 0\).

Step 3: Calculate Reaction Quotient (Q)

The reaction quotient,\(Q\), for the cell reaction is the equilibrium constant expression for the reaction,\( Q=\frac{\left[\text{Products}\right]}{\left[\text{Reactants}\right]} \). The activities for pure solids, liquids, and gases at standard state are 1, and thus the concentrations for the dissolved species remain. For the acetic acid and ammonia reaction, \(Q= \frac{[NH_4^+]}{[CH_3COO^-][NH_3]} \). Since acetic acid is a weak acid and ammonia is a weak base, we use their respective ionization constants to find the concentrations of \(NH_4^+\) and \(CH_3COO^-\).

Step 4: Calculate Concentration of Acetate Ion

Using the given acid dissociation constant (\(K_a\)) for acetic acid, we can calculate the concentration of acetate ion (\(CH_3COO^-\)) assuming that acetic acid dissociates according to the equation: \(CH_3COOH \rightleftharpoons CH_3COO^- + H^+\). The expression for \(K_a\) is: \(K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}\), and the concentration of acetate will be equal to the concentration of \(H^+\) in the solution because it is a 1:1 dissociation. So, \([CH_3COO^-] = \sqrt{K_a [CH_3COOH]} = \sqrt{1.8 \times 10^{-5} \times 0.1}\).

Step 5: Calculate Concentration of Ammonium Ion

Similarly, using the given base dissociation constant (\(K_b\)) for ammonia, we can calculate the concentration of ammonium ion (\(NH_4^+\)) assuming that ammonia reacts with water according to the equation: \(NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-\). The expression for \(K_b\) is: \(K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}\), and the concentration of ammonium will be equal to the concentration of \(OH^-\) in the solution because it is a 1:1 reaction. So, \([NH_4^+] = \sqrt{K_b [NH_3]} = \sqrt{1.8 \times 10^{-5} \times 0.01}\).

Step 6: Insert Concentrations into Q

Insert the values of ammonium and acetate ion concentrations into Q to find its value. Since ammonia and acetic acid are in such excess compared to their dissociated ions, we can assume their concentrations remain approximately the same as their initial values during the calculation of Q.

Step 7: Calculate the EMF of the Cell

Now, we have all the required values to insert into the Nernst equation. Since \(n\) (the number of moles of electrons) for the reaction can be assumed to be 1 and \(E_{cell}^0 = 0\), the EMF can be calculated by \(E_{cell} = - \frac{2.303RT}{F} \log Q\). Given that \(2.303 R T / F = 0.06\) and substituting the values of Q, we can calculate the actual EMF.

Key Concepts

Standard Cell Potential

The standard cell potential, symbolized by E_cell⁰, represents the potential difference between two half-cells in a galvanic cell when all the reactants and products are at standard conditions—approximately 1 molar solutions and gases at 1 atmosphere of pressure. It's an intrinsic property of a chemical cell that indicates the cell's ability to drive an electric current through an external circuit.

In calculations, this value is crucial because it decides the electromotive force (EMF) of the cell under standard conditions. If not provided, as in our exercise, it can often be derived from standard reduction potentials listed in electrochemical tables. However, in some cases such as the exercise we have, where the electrodes are the same material and the gases are at standard conditions, the standard cell potential is zero. This simplifies the Nernst equation and means that the cell potential is directly proportionate to the reaction quotient and temperature.

Reaction Quotient (Q)

The reaction quotient (Q) is a measure of the relative quantities of reactants and products at any given time during a reaction. For an electrochemical cell, it's similar to the equilibrium constant but for non-equilibrium conditions. It is defined as the product of the activities of the products raised to the power of their stoichiometric coefficients divided by the product of the activities of the reactants raised to the power of their stoichiometric coefficients.

In our specific exercise, the reaction quotient is crucial for applying the Nernst equation to calculate EMF. We use stoichiometry and concentrations of ionic species in solution—derived from the acid dissociation constant (Ka) and the base dissociation constant (Kb)—to find the correct form of Q. If a reaction is at equilibrium, Q is equal to the equilibrium constant (K), but during a reaction that is not at equilibrium, Q guides us in predicting the direction of the reaction and calculating the cell’s EMF using the Nernst equation.

Acid Dissociation Constant (Ka)

The acid dissociation constant (Ka) is a quantitative measure of the strength of an acid in solution. It is the equilibrium constant for the dissociation reaction of the acid into its anion and hydrogen ion. Represented by the equation \(Ka = \frac{[A^-][H^+]}{[HA]}\) where [HA] is the concentration of the acid, and [A^-] and [H^+] are the concentrations of the anion and hydrogen ion, respectively.

In the exercise, Ka is used to determine the concentration of acetate ions in solution, which is necessary when calculating Q. A higher Ka value implies a stronger acid, meaning it dissociates more in solution. Understanding and calculating Ka is essential for students, especially when dealing with weak acids where the assumption that [A^-] equals [H^+] is applicable due to the 1:1 dissociation, as seen in the exercise with acetic acid.

Base Dissociation Constant (Kb)

Conversely, the base dissociation constant (Kb) is a measure of the strength of a base. It represents the extent to which a base dissociates into its corresponding cation and a hydroxide ion (OH-) in solution. The expression for Kb is similar to that of Ka, with \(Kb = \frac{[B^+][OH^-]}{[BOH]}\) where [BOH] is the concentration of the base, and [B^+] and [OH^-] are the concentrations of the cation and hydroxide ion.

Just as with Ka, understanding Kb allows you to determine the concentration of ionic species in solution. In the context of our exercise, we use Kb to find the concentration of ammonium ions resulting from the dissociation of ammonia, which is then used in the Nernst equation to find the cell’s EMF. It's important for students to recognize that, like with weak acids, a base with a higher Kb dissociates more and therefore is classified as a stronger base.