Problem 57 - Chemistry: An Atoms-Focused Approach

Question

The oxidation of \(\mathrm{NH}_{4}^{+}\) to \(\mathrm{NO}_{3}^{-}\) in acid solution is described by the following equation: \(\mathrm{NH}_{4}^{+}(a q)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{NO}_{3}^{-}(a q)+2 \mathrm{H}^{+}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell)\) a. Calculate \(E^{\circ}\) for the overall reaction. b. If the reaction is in equilibrium with air \(\left(P_{\mathrm{O}_{1}}=0.21 \mathrm{atm}\right)\) at pH \(5.60,\) what is the ratio of \(\left[\mathrm{NO}_{3}^{-}\right]\) to \(\left[\mathrm{NH}_{4}^{+}\right]\) at \(298 \mathrm{K} ?\)

Step-by-Step Solution

Answer

The ratio of [NO3-] to [NH4+] at equilibrium is approximately 1.54 × 10^(-5).

Step 1: Identify the Half-Reactions

Identify the half-reactions involved in the overall reaction. The half-reactions are: 1. Oxidation of \(\mathrm{NH}_{4}^{+}\): \(\mathrm{NH}_{4}^{+} \rightarrow \mathrm{NO}_{3}^{-} + 2 \mathrm{H}^{+} + 3 e^{-}\). 2. Reduction of \(\mathrm{O}_{2}\): \(\mathrm{O}_{2} + 4 \mathrm{H}^{+} + 4 e^{-} \rightarrow 2 \mathrm{H}_{2}\mathrm{O}\).

Step 2: Find the Standard Reduction Potentials

Look up the standard reduction potentials for the half-reactions: 1. \(E^{\circ}_{\mathrm{NH}_{4}^{+} \rightarrow \mathrm{NO}_{3}^{-}} = -0.8\) V 2. \(E^{\circ}_{\mathrm{O}_{2} \rightarrow \mathrm{H}_{2}\mathrm{O}} = 1.23\) V

Step 3: Calculate the Standard Cell Potential

Add the standard reduction potentials of the half-reactions to find the standard cell potential: \(E^{\circ}_{cell} = E^{\circ}_{\mathrm{NH}_{4}^{+} \rightarrow \mathrm{NO}_{3}^{-}} + E^{\circ}_{\mathrm{O}_{2} \rightarrow \mathrm{H}_{2}\mathrm{O}} = -0.8\,\text{V} + 1.23\,\text{V} = 0.43\,\text{V}\) So, the standard cell potential for the overall reaction is \(0.43\) V. b. If the reaction is in equilibrium with air \(\left(P_{\mathrm{O}_{1}}=0.21 \mathrm{atm}\right)\) at pH \(5.60,\) what is the ratio of \(\left[\mathrm{NO}_{3}^{-}\right]\) to \(\left[\mathrm{NH}_{4}^{+}\right]\) at \(298 \mathrm{K} ?\)

Step 4: Calculate the E_cell under non-standard conditions

Using the Nernst equation, calculate the \(E_{cell}\) under non-standard conditions: \(E_{cell} = E^{\circ}_{cell} - \frac{RT}{nF} \ln\left(Q\right)\), where \(R\) is the gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of electrons transferred, \(F\) is the Faraday constant, and \(Q\) is the reaction quotient.

Step 5: Express Q in terms of Concentrations

First we need to express the reaction quotient \(Q\) in terms of the concentrations of the species involved in the reaction: \(Q = \frac{\left[\mathrm{NO}_3^-\right]\left[\mathrm{H}^+\right]^2}{\left[\mathrm{NH}_4^+\right]\left(\frac{P_{O_2}}{1\,\text{atm}}\right)^2}\)

Step 6: Substitute values and solve for the ratio of concentrations

Assuming \(E_{cell}\) is \(0\), since the reaction is at equilibrium, we can substitute the given values and solve for the ratio of concentrations: \(0 = 0.43 - \frac{8.314\,\text{J/mol⋅K} \times 298\,\text{K}}{3\,\text{mol} \times 96485\,\text{C/mol}} \ln\left(\frac{\left[\mathrm{NO}_3^-\right]\left(\frac{10^{-5.60}}{\text{M}}\right)^2}{\left[\mathrm{NH}_4^+\right]\left(\frac{0.21}{1\,\text{atm}}\right)^2}\right)\) Solve for \(\frac{\left[\mathrm{NO}_3^-\right]}{\left[\mathrm{NH}_4^+\right]}\) to find the ratio of concentrations at equilibrium. The final answer is: \(\frac{\left[\mathrm{NO}_3^-\right]}{\left[\mathrm{NH}_4^+\right]} \approx \mathbf{1.54} \times 10^{-5}\)

Key Concepts

Standard Reduction Potential

Standard Reduction Potential is crucial in understanding how easily a substance gains electrons, or is reduced, in electrochemical reactions. Each half-reaction in a redox process has a specific standard potential, denoted as \(E^{\circ}\), measured in volts (V). These potentials are usually determined under standard conditions, which means at 25°C, 1 M concentration, and 1 atm pressure.

For instance, in the oxidation of \(\mathrm{NH}_4^+\) to \(\mathrm{NO}_3^-\) seen in our exercise, the standard reduction potentials for the involved half-reactions help us calculate the overall cell potential. By adding the standard potentials of the oxidation and reduction reactions, we deduce that the standard cell potential for this reaction is 0.43 V. This value is a key indicator of the reaction's thermodynamic feasibility.

In summary, when we look at the standard reduction potential, we're essentially evaluating how favorable a particular redox reaction is under specific conditions commonly employed in experiments.

Nernst Equation

The Nernst Equation is an essential tool in electrochemistry, allowing us to calculate the cell potential under non-standard conditions. It adjusts the standard cell potential \(E^{\circ}_{cell}\) using the effects of concentration and temperature changes.

The equation is expressed as: \[E_{cell} = E^{\circ}_{cell} - \frac{RT}{nF} \ln(Q)\] Here,

• \(R\) represents the gas constant,
• \(T\) is the temperature in Kelvin,
• \(n\) is the number of moles of electrons transferred,
• \(F\) denotes the Faraday constant, and
• \(Q\) is the reaction quotient.

In our exercise, using the Nernst Equation enabled us to understand how changes in concentration—brought about by varying pH and the presence of atmospheric oxygen—affect the cell potential. By setting \(E_{cell}\) to zero for the equilibrium condition, we could solve for the concentrations of reactants and products, giving us insight into the reaction's dynamics under given circumstances.

Reaction Quotient

The Reaction Quotient \(Q\) is a dimensionless number that determines the direction the reaction will proceed to reach equilibrium. It's calculated similarly to the equilibrium constant \(K\), using the concentrations or partial pressures of the reactants and products at any given moment, not just equilibrium.

For our specific redox reaction, the expression for \(Q\) can be determined as: \[Q = \frac{[\mathrm{NO}_3^-][\mathrm{H}^+]^2}{[\mathrm{NH}_4^+](\frac{P_{O_2}}{1\,\text{atm}})^2}\]Where,

• \([\mathrm{NO}_3^-]\) and \([\mathrm{NH}_4^+]\) are the concentrations of nitrate and ammonium ions, respectively,
• \([\mathrm{H}^+]\) reflects the effect of the pH on the system, and
• \(P_{O_2}\) is the partial pressure of oxygen.

In this exercise, understanding \(Q\) is foundational in applying the Nernst equation. By assessing how \(Q\) influences whether the reaction favors the formation of products or reactants, students can predict the direction in which the reaction is inclined at a given state.

Electrochemistry

Electrochemistry bridges the gap between chemistry and electricity by studying the interchange of chemical and electrical energy. This field is at the heart of many technological applications such as batteries, fuel cells, and sensors.

In the context of our exercise, electrochemistry explains the electron transfer that occurs during the oxidation-reduction process between \(\mathrm{NH}_4^+\) and \(\mathrm{O}_2\) leading to the formation of \(\mathrm{NO}_3^-\). These reactions are based on the principles of electrochemical cells, where oxidation occurs at one electrode and reduction at the other, generating a flow of electrons and thus electrical energy.

Understanding electrochemistry allows us to calculate potential differences across cells, comprehend the role of ions and electrons, and determine how these reactions can be harnessed for energy storage or conversion. This knowledge is crucial for industries focusing on sustainable energy and green technology. By comprehending these electrochemical processes, students can appreciate the complex interplay between chemistry and electricity, paving the way for further innovation and discovery in the field.