Problem 67 - Chemistry: The Central Science in SI Units, Global Edition

Question

A voltaic cell utilizes the following reaction: \(4 \mathrm{Fe}^{2+}(a q)+\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q) \longrightarrow 4 \mathrm{Fe}^{3+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\) (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Fe}^{2+}\right]=1.3 \mathrm{M},\left[\mathrm{Fe}^{3+}\right]=\) \(0.010 \mathrm{M}, P_{\mathrm{O}_{2}}=50.7 \mathrm{kPa},\) and the \(\mathrm{pH}\) of the solution in the cathode half-cell is \(3.50 ?\)

Step-by-Step Solution

Answer

(a) The emf of this cell under standard conditions is: \(E°_{cell} = 0.46 V\) (b) The emf of this cell under non-standard conditions is: \(E \approx 0.99 V\)

Step 1: Write the balanced half-reactions

First, we need to identify the half-reactions for the oxidation and reduction processes: Oxidation half-reaction: \[\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^-\] Reduction half-reaction: \[\frac{1}{2}\text{O}_2(g) + 2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2\text{O}(l)\] Now, let's multiply the oxidation half-reaction by 4 to have the same number of \(e^-\) as in the reduction half-reaction, which gives us the overall balanced redox reaction: \[4 \text{Fe}^{2+}(aq) + \text{O}_{2}(g) + 4 \text{H}^{+}(aq) \rightarrow 4 \text{Fe}^{3+}(aq) + 2 \text{H}_{2}\text{O}(l)\]

Step 2: Calculate E° of the cell under standard conditions

E° of the cell can be calculated by subtracting E° of oxidation (anode) from E° of reduction (cathode): \[E°_{cell} = E°_{red} - E°_{ox}\] For both half reactions, E° of Fe\(^{3+}\)/Fe\(^{2+}\): \(0.77 V\) E° of O\(_2\)/H\(_2\)O: \(1.23 V\) Now, we can calculate E° of the cell: \[E°_{cell} = 1.23 V - 0.77 V = 0.46 V\]

Step 3: Calculate the emf under non-standard conditions using the Nernst equation

The Nernst equation allows us to find the emf under non-standard conditions: \[E = E° - \frac{RT}{nF} \ln(Q)\] Here, R is the universal gas constant, T is the temperature, n is the number of electrons transferred in the redox reaction, F is the Faraday constant, and Q is the reaction quotient. We will use the following values: T = 298 K (approximately 25°C, room temperature) R = 8.314 J/molK (gas constant) F = 96,485 C/mol (Faraday constant) n = 4 electrons (as determined in Step 1) Now, let's write the expression for the reaction quotient Q: \[Q=\frac{\left[\text{Fe}^{3+}\right]^4}{(\left[\text{Fe}^{2+}\right]^4)([\text{O}_2])}\] We have the following given values: \(\left[\text{Fe}^{2+}\right] = 1.3 \text{M}\) \(\left[\text{Fe}^{3+}\right] = 0.010 \text{M}\) \(P_{\text{O}_2} = 50.7 \text{kPa}\) We can convert the partial pressure of O\(_2\) to concentration using Henry's Law: \([O_2] = \frac{P_{O_2}}{k_H}\) Given that for O\(_2\) at 25°C, \(k_H = 769 \text{kPa}/\text{M}\), \([\text{O}_{2}] = \frac{50.7 \text{kPa}}{769 \text{kPa}/\text{M}} = 0.066 \text{M}\) Now, we can calculate Q: \[Q=\frac{(0.010 \text{M})^4}{(1.3 \text{M})^4(0.066 \text{M})} = 1.16 \times 10^{-11}\] Using the values above and the Nernst equation, we can calculate the emf under non-standard conditions: \[E = 0.46 V - \frac{(8.314 \text{J/molK})(298 \text{K})}{(4)(96,485 \text{C/mol})} \ln(1.16 \times 10^{-11})\] \[E \approx 0.99 V\]

Step 4: Final answers:

(a) The emf of this cell under standard conditions is: \(E°_{cell} = 0.46 V\) (b) The emf of this cell under non-standard conditions is: \(E \approx 0.99 V\)

Key Concepts

Electromotive Force (EMF)

Electromotive Force, often abbreviated as EMF, is a fundamental concept when studying voltaic cells like those described in the exercise. EMF refers to the voltage generated by a voltaic cell when no current is flowing through the circuit. This potential difference is the driving force that pushes electrons through the circuit from the anode to the cathode. In chemical terms, it is the difference in potential energy between two electrodes.

• Under standard conditions, the EMF of a cell is measured when reactants and products are in their standard states, typically 1 M concentrations and 1 atm pressure.
• Standard EMF, denoted as E°_cell, is calculated using standard reduction potentials, which are determined experimentally.
• EMF can also be influenced by temperature, concentration of reactants/products, and pressure when not in standard conditions.

Understanding EMF helps in determining the cell's ability to perform electrical work. It also provides insight into the feasibility of the reaction. A positive EMF indicates a spontaneous reaction, whereas a negative EMF implies a non-spontaneous one.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, or redox reactions, are the backbone of how voltaic cells generate electricity. These involve the transfer of electrons between species.

• Oxidation is the loss of electrons, and in voltaic cells, it occurs at the anode. In this exercise, Fe²⁺ is oxidized to Fe³⁺.
• Reduction is the gain of electrons, occurring at the cathode. Here, O₂ is reduced to form H₂O.

These half-reactions combine to form a complete redox reaction. The overall cell potential is derived from the difference in potential energy, or voltage, between the cathode and anode. This potential energy difference is a result of different affinities for electrons.
Recognizing the relationship between the flow of electrons and chemical changes during these reactions is crucial for understanding how the cell generates electrical energy.

Nernst Equation

The Nernst Equation is an essential tool in electrochemistry. It allows for calculating the EMF of a cell under non-standard conditions, which is often the case in practical applications. This equation adjusts the standard EMF by considering real-world conditions like temperature, pressure, and concentration of reactants and products.

• The equation is: \[ E = E° - \frac{RT}{nF} \ln(Q) \]where E is the cell potential under specific conditions, E° is the standard cell potential, R is the gas constant (8.314 J/molK), T is the temperature in Kelvin, n is the number of moles of electrons transferred, and F is the Faraday constant (96485 C/mol).
• The reaction quotient, Q, expresses the ratio of product concentrations to reactant concentrations, each raised to the power of their coefficients in the balanced equation.

Using the Nernst Equation, we can evaluate how variations in environmental conditions impact the electromotive force, thereby providing a more accurate prediction of cell behavior. This is essential for designing and optimizing batteries and other electrochemical devices.