Textbook Question
A voltaic cell utilizes the following reaction: Al1s2 + 3 Ag+1aq2 ¡ Al3+1aq2 + 3 Ag1s2 What is the effect on the cell emf of each of the following changes? (a) Water is added to the anode half-cell, diluting the solution.
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Ch.20 - Electrochemistry
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232
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Table of contents
- Ch.1 - Introduction: Matter, Energy, and Measurement140
- Ch.2 - Atoms, Molecules, and Ions192
- Ch.3 - Chemical Reactions and Reaction Stoichiometry172
- Ch.4 - Reactions in Aqueous Solution156
- Ch.5 - Thermochemistry151
- Ch.6 - Electronic Structure of Atoms137
- Ch.7 - Periodic Properties of the Elements127
- Ch.8 - Basic Concepts of Chemical Bonding143
- Ch.9 - Molecular Geometry and Bonding Theories167
- Ch.10 - Gases147
- Ch.11 - Liquids and Intermolecular Forces97
- Ch.12 - Solids and Modern Materials129
- Ch.13 - Properties of Solutions126
- Ch.14 - Chemical Kinetics175
- Ch.15 - Chemical Equilibrium107
- Ch.16 - Acid-Base Equilibria127
- Ch.17 - Additional Aspects of Aqueous Equilibria133
- Ch.18 - Chemistry of the Environment88
- Ch.19 - Chemical Thermodynamics140
- Ch.20 - Electrochemistry137
- Ch.21 - Nuclear Chemistry99
- Ch.22 - Chemistry of the Nonmetals66
- Ch.23 - Transition Metals and Coordination Chemistry69
- Ch.24 - The Chemistry of Life: Organic and Biological Chemistry11
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232Not the one you use?Change textbook
Chapter 20, Problem 67b
A voltaic cell utilizes the following reaction: 4 Fe2+1aq2 + O21g2 + 4 H+1aq2 ¡ 4 Fe3+1aq2 + 2 H2O1l2 (b) What is the emf of this cell when 3Fe2+4 = 1.3 M, 3Fe3+4= 0.010 M, PO2 = 0.50 atm, and the pH of the solution in the cathode half-cell is 3.50?
Verified step by step guidance1
Identify the half-reactions for the cathode and anode. For the given reaction, the oxidation at the anode is: \(4 \text{Fe}^{2+} \rightarrow 4 \text{Fe}^{3+} + 4e^-\) and the reduction at the cathode is: \(\text{O}_2 + 4 \text{H}^+ + 4e^- \rightarrow 2 \text{H}_2\text{O}\).
Write the Nernst equation for the overall cell reaction. The Nernst equation is given by: \(E = E^\circ - \frac{0.0592}{n} \log Q\), where \(E^\circ\) is the standard cell potential, \(n\) is the number of moles of electrons transferred, and \(Q\) is the reaction quotient.
Calculate the reaction quotient, \(Q\), using the given concentrations and pressures. For the reaction, \(Q = \frac{[\text{Fe}^{3+}]^4 \cdot (\text{P}_{\text{O}_2})^{1/2}}{[\text{Fe}^{2+}]^4 \cdot [\text{H}^+]^4}\). Substitute the values \([\text{Fe}^{2+}] = 1.3 \text{ M}\), \([\text{Fe}^{3+}] = 0.010 \text{ M}\), \(\text{P}_{\text{O}_2} = 0.50 \text{ atm}\), and \([\text{H}^+] = 10^{-3.50} \text{ M}\) into the equation.
Determine the number of electrons transferred, \(n\), in the balanced equation. From the half-reactions, you can see that 4 electrons are transferred.
Substitute the values into the Nernst equation to find the emf of the cell. Remember to use the standard emf, \(E^\circ\), for the reaction which can be found in standard reduction potential tables.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Electrochemical Cells
Electrochemical cells, such as voltaic cells, convert chemical energy into electrical energy through redox reactions. In these cells, oxidation occurs at the anode and reduction at the cathode, allowing for the flow of electrons through an external circuit. Understanding the components and functioning of these cells is essential for calculating their electromotive force (emf).
Nernst Equation
The Nernst equation relates the cell potential (emf) to the concentrations of the reactants and products involved in the electrochemical reaction. It accounts for non-standard conditions by incorporating the reaction quotient and temperature, allowing for the calculation of emf under varying concentrations and pressures. This equation is crucial for determining the emf in the given voltaic cell scenario.
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pH and Hydrogen Ion Concentration
pH is a measure of the hydrogen ion concentration in a solution, which influences the equilibrium of redox reactions in electrochemical cells. In the context of the given question, the pH of the cathode half-cell affects the concentration of H+ ions, thereby impacting the overall cell potential. Understanding the relationship between pH and hydrogen ion concentration is vital for accurate emf calculations.
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Related Practice
Textbook Question
A voltaic cell utilizes the following reaction: (b) What is the emf for this cell when 3Fe3+4 = 3.50 M, PH2= 0.95 atm, 3Fe2+4 = 0.0010 M, and the pH in both half-cells is 4.00?
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Textbook Question
A voltaic cell utilizes the following reaction: 2 Fe3+1aq2 + H21g2 ¡ 2 Fe2+1aq2 + 2 H+1aq2 (a) What is the emf of this cell under standard conditions?
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