Domanda del libro di testo
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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Ch.20 - Electrochemistry
Brown15th EditionChemistry: The Central ScienceISBN: 9780137542970
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TOC.tableOfContents
- Ch.1 - Introduction: Matter, Energy, and Measurement146
- Ch.2 - Atoms, Molecules, and Ions200
- Ch.3 - Chemical Reactions and Reaction Stoichiometry176
- Ch.4 - Reactions in Aqueous Solution155
- Ch.5 - Thermochemistry126
- Ch.6 - Electronic Structure of Atoms131
- Ch.7 - Periodic Properties of the Elements126
- Ch.8 - Basic Concepts of Chemical Bonding123
- Ch.9 - Molecular Geometry and Bonding Theories143
- Ch.10 - Gases147
- Ch.11 - Liquids and Intermolecular Forces91
- Ch.12 - Solids and Modern Materials122
- Ch.13 - Properties of Solutions126
- Ch.14 - Chemical Kinetics174
- Ch.15 - Chemical Equilibrium101
- Ch.16 - Acid-Base Equilibria139
- Ch.17 - Additional Aspects of Aqueous Equilibria146
- Ch.18 - Chemistry of the Environment72
- Ch.19 - Chemical Thermodynamics129
- Ch.20 - Electrochemistry142
- Ch.21 - Nuclear Chemistry101
- Ch.22 - Chemistry of the Nonmetals68
- Ch.23 - Transition Metals and Coordination Chemistry66
- Ch.24 - The Chemistry of Life: Organic and Biological Chemistry10
Brown15th EditionChemistry: The Central ScienceISBN: 9780137542970Non è quello che usi tu?Cambia libro di testo
Capitolo 20, Problema 71b
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?
Guida verificata passo dopo passo1
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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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).
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Percorso guidato
02:46
Electrochemical Cells
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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Percorso guidato
01:17The Nernst Equation
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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Percorso guidato
06:25Hydronium Ion Concentration Example
Pratica correlata
Domanda del libro di testo
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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Domanda del libro di testo
A voltaic cell is constructed that uses the following reaction and operates at 298 K: Zn(s) + Ni2+(aq) → Zn2+(aq) + Ni(s) (b) What is the emf of this cell when [Ni2+] = 3.00 M and [Zn2+] = 0.100 M? (c) What is the emf of the cell when [Ni2+] = 0.200 M and [Zn2+] = 0.900 M?
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Domanda del libro di testo
A voltaic cell utilizes the following reaction: 4 Fe2+1aq2 + O21g2 + 4 H+1aq2 ¡ 4 Fe3+1aq2 + 2 H2O1l2 (a) What is the emf of this cell under standard conditions?
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