Textbook Question
Consider the following table of standard electrode potentials for a series of hypothetical reactions in an aqueous solution: reduction half-reaction E °(V) (c) Which substance(s) can oxidize C2+?
Ch.20 - Electrochemistry
Chapter 20, Problem 7
A voltaic cell is constructed with two silver–silver chloride electrodes, each of which is based on the following half-reaction: AgCl(s) + e- → Ag(s) + Cl-(aq). The two half-cells have [Cl-] = 0.0150 M and [Cl-] = 2.55 M, respectively. (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether [Cl-] will increase, decrease, or stay the same as the cell operates.
Verified step by step guidance1
Step 1: Identify the half-reaction and note that both electrodes are based on the same half-reaction: \( \text{AgCl(s) + e}^- \rightarrow \text{Ag(s) + Cl}^- \text{(aq)} \).
Step 2: Recognize that the cell is a concentration cell, where the same half-reaction occurs at both electrodes but with different concentrations of \( \text{Cl}^- \).
Step 3: Use the Nernst equation to calculate the cell emf: \( E = E^\circ - \frac{RT}{nF} \ln \frac{[\text{Cl}^-]_{\text{cathode}}}{[\text{Cl}^-]_{\text{anode}}} \). Since \( E^\circ = 0 \) for a concentration cell, simplify to \( E = - \frac{RT}{nF} \ln \frac{[\text{Cl}^-]_{\text{cathode}}}{[\text{Cl}^-]_{\text{anode}}} \).
Step 4: Determine which electrode is the anode and which is the cathode. The anode is where oxidation occurs (lower concentration of \( \text{Cl}^- \)), and the cathode is where reduction occurs (higher concentration of \( \text{Cl}^- \)).
Step 5: Predict the change in \( [\text{Cl}^-] \) for each electrode as the cell operates. At the anode, \( [\text{Cl}^-] \) will increase as \( \text{AgCl} \) dissolves, and at the cathode, \( [\text{Cl}^-] \) will decrease as \( \text{AgCl} \) precipitates.
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Nernst Equation
The Nernst Equation relates the cell potential (emf) of an electrochemical cell to the concentrations of the reactants and products involved in the half-reactions. It is expressed as E = E° - (RT/nF) ln(Q), where E° 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. This equation allows us to calculate the emf under non-standard conditions, such as varying ion concentrations.
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The Nernst Equation
Electrode Reactions
In a voltaic cell, each electrode undergoes a specific half-reaction that involves the transfer of electrons. For the silver-silver chloride electrodes, the half-reaction involves the reduction of AgCl to Ag and the release of Cl- ions. Understanding these half-reactions is crucial for predicting how the concentrations of the reactants and products will change as the cell operates, which directly affects the cell's emf and overall performance.
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Concentration Effects on Equilibrium
According to Le Chatelier's principle, if a system at equilibrium is disturbed by changing the concentration of a reactant or product, the system will shift in a direction that counteracts the change. In the context of the voltaic cell, as the cell operates, the concentration of Cl- ions will change in each half-cell, affecting the equilibrium of the half-reactions. This principle helps predict whether the concentration of Cl- will increase, decrease, or remain constant during the operation of the cell.
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Related Practice
Textbook Question
Consider the following voltaic cell:
(c) What is the change in the cell voltage when the ion concentrations in the cathode half-cell are increased by a factor of 10?
Textbook Question
The electrodes in a silver oxide battery are silver oxide 1Ag2O2 and zinc (b) Which battery do you think has an energy density most similar to the silver oxide battery: a Li-ion battery, a nickel– cadmium battery, or a lead–acid battery? [Section 20.7]
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