Chapter 11: Problem 3
Calculate the activity of \(\mathrm{KI}, x\), in the following electrochemical cell if the potential is \(+0.294 \mathrm{~V}\) $$ \mathrm{Ag}\left|\mathrm{AgCl}(s), \mathrm{NaCl}(a q, 0.1) \| \mathrm{KI}(a q, x), \mathrm{I}_{2}(s)\right| \mathrm{Pt} $$
Short Answer
Expert verified
The activity of KI is approximately 0.123.
Step by step solution
01
Identify the Electrochemical Cell Components
The electrochemical cell given is: \( \mathrm{Ag} | \mathrm{AgCl}(s), \mathrm{NaCl}(aq, 0.1) || \mathrm{KI}(aq, x), \mathrm{I}_2(s) | \mathrm{Pt} \). It involves two half-cells: Ag/AgCl and Iodine/Iodide. We are tasked with finding the activity of KI, denoted as \( x \).
02
Write Half-Reactions
Write down the half-reactions for each half-cell. The left half-cell (Ag/AgCl) involves the reaction \( \mathrm{AgCl}(s) + e^- \rightarrow \mathrm{Ag}(s) + \mathrm{Cl}^- \). The right half-cell (Iodine/Iodide) involves the reaction \( \mathrm{I}_2(s) + 2e^- \rightarrow 2\mathrm{I}^- \).
03
Write Nernst Equation
The Nernst equation for the cell reaction at 298 K is given by: \[ E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q \] where \( E_{cell} = 0.294 \) V, \( E^0_{cell} \) is the standard cell potential, \( Q \) is the reaction quotient for the cell, \( R = 8.314 \) J/mol·K, \( T = 298 \) K, \( n \) is the number of moles of electrons transferred (which is 2 in this reaction), and \( F = 96485 \) C/mol is the Faraday constant.
04
Calculate Standard Cell Potential
The standard cell potential, \( E^0_{cell} \), is calculated from:\[ E^0_{cell} = E^0_{I_2/I^-} - E^0_{AgCl/Ag, Cl^-} \] Use standard reduction potentials: \( E^0_{I_2/I^-} = 0.54 \) V and \( E^0_{AgCl/Ag, Cl^-} = 0.222 \) V. Calculate: \[ E^0_{cell} = 0.54 - 0.222 = 0.318 \text{ V} \]
05
Calculate Reaction Quotient (Q)
From the cell reaction, the reaction quotient \( Q \) can be expressed as: \[ Q = \frac{[\mathrm{Cl}^-]}{[\mathrm{I}^-]^2} \]Given \( [\mathrm{Cl}^-] = 0.1 \) M from NaCl, and \( [\mathrm{I}^-] = x \). Thus, \[ Q = \frac{0.1}{x^2} \]
06
Solve Nernst Equation for x
Substitute known values into the Nernst equation:\[ 0.294 = 0.318 - \frac{(8.314)(298)}{(2)(96485)} \ln \left( \frac{0.1}{x^2} \right) \] Simplify to find \( x \): \[ \ln \left( \frac{0.1}{x^2} \right) = \frac{0.318 - 0.294}{0.01273} \]\[ \ln \left( \frac{0.1}{x^2} \right) = 1.89 \] \[ \frac{0.1}{x^2} = e^{1.89} = 6.626 \] Solving gives \( x^2 = \frac{0.1}{6.626} \), hence \( x = \sqrt{\frac{0.1}{6.626}} \).
07
Calculate the Activity of KI, x
Compute \( x \) using the derived expression: \[ x = \sqrt{\frac{0.1}{6.626}} \approx 0.123 \] Thus, the activity of \( \mathrm{KI} \) in the solution is approximately \( 0.123 \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Nernst Equation
The Nernst Equation is a powerful tool in electrochemistry. It allows us to calculate the cell potential of an electrochemical cell under non-standard conditions. The equation is expressed as: \[ E_{cell} = E_{cell}^0 - \frac{RT}{nF} \ln Q \] Where:
- \( E_{cell} \) is the cell potential under non-standard conditions.
- \( E_{cell}^0 \) is the standard cell potential.
- \( R \) is the universal gas constant, \( 8.314 \) J/mol·K.
- \( T \) is the temperature in Kelvin.
- \( n \) is the number of moles of electrons transferred.
- \( F \) is Faraday's constant, \( 96485 \) C/mol.
- \( Q \) is the reaction quotient.
Standard Cell Potential
The concept of standard cell potential, denoted as \( E^0_{cell} \), is essential in understanding how spontaneous a reaction is under standard conditions. It is the potential difference between two electrodes when the concentrations of all species are at standard states (usually 1 M for aqueous solutions and 1 atm pressure for gases). Under these conditions, the cell potential is calculated using standard reduction potentials from a given table. For the sample problem, the standard cell potential is determined by the difference between the reduction potentials of the two half-reactions: \[ E^0_{cell} = E^0_{\text{I}_2/\text{I}^-} - E^0_{\text{AgCl/Ag, Cl}^-} \]With \( E^0_{\text{I}_2/\text{I}^-} = 0.54 \text{ V} \) and \( E^0_{\text{AgCl/Ag, Cl}^-} = 0.222 \text{ V} \), the standard cell potential for the reaction is calculated as \( 0.318 \text{ V} \). This value indicates the cell's ability to perform work and drive a reaction.
Reaction Quotient
The reaction quotient, \( Q \), is an expression that relates the concentrations or pressures of the reactants and products at any point in a reaction. It is similar to the equilibrium constant \( K \), but is used when the system is not at equilibrium. The general form of the reaction quotient is: \[ Q = \frac{\text{products}}{\text{reactants}} \]In our electrochemical cell problem, the reaction quotient is specifically for the cell reaction: \[ Q = \frac{[\text{Cl}^-]}{[\text{I}^-]^2} \] where the concentration of chloride ions \([\text{Cl}^-] = 0.1 \text{ M}\) and \([\text{I}^-]\) is represented by \( x \), the activity of KI we need to calculate. Using the reaction quotient helps determine how far the reaction has shifted from standard state conditions. Knowing \( Q \) allows for a more accurate calculation of cell potential using the Nernst equation.
Half-Reactions
Understanding half-reactions is crucial to grasping how electrochemical cells operate. Each half-reaction represents either a reduction or an oxidation process in the cell. In this exercise, the electrochemical cell consists of two half-reactions:
- The Ag/AgCl half-cell undergoes the reduction: \( \text{AgCl(s)} + e^- \rightarrow \text{Ag(s)} + \text{Cl}^- \).
- The Iodine/Iodide half-cell undergoes the reduction process: \( \text{I}_2(s) + 2e^- \rightarrow 2\text{I}^- \).