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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 \( 0.123 \).

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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.
This equation gives insight into how the concentration of ions affects the potential of a cell. By adjusting for conditions like temperature and concentration, the Nernst equation helps predict the direction of electrochemical reactions.
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: \( \text{I}_2(s) + 2e^- \rightarrow 2\text{I}^- \).
Each half-reaction's potential, known as the half-cell potential, contributes to the overall cell potential. By adding or subtracting these values, you can derive the standard cell potential \( E^0_{cell} \) for the entire electrochemical setup. Understanding half-reactions helps in balancing chemical equations, allowing for the understanding of electron flow during redox reactions.

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Most popular questions from this chapter

The purity of a sample of \(\mathrm{K}_{3} \mathrm{Fe}(\mathrm{CN})_{6}\) is determined using linear-potential scan hydrodynamic voltammetry at a glassy carbon electrode. The following data were obtained for a set of external calibration standards. \begin{tabular}{cc} {\(\left[\mathrm{K}_{3} \mathrm{Fe}(\mathrm{CN})_{6}\right](\mathrm{mM})\)} & limiting current \((\mu \mathrm{A})\) \\ \hline 2.0 & 127 \\ 4.0 & 252 \\ 6.0 & 376 \\ 8.0 & 500 \\ 10.0 & 624 \end{tabular} A sample of impure \(\mathrm{K}_{3} \mathrm{Fe}(\mathrm{CN})_{6}\) is prepared for analysis by diluting a \(0.246-\mathrm{g}\) portion to volume in a \(100-\mathrm{mL}\) volumetric flask. The limiting current for the sample is \(444 \mu \mathrm{A}\). Report the purity of this sample of \(\mathrm{K}_{3} \mathrm{Fe}(\mathrm{CN})_{6}\)

What reaction prevents us from using \(Z n\) as an electrode of the first kind in an acidic solution? Which other metals do you expect to behave in the same manner as Zn when immersed in an acidic solution?

One method for determining whether an individual recently fired a gun is to look for traces of antimony in residue collected from the individual's hands. Anodic stripping voltammetry at a mercury film electrode is ideally suited for this analysis. In a typical analysis a sample is collected from a suspect using a cotton-tipped swab wetted with \(5 \%\) \(\mathrm{v} / \mathrm{v} \mathrm{HNO}_{3} .\) After returning to the lab, the swab is placed in a vial that contains \(5.0 \mathrm{~mL}\) of \(4 \mathrm{M}\) HCl that is \(0.02 \mathrm{M}\) in hydrazine sulfate. After soaking the swab, a \(4.0-\mathrm{mL}\) portion of the solution is transferred to an electrochemical cell along with \(100 \mu \mathrm{L}\) of \(0.01 \mathrm{M} \mathrm{HgCl}_{2}\). After depositing the thin film of mercury and the antimony, the stripping step gives a peak current of \(0.38 \mu \mathrm{A}\). After adding a standard addition of \(100 \mu \mathrm{L}\) of \(5.00 \times 10^{2}\) ppb Sb, the peak current increases to \(1.14 \mu\) A. How many nanograms of Sb were collected from the suspect's hand?

The concentration of \(\mathrm{Cu}^{2+}\) in seawater is determined by anodic stripping voltammetry at a hanging mercury drop electrode after first releasing any copper bound to organic matter. To a \(20.00-\mathrm{mL}\) sample of seawater is added \(1 \mathrm{~mL}\) of \(0.05 \mathrm{M} \mathrm{HNO}_{3}\) and \(1 \mathrm{~mL}\) of \(0.1 \% \mathrm{H}_{2} \mathrm{O}_{2} .\) The sample is irradiated with UV light for \(8 \mathrm{hr}\) and then diluted to volume in a \(25-\mathrm{mL}\) volumetric flask. Deposition of \(\mathrm{Cu}^{2+}\) takes place at \(-0.3 \mathrm{~V}\) versus an SCE for 10 min, producing a peak current of 26.1 (arbitrary units). A second \(20.00-\mathrm{mL}\) sample of the seawater is treated identically, except that \(0.1 \mathrm{~mL}\) of a \(5.00 \mu \mathrm{M}\) solution of \(\mathrm{Cu}^{2+}\) is added, producing a peak current of 38.4 (arbitrary units). Report the concentration of \(\mathrm{Cu}^{2+}\) in the seawater in \(\mathrm{mg} / \mathrm{L}\)

The standard-state reduction potential for \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\) is \(+0.342 \mathrm{~V}\) versus the SHE. Given that \(\mathrm{Cu}^{2+}\) forms a very stable complex with the ligand EDTA, do you expect that the standard- state reduction potential for \(\mathrm{Cu}(\mathrm{EDTA})^{2-}\) is greater than \(+0.342 \mathrm{~V}\), less than \(+0.342 \mathrm{~V}\), or equal to \(+0.342 \mathrm{~V}\) ? Explain your reasoning.

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