Problem 23
Question
When iodine is added to an aqueous solution of iodide ion, the \(I_{3}^{-}\) ion is formed, according to the reaction below: $$\mathrm{I}_{2}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{I}_{3}^{-}(\mathrm{aq})$$ The equilibrium constant for the reaction above is \(K=7.7 \times 10^{2}\) at \(25^{\circ} \mathrm{C}\) (a) What is \(E^{\circ}\) for the reaction above? (b) If a 0.0010 mol sample of \(I_{2}\) is added to 1.0 L of \(0.0050 \mathrm{M} \mathrm{NaI}(\mathrm{aq})\) at \(25^{\circ} \mathrm{C},\) then what fraction of the \(\mathrm{I}_{2}\) remains unreacted at equilibrium?
Step-by-Step Solution
Verified Answer
a) The standard cell potential, \(E^{\circ}\), for the reaction is 0.536 V. b) The fraction of \(I_{2}\) that remains unreacted at equilibrium is 0.094.
1Step 1: Calculate E°
First calculate the standard electrode potential (E°) using the Nernst equation: \(E=E^{\circ}- \frac{0.05916}{n} \log K\). For the given reaction, reduction of \(I_{2}\) to \(I_{3}^{-}\), the number of electron transfers (n) is 2 and the equilibrium constant (K) is given as \(7.7 \times 10^{2}\). Since at equilibrium E is 0, substitute these values into the Nernst equation to find E°.
2Step 2: Set Up ICE Table
Set up an ICE (Initial, Change, Equilibrium) table to find the equilibrium concentrations of the species in the reaction. Initially, the concentration of \(I_{2}\) is negligible since it is assumed to be completely dissolved in the aqueous solution, the concentration of \(I^{-}\) is \(0.0050 M\), and \(I_{3}^{-}\) is 0 because it has not been formed yet. Because 1 mole of \(I_{2}\) reacts with 1 mole of \(I^{-}\) to form 1 mole of \(I_{3}^{-}\), the change in concentrations at equilibrium can be represented by -x for \(I_{2}\) and \(I^{-}\) and +x for \(I_{3}^{-}\).
3Step 3: Expression of K and Calculation of x
Write the expression for the equilibrium constant (K), which is [\(I_{3}^{-}\)]/[\(I_{2}\)][\(I^{-}\)]. Substitute the equilibrium concentrations from the ICE table to get \(7.7 \times 10^{2}=\frac{x}{(0.0010-x)(0.0050-x)}\). Solving this quadratic equation will yield the equilibrium concentration of \(I_{2}\), which can be used to calculate the fraction of \(I_{2}\) that remains unreacted.
Key Concepts
Iodine ChemistryNernst EquationEquilibrium ConstantElectrode Potential
Iodine Chemistry
Iodine chemistry encompasses the behavior and reactions of iodine molecules (\(I_2\)) in solutions. As a halogen, iodine is known for forming various compounds, especially with iodide ions (\(I^-\)). When iodine is added to an aqueous solution, it can react with iodide ions to form triiodide ions (\(I_3^-\)). This process is a classic example of a chemical equilibrium reaction, where iodine molecules, iodide ions, and triiodide ions exist in a dynamic state of balance.
- Triiodide Formation: The reaction converts \(I_2\) and \(I^-\) to \(I_3^-\), represented by: \(\mathrm{I}_{2}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{I}_{3}^{-}(\mathrm{aq})\).
- Aqueous Solutions: In water, iodine rarely remains as \(I_2\); it tends to form \(I_3^-\) due to its sparing solubility.
Nernst Equation
The Nernst Equation is a pivotal tool in electrochemistry, relating the electrode potential of a redox reaction to its concentrations and temperature. Its role is integral in determining the cell potential under non-standard conditions.
- Equation: The standard form of the Nernst Equation is: \(E = E^\circ - \frac{0.05916}{n}\log K\).
- Variables: Here, \(E\) is the electrode potential at certain concentrations, \(E^\circ\) is the standard electrode potential, \(n\) is the number of moles of electrons transferred, and \(K\) is the equilibrium constant.
- Application: In the context of the iodine reaction, the Nernst Equation is used to calculate the standard electrode potential (\(E^\circ\)) when the equilibrium constant is known.
Equilibrium Constant
The equilibrium constant (\(K\)) quantifies the ratio of product to reactant concentrations at equilibrium for a reversible chemical reaction. It reflects the extent to which a reaction proceeds.
- Definition: For the reaction \(\mathrm{I}_{2}+\mathrm{I}^{-} \rightleftharpoons \mathrm{I}_{3}^{-}\), the equilibrium constant is expressed as \(K=\frac{[I_3^-]}{[I_2][I^-]}\).
- Significance: A large \(K\) value (\(7.7 \times 10^2\) in this case) indicates that the formation of \(I_3^-\) is favored at equilibrium.
- Calculation: By establishing an ICE (Initial, Change, Equilibrium) table, students can calculate concentration changes and determine the fraction of \(I_2\) left unreacted.
Electrode Potential
Electrode potential, or standard electrode potential (\(E^\circ\)), is a measure of the inherent tendency of a chemical species to be reduced, and it is a vital concept in electrochemistry.
- Relationship with Reactions: It indicates whether a reaction is likely to occur. Positive values often suggest a reaction that tends to proceed.
- In Context: For the iodine reaction, calculating \(E^\circ\) helps predict the overall charge balance at standard conditions.
- Role in Nernst Equation: It serves as a baseline when evaluating how varying reaction conditions affect potential.
Other exercises in this chapter
Problem 21
Predict the geometric structures of (a) \(\mathrm{BrF}_{3} ;\) (b) IF \(_{5}\); (c) \(\mathrm{Cl}_{3} \mathrm{IF}^{-}\). (Central atom underlined.).
View solution Problem 22
Which of the following species has a linear structure: \(\mathrm{ClF}_{2}^{+}, \mathrm{IBrF}^{-}, \mathrm{OCl}_{2}, \mathrm{ClF}_{3},\) or \(\mathrm{SF}_{4} ?\)
View solution Problem 24
The trichloride ion, \(\mathrm{Cl}_{3}^{-1}\), is not very stable in aqueous solution. The equilibrium constant for the following dissociation reaction is 5.5 a
View solution Problem 25
Each of the following compounds decomposes to produce \(\mathrm{O}_{2}(\mathrm{g})\) when heated: \((\mathrm{a}) \mathrm{HgO}(\mathrm{s}) ;\) (b) \(\mathrm{KClO
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