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?

Short 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.

Step-by-step solution

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°.

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 \(I_{3}^{-}\).

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 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.

The study of iodine chemistry is crucial for understanding its role in processes like redox reactions, where iodine transitions between different oxidation states.

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.

The Nernst Equation provides insights into how concentration changes can drive electrochemical processes, making it essential in battery technology and sensors.

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.

Understanding the equilibrium constant is vital for predicting the outcome of complex reactions in both industrial chemistry and biological systems.

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.

Through understanding electrode potentials, students can better grasp how chemical reactions are harnessed in processes like energy production and corrosion control.