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Chapter 22: Problem 23

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

Expert verified
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

01

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°.
02

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 \(I_{3}^{-}\).
03

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.

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Most popular questions from this chapter

To displace \(\mathrm{Br}_{2}\) from an aqueous solution of \(\mathrm{Br}^{-}\) add \((\mathrm{a}) \mathrm{I}_{2}(\mathrm{aq}) ;\) (b) \(\mathrm{Cl}_{2}(\mathrm{aq}) ;\) (c) \(\mathrm{H}_{2}(\mathrm{g}) ;\) (d) \(\mathrm{Cl}^{-}(\mathrm{aq})\) (e) \(\mathrm{I}_{3}^{-}(\mathrm{aq})\).

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