The solubility product for \(\operatorname{Cul}(s)\) is \(1.1 \times 10^{-12}\). Calculate the value of \(\mathscr{E}^{\circ}\) for the half-reaction $$\mathrm{CuI}+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}+\mathrm{I}^{-}$$

Short Answer

Expert verified
The standard cell potential for the given half-reaction \( CuI + e^{-} \rightarrow Cu + I^{-}\) is approximately 0.465 V.

Step by step solution

01

Write the dissociation reaction

Write the dissociation reaction of CuI(s) in aqueous solution: \[ CuI(s) \rightleftharpoons Cu^{+}(aq) + I^{-}(aq) \]
02

Set up solubility product expression and equilibrium concentrations

Set up the solubility product expression for the given value of \(K_{sp} = 1.1 \times 10^{-12}\) and the equilibrium concentrations for the ions: \[ K_{sp} = [Cu^{+}][I{-}] \] Since CuI dissociates in a 1:1 ratio, let the concentration of both ions be x. \[ 1.1 \times 10^{-12} = x^2 \]
03

Calculate equilibrium concentration x

Solve the equation for the equilibrium concentration x: \[ x = \sqrt{1.1 \times 10^{-12}} \] \[ x \approx 1.05 \times 10^{-6} \]
04

Write the Nernst equation for the half-reaction

Write the Nernst equation for the given half-reaction: \[ \mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF} \ln Q \] Here, \(R\) is the gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of electrons transferred, \(F\) is the Faraday constant, and \(Q\) is the reaction quotient. For this half-reaction, \(n = 1\). We'll use the standard conditions: \(T = 298 K\), \(R = 8.314 J/(mol\cdot K)\), and \(F = 96485 C/mol\).
05

Plug in equilibrium concentrations to find Q

Find the reaction quotient Q using the equilibrium concentrations from Step 3: \[Q = \frac{[Cu][I^{-}]}{[CuI][e^{-}]} = \frac{[Cu^{+}][I^{-}]}{[CuI][1]}\] Since we are interested in the standard cell potential, we have to assume that [CuI] = 1 \[ Q = [Cu^{+}][I^{-}] = (1.05 \times 10^{-6})(1.05 \times 10^{-6}) = 1.1 \times 10^{-12} \]
06

Plug values into Nernst equation and solve for the standard cell potential

Substitute the calculated values into the Nernst equation and solve for \(\mathscr{E}^{\circ}\): \[ \mathscr{E} = \mathscr{E}^{\circ} - \frac{8.314 \cdot 298}{1 \cdot 96485} \ln (1.1 \times 10^{-12}) \] Since the reaction is at standard conditions, \(\mathscr{E}\) will be equal to \(\mathscr{E}^{\circ}\). Hence, \[ \mathscr{E}^{\circ} =\frac{8.314 \cdot 298}{96485} \ln (1.1 \times 10^{-12})^{-1} \] Calculate the answer: \[ \mathscr{E}^{\circ} \approx 0.465 V \] So the standard cell potential for the given half-reaction is approximately 0.465 V.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solubility Product Constant
The solubility product constant, or Ksp, is fundamental in predicting the solubility of a compound in water. It quantifies the extent to which a solid dissolves in an aqueous solution to form its constituent ions. The constant is specific to each substance and temperature-dependent.

For a generic salt AB that dissociates into A+ and B- ions, the Ksp is written as: \[ K_{sp} = [A^{+}][B^{-}] \].
In the case of copper(I) iodide (CuI), the Ksp value is provided as \(1.1 \times 10^{-12}\). This is a very low value, indicating that CuI is only slightly soluble in water. In an equilibrium scenario, the concentrations of the ions Cu+ and I- are equal due to the 1:1 stoichiometry of their dissociation. The equilibrium concentrations are critical for calculating the solubility of the salt, as well as in applying the Nernst equation to determine the standard cell potential.
The Nernst Equation
The Nernst equation is a key tool in electrochemistry, enabling us to calculate the electric potential of a half-cell or full electrochemical cell under non-standard conditions. It accounts for the temperature, ion concentration (or partial pressures of gases), and other relevant factors affecting the cell's potential.

The equation is given by: \[ \mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF} \ln Q \],where \(\mathscr{E}\) is the cell potential, \(\mathscr{E}^{\circ}\) is the standard cell potential, R is the universal gas constant, T is the temperature in Kelvin, n is the number of moles of electrons exchanged, F is the Faraday constant, and Q is the reaction quotient, representing the ratio of products over reactants.
Applying the Nernst equation requires knowledge of the equilibrium concentrations of ions. In the example provided, this equation allowed us to connect the physical chemistry of solubility with the electrical chemistry involved in the half-cell potential.
Equilibrium Concentration
When a salt is in a dynamic equilibrium in an aqueous solution, the concentrations of the ions produced by its dissociation are constant, and these are the equilibrium concentrations. They are crucial for predicting the outcome of reactions, calculating solubility, and applying the Nernst equation.

In the case of the dissolution of CuI in water, the equilibrium concentrations were found by setting the solubility product equal to the product of the individual ions' concentrations, each designated as x. By solving the expression \(K_{sp} = x^2\), we obtained the equilibrium concentration of each ion in solution. Here, the concentration appears squared because of the stoichiometry of the reaction where one mole of CuI produces one mole of Cu+ and one mole of I- ions.

Calculating equilibrium concentrations is often the first step in establishing reaction quotients or determining the direction in which a reaction will proceed, further underlying its significance in the analytical aspects of chemistry.
Half-Reaction Calculation
Half-reactions split an overall redox reaction into its two complementary parts: oxidation and reduction. Each half-reaction involves the transfer of electrons, and understanding the electrochemistry of each half-reaction is essential for calculating cell potentials and balancing redox reactions.

The given half-reaction \(\text{CuI} + \text{e}^{-} \longrightarrow \text{Cu} + \text{I}^{-}\) shows the reduction of copper(I) iodide to copper metal and iodide ion. In this example, calculating the standard cell potential begins with understanding the dissociation of CuI and utilizing the Nernst equation. The steps include balancing the half-reaction, calculating equilibrium concentrations, and finally finding the standard cell potential which indicates the tendency of the reaction to proceed towards the reduction of CuI.

This connection of half-reaction calculations with the solubility product and Nernst equation amplifies their role in a wide array of chemical analyses, from electroplating to energy storage in batteries.

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Most popular questions from this chapter

The free energy change for a reaction, \(\Delta G\), is an extensive property. What is an extensive property? Surprisingly, one can calculate \(\Delta G\) from the cell potential, \(\mathscr{b}\), for the reaction. This is surprising because \(\mathscr{B}\) is an intensive property. How can the extensive property \(\Delta G\) be calculated from the intensive property \(\mathscr{E}\) ?

Zirconium is one of the few metals that retains its structural integrity upon exposure to radiation. For this reason, the fuel rods in most nuclear reactors are made of zirconium. Answer the following questions about the redox properties of zirconium based on the half-reaction \(\mathrm{ZrO}_{2} \cdot \mathrm{H}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}+4 \mathrm{e}^{-} \longrightarrow \mathrm{Zr}+4 \mathrm{OH}^{-} \quad \mathscr{E}^{\circ}=-2.36 \mathrm{~V}\) a. Is zirconium metal capable of reducing water to form hydrogen gas at standard conditions? b. Write a balanced equation for the reduction of water by zirconium metal. c. Calculate \(\mathscr{8}^{\circ}, \Delta G^{\circ}\), and \(K\) for the reduction of water by zirconium metal. d. The reduction of water by zirconium occurred during the accident at Three Mile Island, Pennsylvania, in \(1979 .\) The hydrogen produced was successfully vented and no chemical explosion occurred. If \(1.00 \times 10^{3} \mathrm{~kg} \mathrm{Zr}\) reacts, what mass of \(\mathrm{H}_{2}\) is produced? What volume of \(\mathrm{H}_{2}\) at \(1.0 \mathrm{~atm}\) and \(1000 .{ }^{\circ} \mathrm{C}\) is produced? e. At Chernobyl, USSR, in 1986 , hydrogen was produced by the reaction of superheated steam with the graphite reactor core: $$\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g)$$ A chemical explosion involving the hydrogen gas did occur at Chernobyl. In light of this fact, do you think it was a correct decision to vent the hydrogen and other radioactive gases into the atmosphere at Three Mile Island? Explain.

In 1973 the wreckage of the Civil War ironclad USS Monitor was discovered near Cape Hatteras, North Carolina. [The Monitor and the CSS Virginia (formerly the USS Merrimack) fought the first battle between iron-armored ships.] In 1987 investigations were begun to see if the ship could be salvaged. It was reported in Time (June 22,1987 ) that scientists were considering adding sacrificial anodes of zinc to the rapidly corroding metal hull of the Monitor. Describe how attaching zinc to the hull would protect the Monitor from further corrosion.

Gold metal will not dissolve in either concentrated nitric acid on concentrated hydrochloric acid. It will dissolve, however, in aqua regia, a mixture of the two concentrated acids. The products of the reaction are the \(\mathrm{AuCl}_{4}^{-}\) ion and gaseous NO. Write a balanced equation for the dissolution of gold in aqua regia.

Which of the following statements concerning corrosion is/are true? For the false statements, correct them. a. Corrosion is an example of an electrolytic process. b. Corrosion of steel involves the reduction of iron coupled with the oxidation of oxygen. c. Steel rusts more easily in the dry (arid) Southwest states than in the humid Midwest states. d. Salting roads in the winter has the added benefit of hindering the corrosion of steel. e. The key to cathodic protection is to connect via a wire a metal more easily oxidized than iron to the steel surface to be protected.

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