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Chapter 19: Problem 25

Under standard-state conditions, what spontaneous reaction will occur in aqueous solution among the ions \(\mathrm{Ce}^{4+}, \mathrm{Ce}^{3+}, \mathrm{Fe}^{3+},\) and \(\mathrm{Fe}^{2+}\) ? Calculate \(\Delta G^{\circ}\) and \(K_{\mathrm{c}}\) for the reaction.

Short Answer

Expert verified
Based on the reduction potentials, the spontaneous reaction is \( Ce^{4+} + Fe^{2+} → Ce^{3+} + Fe^{3+} \). We calculate the standard Gibbs free energy and equilibrium constant for this reaction using the provided formulas with the known constants and temperatures.

Step by step solution

01

Identify the Oxidizer and Reducer

From the redox table, we know that Ce⁴⁺ has a greater reduction potential than Fe³⁺. This means Ce⁴⁺ is the stronger oxidant, and it gets reduced, meaning it will gain electrons and convert to Ce³⁺. On the other hand, Fe²⁺ has a lesser reduction potential than Fe³⁺, making it a stronger reductant, which gets oxidized to Fe³⁺. The overall redox reaction becomes:\( Ce^{4+} + Fe^{2+} \rightarrow Ce^{3+} + Fe^{3+} \)
02

Determine the Gibbs Free Energy

The standard state Gibbs free energy for a reaction at constant temperature T and pressure P can be calculated using the equation:\[ \Delta G = -nFE \]In the equation, n is total moles of electrons transferred (which is 1 in this case) and F is the Faraday constant with a value of 96,500 coulombs per mole. E is the standard electrode potential, which can be calculated as the sum of the reduction potential of oxidizing agent (Ce⁴⁺) and oxidizing potential of reducing agent (Fe²⁺). Once we have E, we can plug into the above formula to get \(\Delta G^{\circ}\).
03

Calculate the Equilibrium Constant

Next, calculate the equilibrium constant, \(K_{c}\), from the standard Gibbs free energy. We use the relationship between these quantities given as:\[ \Delta G = -RT \ln K \]Here, R is the universal gas constant (8.314 J/(K mol)), and T is the temperature in Kelvin. Rearranging for K, we get\[ K = e^{(-\Delta G/RT)} \]Calculate \(K_{c}\) well from the obtained value of \(\Delta G^{\circ}\).

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Most popular questions from this chapter

Tarnished silver contains \(\mathrm{Ag}_{2} \mathrm{~S}\). The tarnish can be removed by placing silverware in an aluminum pan containing an inert electrolyte solution, such as \(\mathrm{NaCl}\). Explain the electrochemical principle for this procedure. [The standard reduction potential for the half- cell reaction \(\mathrm{Ag}_{2} \mathrm{~S}(s)+2 e^{-} \rightarrow 2 \mathrm{Ag}(s)+\mathrm{S}^{2-}(a q)\) is \(-0.71 \mathrm{~V} .]\)

The magnitudes (but not the signs) of the standard reduction potentials of two metals \(X\) and \(Y\) are $$ \begin{array}{ll} \mathrm{Y}^{2+}+2 e^{-} \longrightarrow \mathrm{Y} & \left|E^{\circ}\right|=0.34 \mathrm{~V} \\ \mathrm{X}^{2+}+2 e^{-} \longrightarrow \mathrm{X} & \left|E^{\circ}\right|=0.25 \mathrm{~V} \end{array} $$ where the \(\|\) notation denotes that only the magnitude (but not the sign) of the \(E^{\circ}\) value is shown. When the half-cells of \(X\) and \(Y\) are connected, electrons flow from \(X\) to \(Y\). When \(X\) is connected to a SHE, electrons flow from \(X\) to SHE. (a) Are the \(E^{\circ}\) values of the half- reactions positive or negative? (b) What is the standard emf of a cell made up of \(X\) and \(Y ?\)

"Galvanized iron" is steel sheet that has been coated with zinc; "tin" cans are made of steel sheet coated with tin. Discuss the functions of these coatings and the electrochemistry of the corrosion reactions that occur if an electrolyte contacts the scratched surface of a galvanized iron sheet or a tin can.

A galvanic cell is constructed by immersing a piece of copper wire in \(25.0 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{CuSO}_{4}\) solution and a zinc strip in \(25.0 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{ZnSO}_{4}\) solution. (a) Calculate the emf of the cell at \(25^{\circ} \mathrm{C}\) and predict what would happen if a small amount of concentrated \(\mathrm{NH}_{3}\) solution were added to (i) the \(\mathrm{CuSO}_{4}\) solution and (ii) the \(\mathrm{ZnSO}_{4}\) solution. Assume that the volume in each compartment remains constant at \(25.0 \mathrm{~mL}\). (b) In a separate experiment, \(25.0 \mathrm{~mL}\) of \(3.00 M \mathrm{NH}_{3}\) are added to the \(\mathrm{CuSO}_{4}\) solution. If the emf of the cell is \(0.68 \mathrm{~V},\) calculate the formation constant \(\left(K_{\mathrm{f}}\right)\) of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\).

What is the function of a salt bridge? What kind of electrolyte should be used in a salt bridge?
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