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Chapter 20: Problem 57

A cell has a standard cell potential of \(+0.257 \mathrm{~V}\) at $298 \mathrm{~K}$. What is the value of the equilibrium constant for the reaction \((\mathbf{a})\) if \(n=1 ?(\mathbf{b})\) if \(n=2 ?(\mathbf{c})\) if \(n=3 ?\)

Short Answer

Expert verified
For the given standard cell potential of +0.257 V and temperature of 298 K: (a) For n=1, the equilibrium constant \(K ≈ 9.33 \times 10^{10}\) (b) For n=2, the equilibrium constant \(K ≈ 8.69 \times 10^{21}\) (c) For n=3, the equilibrium constant \(K ≈ 7.67 \times 10^{32}\)

Step by step solution

01

Recap the Nernst Equation

The Nernst equation relates the standard cell potential (E°), the equilibrium constant (K), temperature (T), and the number of electron transferred (n) in a redox reaction: \[E° = \frac{RT}{nF} \ln K\] where R is the universal gas constant (R = 8.314 J/mol·K) and F is the Faraday constant (F = 96,485 C/mol).
02

Calculate K for n = 1

Given E° = +0.257 V, T = 298 K, and n = 1, we can find the equilibrium constant K using the Nernst equation: \[K = e^{\frac{nFE°}{RT}}\] For n = 1: \[K = e^{\frac{1(96,485 C/mol)(0.257 V)}{(8.314 J/mol·K)(298 K)}}\] \[K = e^{25.277}\] \(K ≈ 9.33 \times 10^{10}\)
03

Calculate K for n = 2

Now we will find the equilibrium constant K for n = 2. For n = 2: \[K = e^{\frac{2(96,485 C/mol)(0.257 V)}{(8.314 J/mol·K)(298 K)}}\] \[K = e^{50.555}\] \(K ≈ 8.69 \times 10^{21}\)
04

Calculate K for n = 3

Finally, we will find the equilibrium constant K for n = 3. For n = 3: \[K = e^{\frac{3(96,485 C/mol)(0.257 V)}{(8.314 J/mol·K)(298 K)}}\] \[K = e^{75.832}\] \(K ≈ 7.67 \times 10^{32}\)
05

Summarize the Results

For the given standard cell potential of +0.257 V and temperature of 298 K: (a) For n=1, the equilibrium constant K ≈ 9.33 × 10^{10} (b) For n=2, the equilibrium constant K ≈ 8.69 × 10^{21} (c) For n=3, the equilibrium constant K ≈ 7.67 × 10^{32}

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Most popular questions from this chapter

(a) Write the reactions for the discharge and charge of a nickel-cadmium (nicad) rechargeable battery. (b) Given the following reduction potentials, calculate the standard emf of the cell: $$ \begin{aligned} \mathrm{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)+2 \mathrm{OH}^{-}(a q) & \\ E_{\mathrm{red}}^{\circ} &=-0.76 \mathrm{~V} \\ \mathrm{NiO}(\mathrm{OH})(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}^{\circ} &=+0.49 \mathrm{~V} \end{aligned} $$ (c) A typical nicad voltaic cell generates an emf of \(+1.30 \mathrm{~V}\). Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value.

Hydrogen gas has the potential for use as a clean fuel in reaction with oxygen. The relevant reaction is $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) $$ Consider two possible ways of utilizing this reaction as an electrical energy source: (i) Hydrogen and oxygen gases are combusted and used to drive a generator, much as coal is currently used in the electric power industry; (ii) hydrogen and oxygen gases are used to generate electricity directly by using fuel cells that operate at \(85^{\circ} \mathrm{C} .\) (a) Use data in Appendix \(\mathrm{C}\) to calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the reaction. We will assume that these values do not change appreciably with temperature. (b) Based on the values from part (a), what trend would you expect for the magnitude of \(\Delta G\) for the reaction as the temperature increases? (c) What is the significance of the change in the magnitude of \(\Delta G\) with temperature with respect to the utility of hydrogen as a fuel? (d) Based on the analysis here, would it be more efficient to use the combustion method or the fuel-cell method to generate electrical energy from hydrogen?

From each of the following pairs of substances, use data in Appendix \(\mathrm{E}\) to choose the one that is the stronger reducing agent: (a) \(\mathrm{Al}(s)\) or \(\mathrm{Mg}(s)\) (b) \(\mathrm{Fe}(s)\) or \(\mathrm{Ni}(s)\) (c) \(\mathrm{H}_{2}(g\), acidic solution) or \(\operatorname{Sn}(s)\) (d) \(\mathrm{I}^{-}(a q)\) or \(\mathrm{Br}^{-}(a q)\)

A mixture of copper and gold metals that is subjected to electrorefining contains tellurium as an impurity. The standard reduction potential between tellurium and its lowest common oxidation state, \(\mathrm{Te}^{4+}\), is $$ \mathrm{Te}^{4+}(a q)+4 \mathrm{e}^{-} \longrightarrow \mathrm{Te}(s) \quad E_{\mathrm{red}}^{\circ}=0.57 \mathrm{~V} $$ Given this information, describe the probable fate of tellurium impurities during electrorefining. Do the impurities fall to the bottom of the refining bath, unchanged, as copper is oxidized, or do they go into solution as ions? If they go into solution, do they plate out on the cathode?

Copper corrodes to cuprous oxide, \(\mathrm{Cu}_{2} \mathrm{O},\) or cupric oxide, \(\mathrm{CuO},\) depending on environmental conditions. (a) What is the oxidation state of copper in cuprous oxide? (b) What is the oxidation state of copper in cupric oxide? (c) Copper peroxide is another oxidation product of elemental copper. Suggest a formula for copper peroxide based on its name. (d) Copper(III) oxide is another unusual oxidation product of elemental copper. Suggest a chemical formula for copper(III) oxide.
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