/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 101 A Zn electrode is immersed in a ... [FREE SOLUTION] | 91影视

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Chapter 23: Problem 101

A Zn electrode is immersed in a solution that is \(1.00 \mathrm{M}\) in \(\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) and \(1.00 \mathrm{M}\) in \(\mathrm{NH}_{3}\). When the cathode is a standard hydrogen electrode, the emf of the cell is found to be \(+1.04 \mathrm{~V}\). What is the formation constant for \(\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+} ?\)

Short Answer

Expert verified
The formation constant (Kf) for the [Zn(NH3)4]^2+ complex is approximately 2.92 脳 10^-9.

Step by step solution

01

Write down the half-reactions and the overall reaction

We need to identify the half-reactions and the overall cell reaction before finding the formation constant of [Zn(NH3)4]^2+. The half-reactions for this system are: (1) Zn^2+ + 2e^- 鉄� Zn (Reduction reaction at the Zn electrode) (2) 2H^+ + 2e^- 鉄� H2 (Reduction reaction at the standard hydrogen electrode, SHE) Now, we will write down the overall cell reaction by adding the two half-reactions: Zn^2+ + 2H^+ + 2e^- 鉄� Zn + H2 + 2e^- Zn^2+ + 2H^+ 鉄� Zn + H2
02

Write down the Nernst equation and calculate the concentration of Zn^2+

Now, we will use the Nernst equation to relate the cell potential (EMF) and the concentrations of the various species. The Nernst equation is: E = E掳 - (RT/nF) * ln(Q) Since the given cell potential is +1.04 V, we can find the equilibrium constant (K) using the relation between E, E掳, and K: 1.04 V = 0 V - (RT/2F) * ln(Q) Here, E掳 = 0 V is the standard cell potential, R = 8.314 J/(mol K) is the gas constant, T = 298.15 K (assuming room temperature), F = 96485 C/mol is Faraday's constant, and n = 2 is the number of electrons transferred. Now, let's find the reaction quotient (Q) for the cell reaction: Q = [Zn^2+]/[H^+]^2 Since the solution is 1.00 M in NH3, we can set up an equilibrium expression for the formation constant Kf of the complex, relating Kf, the initial concentration of NH3 and the equilibrium concentration of Zn^2+: Kf = [Zn(NH3)4]^2+ / ([Zn^2+][NH3]^4) We have found the value of Q from Nernst equation above: Q = [Zn^2+]/[H^+]^2 Therefore, Kf = [Zn(NH3)4]^2+ / ([Zn^2+][NH3]^4) = [Zn^2+]/[H^+]^2 We will use this relationship to find the formation constant Kf after finding the concentration of Zn^2+. Solving the Nernst equation for [Zn^2+], we get: 1.04 V = - (RT/2F) * ln(([Zn^2+])/[H^+]^2) Rearranging the equation to isolate [Zn^2+]: [Zn^2+] = [H^+]^2 * exp(-2F(1.04 V)/(RT)) Using R = 8.314 J/(mol K), T = 298.15 K, and F = 96485 C/mol, we find: [Zn^2+] 鈮� 2.92 脳 10^-9 M
03

Calculate the formation constant Kf

Now that we have the concentration of Zn^2+, we can calculate the formation constant Kf for the [Zn(NH3)4]^2+ complex. Using the relationship derived in step 2: Kf = [Zn(NH3)4]^2+/([Zn^2+][NH3]^4) = [Zn^2+]/[H^+]^2 Since the complex and NH3 have equal concentration 1.00 M given by the problem: Kf = [Zn^2+]/[H^+]^2 = (2.92 脳 10^-9 M) / (1.00 M)^2 Kf 鈮� 2.92 脳 10^-9 Hence, the formation constant for the [Zn(NH3)4]^2+ complex is approximately 2.92 脳 10^-9.

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Key Concepts

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

Nernst equation
The Nernst equation is fundamental in electrochemistry and helps us understand how the cell potential is affected by concentration changes. It is given by the formula: \[ E = E^掳 - \frac{RT}{nF} \ln(Q) \]Here, \(E\) is the cell potential under non-standard conditions, \(E^掳\) 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 transferred in the reaction, \(F\) is the Faraday constant, and \(Q\) is the reaction quotient. The Nernst equation demonstrates that the cell potential changes with the logarithm of the reaction quotient, \(Q\).
  • At equilibrium, when \(E = 0\), the Nernst equation relates \(E^掳\) to the equilibrium constant \(K\) through \(K = e^{\frac{nFE^掳}{RT}}\).
  • In practice, it is useful to calculate the cell potential for non-standard conditions based on known concentrations.
It helps chemists predict how galvanic cells will operate in different environments.
standard hydrogen electrode
The standard hydrogen electrode (SHE) is a reference electrode with a defined potential of 0 volts under standard conditions. It consists of a platinum electrode in contact with 1.00 M hydrogen ion solution and a room pressure hydrogen gas atmosphere (1 atm). Hydrogen ions accept electrons at the surface, reducing to form hydrogen gas:\[ 2H^+ + 2e^- \rightarrow H_2 \]
  • The SHE is used to measure the electrode potential of other half-cells, providing a universal baseline for comparison.
  • By connecting different electrodes to SHE, you can determine their standard electrode potentials (E掳 values).
Being a primary reference, the SHE plays a critical role in understanding metal reactivity and designing electrochemical cells.
chemical equilibrium
Chemical equilibrium refers to the state where the rate of the forward reaction equals the rate of the reverse reaction in a closed system. At equilibrium, concentrations of reactants and products remain constant over time, but not necessarily equal. The equilibrium constant \(K\) quantifies the ratio of product to reactant concentrations at equilibrium for a given reaction:\[ K = \frac{[Products]}{[Reactants]} \]
  • For the formation of complexes like [Zn(NH3)4]^2+, the formation constant (a type of equilibrium constant) indicates the stability of the complex.
  • The larger the formation constant, the more stable the complex.
Understanding chemical equilibrium allows predicting how systems respond to changes in concentration, pressure, or temperature.
electrochemistry
Electrochemistry is the branch of chemistry that deals with the relationship between electrical energy and chemical change. It encompasses various processes and applications:
  • Understanding galvanic cells, where spontaneous chemical reactions generate electrical energy.
  • Using electrolytic cells, where electrical energy is used to drive non-spontaneous reactions.
  • Applications such as batteries, electroplating, and corrosion analysis.
In electrochemistry, reversible chemical reactions can transfer electrons, creating a flow of current when connected in a circuit. By studying electrochemical potentials and kinetics, scientists can design efficient energy storage devices and understand processes like corrosion or metal deposition.

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Most popular questions from this chapter

The lobes of which \(d\) orbitals point directly between the ligands in (a) octahedral geometry, (b) tetrahedral geometry?

A four-coordinate complex \(\mathrm{MA}_{2} \mathrm{~B}_{2}\) is prepared and found to have two different isomers. Is it possible to determine from this information whether the complex is square planar or tetrahedral? If so, which is it?

Indicate the coordination number and the oxidation number of the metal for each of the following complexes: (a) \(\mathrm{K}_{2} \mathrm{PtCl}_{4}\) (b) \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right] \mathrm{Br}_{2}\) (c) \(\mathrm{OsO}_{4}\) (d) \(\left[\mathrm{Mn}(\mathrm{en})_{3}\right]\left(\mathrm{NO}_{3}\right)_{2}\) (e) \(\left[\mathrm{Cr}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{3}\) (f) \(\left[\mathrm{Zn}(\mathrm{bipy})_{2}\right]\left(\mathrm{ClO}_{4}\right)_{2}\)

The complexes \(\left[\mathrm{CrBr}_{6}\right]^{3-}\) and \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) are both known. (a) Draw the \(d\) -orbital energy-level diagram for octahedral \(\mathrm{Cr}(\mathrm{III})\) complexes. (b) What gives rise to the colors of these complexes? (c) Which of the two complexes would you expect to absorb light of higher energy?

Explain why the transition metals in periods 5 and 6 have nearly identical radii in each group.
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