/*! 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 55 Using the standard reduction pot... [FREE SOLUTION] | 91影视

91影视

Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at \(298 \mathrm{K} :\) $$ \begin{array}{l}{\text { (a) } \mathrm{Fe}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{Ni}(s)} \\ {\text { (b) } \mathrm{Co}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Co}^{2+}(a q)+\mathrm{H}_{2}(g)} \\ {\text { (c) } 10 \mathrm{Br}^{-}(a q)+2 \mathrm{MnO}_{4}^{-}(a q)+16 \mathrm{H}^{+}(a q) \rightarrow} \\ {\quad 2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l)+5 \mathrm{Br}_{2}(l)}\end{array} $$

Short Answer

Expert verified
The equilibrium constants for the given reactions at 298 K are: a) Fe(s) + Ni虏鈦(aq) 鈫 Fe虏鈦(aq) + Ni(s): \(K \approx 5.17 \times 10^{-15}\) b) Co(s) + 2 H鈦(aq) 鈫 Co虏鈦(aq) + H鈧(g): \(K \approx 2.57 \times 10^{-5}\) c) 10 Br鈦(aq) + 2 MnO鈧勨伝(aq) + 16 H鈦(aq) 鈫 2 Mn虏鈦(aq) + 8 H鈧侽(l) + 5 Br鈧(l): \(K \approx 2.73 \times 10^{-5}\)

Step by step solution

01

Reaction (a): Fe(s) + Ni虏鈦(aq) 鈫 Fe虏鈦(aq) + Ni(s)

First, let's identify the half-reactions and their standard reduction potentials from Appendix E. The half-reactions and their potentials will be: - Fe虏鈦(aq) + 2e鈦 鈫 Fe(s), E掳 = -0.44 V - Ni虏鈦(aq) + 2e鈦 鈫 Ni(s), E掳 = -0.25 V Step 1: Calculate the standard cell potential The given reaction is essentially the reverse of the Fe half-reaction, making that half-reaction an oxidation process. Therefore, the standard reduction potentials should be calculated by reversing the sign of the Fe half-reaction and summing both potentials: \[E^0 = (-0.44) + (-0.25) = -0.69\,V\] Step 2: Calculate the equilibrium constant Now, we can use the Nernst equation to calculate K: \[E^0 = \frac{RT}{nF}\ln{K}\] Using the values at 298 K: R = 8.314 J/(mol路K), T = 298 K, n = 2 (two electrons are exchanged in both half-reactions), and F = 96485 C/mol, we get: \[-0.69 = \frac{(8.314)(298)}{(2)(96485)}\ln{K}\] Solve for K: \[\ln{K} = -32.96\] \[K = e^{-32.96} \approx 5.17 \times 10^{-15}\]
02

Reaction (b): Co(s) + 2 H鈦(aq) 鈫 Co虏鈦(aq) + H鈧(g)

First, let's identify the half-reactions and their standard reduction potentials from Appendix E. The half-reactions and their potentials will be: - Co虏鈦(aq) + 2e鈦 鈫 Co(s), E掳 = -0.28 V - 2H鈦(aq) + 2e鈦 鈫 H鈧(g), E掳 = 0.00 V Step 1: Calculate the standard cell potential The given reaction is essentially the reverse of the Co half-reaction, making that half-reaction an oxidation process. Therefore, the standard reduction potentials should be calculated by reversing the sign of the Co half-reaction and summing both potentials: \[E^0 = (-0.28) + (0.00) = -0.28\,V\] Step 2: Calculate the equilibrium constant Now, using the Nernst equation to calculate K: \[E^0 = \frac{RT}{nF}\ln{K}\] Using the values at 298 K: R = 8.314 J/(mol路K), T = 298 K, n = 2 (two electrons are exchanged in both half-reactions), and F = 96485 C/mol, we get: \[-0.28 = \frac{(8.314)(298)}{(2)(96485)}\ln{K}\] Solve for K: \[\ln{K} = -10.56\] \[K = e^{-10.56} \approx 2.57 \times 10^{-5}\]
03

Reaction (c): 10 Br鈦(aq) + 2 MnO鈧勨伝(aq) + 16 H鈦(aq) 鈫 2 Mn虏鈦(aq) + 8 H鈧侽(l) + 5 Br鈧(l)

First, let's identify the half-reactions and their standard reduction potentials from Appendix E. The half-reactions and their potentials will be: - 2MnO鈧勨伝(aq) + 8H鈧侽(l) + 10e鈦 鈫 2Mn虏鈦(aq) + 16H鈦(aq) + (-1.51) V - 5Br鈧(l) + 10e鈦 鈫 10Br鈦(aq), E掳 = 1.07 V Step 1: Calculate the standard cell potential For this reaction, we simply have to sum both potentials: \[E^0 = (-1.51) + (1.07) = -0.44\,V\] Step 2: Calculate the equilibrium constant Now, using the Nernst equation to calculate K: \[E^0 = \frac{RT}{nF}\ln{K}\] Using the values at 298 K: R = 8.314 J/(mol路K), T = 298 K, n = 10 (ten electrons are exchanged in both half-reactions), and F = 96485 C/mol, we get: \[-0.44 = \frac{(8.314)(298)}{(10)(96485)}\ln{K}\] Solve for K: \[\ln{K} = -10.52\] \[K = e^{-10.52} \approx 2.73 \times 10^{-5}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Equilibrium Constant Calculation
Understanding how to calculate an equilibrium constant (K) for a chemical reaction is critical in the field of chemistry, particularly when studying reactions in an electrochemical context. The value of K provides insight into the extent to which a reaction will proceed under standard conditions.

In the case of redox reactions, the equilibrium constant can be calculated using cell potentials and the Nernst equation. As shown in the textbook solution, after obtaining the standard reduction potentials for the half-reactions, we calculate the standard cell potential (\(E^0\)) which is the sum of the potentials for each half-reaction involved in the overall reaction.

However, when a half-reaction is reversed (as in oxidation), its potential's sign is also reversed. Once we have our standard cell potential, we can use the equation derived from the Nernst equation for the reaction at standard conditions:\[\begin{equation}E^0 = \frac{RT}{nF}\ln{K}\end{equation}\]where R is the ideal gas constant, T is temperature in Kelvin, n is the number of moles of electrons exchanged, and F is the Faraday constant. The calculated value of \(K\) gives us the equilibrium constant, indicating how far the reaction goes towards products or reactants.
Nernst Equation
The Nernst equation is a fundamental equation in electrochemistry, providing a connection between the electrochemical potential of a cell and the concentrations of reactants and products. It is expressed as:\[\begin{equation}E = E^0 - \frac{RT}{nF}\ln{Q}\end{equation}\]where \(E\) is the cell potential, \(E^0\) is the standard cell potential, \(R\) is the ideal 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.

At equilibrium, \(E\) becomes zero, leading to the simplified equation used for equilibrium constant calculation, as \(Q\) becomes \(K\). The textbook solutions highlight the application of the Nernst equation in obtaining the equilibrium constant. In simple terms, the Nernst equation adjusts the standard cell potential to reflect the actual conditions of a reaction taking place at concentrations different from standard conditions.
Electrochemistry
Electrochemistry is the branch of chemistry that deals with the relationship between electrical energy and chemical changes. It's an interdisciplinary realm that crosses over into physics and materials science, underpinning technologies such as batteries, fuel cells, and electroplating.

The standard reduction potentials of various half-reactions are cornerstone elements in electrochemistry. These potentials, measured in volts (V), indicate a substance's tendency to gain electrons and be reduced. They are determined under standard conditions (1 M concentration, 1 atm pressure, and 298 K), and provide a scale against which all redox reactions can be compared.

In the given solutions, the reactions' equilibrium constants were calculated using standard reduction potentials, highlighting the intricacies of electrochemical reactions. These constants reflect the thermodynamic feasibility of reactions, and emphasizing their concept enhances understanding in fundamental electrochemistry.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A cell has a standard cell potential of \(+0.177 \mathrm{V}\) at 298 \(\mathrm{K}\) . What is the value of the equilibrium constant for the reaction ( a ) if \(n=1 ?(\mathbf{b})\) if \(n=2 ?(\mathbf{c})\) if \(n=3 ?\)

Magnesium is obtained by electrolysis of molten \(\mathrm{MgCl}_{2}\) . (a) Why is an aqueous solution of MgCl_ not used in the electrolysis? (b) Several cells are connected in parallel by very large copper bars that convey current to the cells. Assuming that the cells are 96\(\%\) efficient in producing the desired products in electrolysis, what mass of Mg is formed by passing a current of \(97,000\) A for a period of 24 \(\mathrm{h} ?\)

A 1 M solution of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) is placed in a beaker with a strip of Cu metal. A 1 \(\mathrm{M}\) solution of \(\mathrm{SnSO}_{4}\) is placed in a second beaker with a strip of Sn metal. A salt bridge connects the two beakers, and wires to a voltmeter link two metal electrodes. (a) Which electrode serves as the anode, and which as the cathode? (b) Which electrode gains mass, and which loses mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?

(a) How many coulombs are required to plate a layer of chromium metal 0.25 \(\mathrm{mm}\) thick on an auto bumper with a total area of 0.32 \(\mathrm{m}^{2}\) from a solution containing \(\mathrm{CrO}_{4}^{2-}\) ? The density of chromium metal is 7.20 \(\mathrm{g} / \mathrm{cm}^{3} .\) (b) What current flow is required for this electroplating if the bumper is to be plated in 10.0 s? (c) If the external source has an emf of \(+6.0 \mathrm{V}\) and the electrolytic cell is 65\(\%\) efficient, how much electrical power is expended to electroplate the bumper?

Iron corrodes to produce rust, \(\mathrm{Fe}_{2} \mathrm{O}_{3},\) but other corrosion products that can form are \(\mathrm{Fe}(\mathrm{O})(\mathrm{OH}),\) iron oxyhydroxide, and magnetite, \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) . (a) What is the oxidation number of Fe in iron oxyhydroxide, assuming oxygen's oxidation number is \(-2 ?\) (b) The oxidation number for Fe in magnetite was controversial for a long time. If we assume that oxygen's oxidation number is \(-2,\) and Fe has a unique oxidation number, what is the oxidation number for Fe in magnetite? (c) It turns out that there are two different kinds of Fe in magnetite that have different oxidation numbers. Suggest what these oxidation numbers are and what their relative stoichiometry must be, assuming oxygen's oxidation number is - 2 .

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.