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

Short Answer

Expert verified
The equilibrium constants for the given reactions at 298 K are as follows: (a) K 鈮 \(10^6\) (b) K 鈮 \(10^9\) (c) K 鈮 \(10^{75}\)

Step by step solution

01

Determine the standard reduction potentials

From Appendix E2, we find the standard reduction potentials for Fe虏鈦 and Ni虏鈦: Fe虏鈦 + 2e鈦 鈫 Fe E掳(Fe虏鈦/Fe) = -0.44 V Ni虏鈦 + 2e鈦 鈫 Ni E掳(Ni虏鈦/Ni) = -0.26 V
02

Calculate the standard cell potential (E掳)

Since Fe is being oxidized and Ni虏鈦 is being reduced, we have: E掳 = E掳(Ni虏鈦/Ni) - E掳(Fe虏鈦/Fe) = -0.26 - (-0.44) = 0.18 V
03

Calculate the equilibrium constant (K)

Using the modified Nernst equation, we find the equilibrium constant for reaction (a): K = 10^\frac{nE掳}{0.0592} = 10^\frac{2(0.18)}{0.0592} = 10^6 K 鈮 10^6 (b) Co(s) + 2 H鈦(aq) 鈫 Co虏鈦(aq) + H鈧(g)
04

Determine the standard reduction potentials

From Appendix E2: Co虏鈦 + 2e鈦 鈫 Co E掳(Co虏鈦/Co) = -0.28 V 2H鈦 + 2e鈦 鈫 H鈧 E掳(H鈧/H鈦) = 0 V
05

Calculate the standard cell potential (E掳)

Since Co is being oxidized and H鈦 is being reduced, we have: E掳 = E掳(H鈧/H鈦) - E掳(Co虏鈦/Co) = 0 - (-0.28) = 0.28 V
06

Calculate the equilibrium constant (K)

Using the modified Nernst equation for reaction (b): K = 10^\frac{nE掳}{0.0592} = 10^\frac{2(0.28)}{0.0592} 鈮 10^9 K 鈮 10^9 (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)
07

Determine the standard reduction potentials

From Appendix E2: 2MnO鈧勨伝 + 16H鈦 + 10e鈦 鈫 2 Mn虏鈦 + 8 H鈧侽 E掳(MnO鈧勨伝/Mn虏鈦) = 1.51 V 2Br鈦 鈫 Br鈧 + 2e鈦 E掳(Br鈧/Br鈦) = 1.07 V
08

Calculate the standard cell potential (E掳)

Since Br鈦 is being oxidized and MnO鈧勨伝 is being reduced, we have: E掳 = E掳(MnO鈧勨伝/Mn虏鈦) - E掳(Br鈧/Br鈦) = 1.51 - 1.07 = 0.44 V
09

Calculate the equilibrium constant (K)

Using the modified Nernst equation for reaction (c): K = 10^\frac{nE掳}{0.0592} = 10^\frac{10(0.44)}{0.0592} 鈮 10^75 K 鈮 10^75 So, the equilibrium constant for each reaction at 298 K is: (a) K 鈮 10^6 (b) K 鈮 10^9 (c) K 鈮 10^75

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Key Concepts

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

Standard Reduction Potential
The standard reduction potential, often symbolized as E掳, is a measure of the tendency of a chemical species to acquire electrons and thereby be reduced. It is a fundamental concept in electrochemistry, indicating how easily an atom or molecule gains electrons. In standard conditions, which mean a temperature of 298 K, a 1M concentration, and a pressure of 1 atm, the potential is measured in volts (V).

Each half-cell in an electrochemical reaction has its own standard reduction potential. The more positive the standard reduction potential, the higher the affinity for electrons, or the greater the tendency to be reduced. For example, in the provided exercise, we see that Fe2+ has an E掳 value of -0.44 V, whereas Ni2+ has -0.26 V, indicating that nickel is more readily reduced than iron in their respective reactions.

To calculate the potential for the full cell reaction, we take the difference between the E掳 values of the cathode (where reduction takes place) and the anode (where oxidation occurs). This gives us the standard cell potential (E掳 cell), which is positive if the reaction is spontaneous, leading us to the relevance of such values in determining the direction of electron flow in electrochemical cells.

Nernst Equation
The Nernst equation is a key mathematical expression in electrochemistry that enables us to predict the cell potential under non-standard conditions. It accounts for the effect of temperature, concentration, and gas pressure on the potential of an electrochemical cell.

Formula

The generic form of the Nernst equation is:
\[ E = E掳 - \frac{RT}{nF} ln\frac{[Red]}{[Ox]} \]
In this equation, E is the cell potential, E掳 is the standard cell potential, R is the gas constant, T is the temperature in kelvin, n is the number of moles of electrons transferred in the half-reaction, F is the Faraday constant, and [Red] and [Ox] are the concentrations of the reduced and oxidized species, respectively.

For the equilibrium constant calculation, the modified Nernst equation is used because the concentrations of species are assumed to be at equilibrium. Thus, it simplifies to:
\[ E = \frac{0.0592}{n} \log K \]
Where K is the equilibrium constant of the reaction. This modified Nernst equation is what's utilized in our textbook solution to connect the standard cell potentials to the equilibrium constants (K) of the reactions in question.

Electrochemistry
Electrochemistry is the branch of chemistry that deals with the interconversion of chemical energy and electrical energy. This field encompasses a variety of phenomena such as the behavior of electrolytes in solutions, electrochemical cell design, and the kinetics of electrochemical processes.

Central to electrochemistry are redox (reduction-oxidation) reactions, where electrons are transferred between species, resulting in changes in oxidation states. This transfer of electrons is harnessed in various applications, from batteries and fuel cells to metal plating and corrosion prevention.

Understanding the concepts such as standard reduction potentials and the Nernst equation is crucial for mastering electrochemistry, as they allow us to predict the feasibility and direction of redox reactions. Moreover, they help in calculating important parameters such as cell potential and equilibrium constants, which are necessary to design and analyze electrochemical cells efficiently.

Galvanic Cells
A galvanic cell, also known as a voltaic cell, is a type of electrochemical cell that generates electrical energy from spontaneous redox reactions. It consists of two different metals connected by a salt bridge or a porous partition, within which occurs the flow of ions.

Components and Functioning

Galvanic cells have two electrodes: an anode (where oxidation occurs) and a cathode (where reduction takes place). The flow of electrons from the anode to the cathode through an external circuit is what generates electric current.

The standard reduction potentials of the electrode materials are pivotal in determining the voltage of the cell. For instance, in the exercise above, nickel and iron formed the electrodes in one of the galvanic cells. The cell potential calculated from their standard reduction potentials tells us that the Fe/Ni cell will produce 0.18 V of electrical potential under standard conditions.

It's the understanding of these components and how they work together that not only permits the calculation of the equilibrium constant for a given electrochemical reaction but also predicts how much voltage the cell can produce, which is important in practical applications like batteries.

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Most popular questions from this chapter

By using the data in Appendix \(\mathrm{E}\), determine whether each of the following substances is likely to serve as an oxidant or a reductant: (a) \(\mathrm{Cl}_{2}(g),\) (b) \(\mathrm{MnO}_{4}^{-}(a q,\) acidic solution) (c) \(\mathrm{Ba}(s),\) (d) \(\mathrm{Zn}(s)\)

Some years ago a unique proposal was made to raise the Titanic. The plan involved placing pontoons within the ship using a surface-controlled submarine-type vessel. The pontoons would contain cathodes and would be filled with hydrogen gas formed by the electrolysis of water. It has been estimated that it would require about \(7 \times 10^{8} \mathrm{~mol}\) of \(\mathrm{H}_{2}\) to provide the buoyancy to lift the ship (J. Chem. Educ, \(1973,\) Vol. 50,61 ). (a) How many coulombs of electrical charge would be required? (b) What is the minimum voltage required to generate \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) if the pressure on the gases at the depth of the wreckage ( \(2 \mathrm{mi}\) ) is \(300 \mathrm{~atm} ?\) (c) What is the minimum electrical energy required to raise the Titanic by electrolysis? (d) What is the minimum cost of the electrical energy required to generate the necessary \(\mathrm{H}_{2}\) if the electricity costs 85 cents per kilowatt-hour to generate at the site?

(a) Calculate the mass of Li formed by electrolysis of molten LiCl by a current of \(7.5 \times 10^{4}\) A flowing for a period of \(24 \mathrm{~h}\). Assume the electrolytic cell is \(85 \%\) efficient. (b) What is the minimum voltage required to drive the reaction?

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

If the equilibrium constant for a two-electron redox reaction at \(298 \mathrm{~K}\) is \(1.5 \times 10^{-4},\) calculate the corresponding \(\Delta G^{\circ}\) and \(E_{\text {red }}^{\circ}\).

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