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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 \(n=3 ?\)

Short Answer

Expert verified
For the given standard cell potential of +0.177 V at 298 K, the equilibrium constant (K) can be calculated using the Nernst Equation for different values of n. When n=1, K ≈ 1.96×10^(-9); when n=2, K ≈ 3.84×10^(-18); and when n=3, K ≈ 7.61×10^(-27).

Step by step solution

01

Write the Nernst Equation and relationship between E and K

Use the Nernst equation to relate the cell potential, number of electrons transferred, and the reaction quotient to the equilibrium constant: E° = - (\(R * T\)) / (\(n * F\)) * ln K Where E° is the standard cell potential in volts (V), R is the gas constant equal to \(8.314\frac{J}{K\cdot mol}\), T is the absolute temperature in Kelvin (K), n is the number of moles of electrons transferred, F is the Faraday's constant equal to \(9.6485 * 10^4\frac{C}{mol}\), and ln K is the natural logarithm of the equilibrium constant.
02

Solve for ln K

Rearrange the equation to solve for ln K: ln K = - (\(n * F\)) / (\(R * T\)) * E°
03

Calculate K for n=1

Plug in the given values of E°=0.177 V, T=298 K, and n=1 into the equation and calculate K: ln K = - (\(1 * (9.6485 * 10^4\frac{C}{mol})\)) / (\(8.314\frac{J}{K\cdot mol} * 298K\)) * (0.177 V) ln K = -20.099 K = e^(-20.099) K ≈ 1.96×10^(-9) Thus, when n=1, the equilibrium constant (K) is approximately 1.96×10^(-9).
04

Calculate K for n=2

Plug in the given values of E°=0.177 V, T=298 K, and n=2 into the equation and calculate K: ln K = - (\(2 * (9.6485 * 10^4\frac{C}{mol})\)) / (\(8.314\frac{J}{K\cdot mol} * 298K\)) * (0.177 V) ln K = -40.198 K = e^(-40.198) K ≈ 3.84×10^(-18) Thus, when n=2, the equilibrium constant (K) is approximately 3.84×10^(-18).
05

Calculate K for n=3

Plug in the given values of E°=0.177 V, T=298 K, and n=3 into the equation and calculate K: ln K = - (\(3 * (9.6485 * 10^4\frac{C}{mol})\)) / (\(8.314\frac{J}{K\cdot mol} * 298K\)) * (0.177 V) ln K = -60.297 K = e^(-60.297) K ≈ 7.61×10^(-27) Thus, when n=3, the equilibrium constant (K) is approximately 7.61×10^(-27).

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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 essential in electrochemistry. It helps us understand how a chemical reaction behaves under different conditions. The equation relates the standard cell potential, number of electrons transferred, temperature, and reaction quotient to the cell potential.
It is represented by the formula: \[ E = E^0 - \left( \frac{RT}{nF} \right) \ln Q \] Here:
  • \( E \) is the cell potential under non-standard conditions.
  • \( E^0 \) is the standard cell potential.
  • \( R \) is the universal gas constant \( 8.314 \frac{J}{K \cdot mol} \).
  • \( T \) is the temperature in Kelvin.
  • \( n \) represents the moles of electrons transferred in the reaction.
  • \( F \) is Faraday's constant \( 9.6485 \times 10^4 \frac{C}{mol} \).
  • \( Q \) is the reaction quotient.
Using the Nernst equation, we can predict how a change in concentration, pressure, or temperature can affect the cell potential. It also allows us to calculate the equilibrium constant \( K \) when the reaction is at equilibrium, where \( Q = K \). This makes it a powerful tool in electrochemical calculations.
equilibrium constant
The equilibrium constant, denoted as \( K \), is a crucial concept in chemistry, indicating the ratio of concentrations of products to reactants at equilibrium. A high value of \( K \) suggests that the reaction favors the formation of products, while a low \( K \) value indicates that reactants are favored.

In electrochemistry, \( K \) can be determined using the Nernst equation once the standard cell potential \( E^0 \) is known. The relationship can be expressed as:\[ ln K = \left( \frac{nF}{RT} \right) E^0 \]Where:
  • \( n \) is the number of moles of electrons transferred.
  • \( F \) is Faraday's constant.
  • \( R \) is the gas constant.
  • \( T \) is the temperature in Kelvin.
  • \( E^0 \) is the standard cell potential.
At equilibrium, the cell potential \( E \) is zero, and this relationship provides a direct way to calculate \( K \), thus linking thermodynamics and electrochemistry efficiently.
Faraday's constant
Faraday's constant is an essential quantity in electrochemistry. It represents the charge of one mole of electrons. Its value is approximately \( 9.6485 \times 10^4 \frac{C}{mol} \). This constant is named after Michael Faraday, a significant figure in the field of electrochemistry.

Faraday's constant is instrumental when calculating the amount of charge needed to carry out an electrochemical reaction. It allows chemists to connect the amount of substance converted in a reaction to the electrical charge passed through the system.For example:
  • If you have one mole of electrons, the total charge is \( 9.6485 \times 10^4C \).
  • This linkage between moles and charge is critical when using the Nernst equation, as it involves calculating the total change in Gibbs free energy for a reaction.
Thus, the constant is a core component of all electrochemical calculations, providing a bridge between the fields of chemistry and physics.
electrochemistry
Electrochemistry is a branch of chemistry focused on the chemical changes caused by the movement of electrons. It involves the study of reactions in cells where electrons are transferred between substances, generating an electric current. This field has wide applications, including batteries, fuel cells, and electrolysis.

A few key concepts in electrochemistry include:
  • Redox reactions, where oxidation and reduction occur, are central to the study of electrochemistry.
  • Electrochemical cells, like Galvanic and electrolytic cells, help us understand how chemical energy is converted to electrical energy and vice versa.
  • The standard cell potential \( E^0 \) helps predict the direction of electron flow in a cell.
Understanding electrochemistry provides insight into how energy is stored and released, and how different substances can be transformed through electron exchange. It's a dynamic and continually evolving field that has greatly impacted technological advancements and everyday life.

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Most popular questions from this chapter

From each of the following pairs of substances, use data in Appendix E to choose the one that is the stronger reducing agent: $$ \begin{array}{l}{\text { (a) } \mathrm{Fe}(s) \text { or } \mathrm{Mg}(s)} \\\ {\text { (b) } \mathrm{Ca}(s) \text { or } \mathrm{Al}(s)} \\ {\text { (c) } \mathrm{H}_{2}\left(g, \text { acidic solution ) or } \mathrm{H}_{2} \mathrm{S}(g)\right.} \\ {\text { (d) } \mathrm{BrO}_{3}^{-}(a q) \text { or } \mathrm{IO}_{3}^{-}(a q)}\end{array} $$

Using standard reduction potentials (Appendix E), calculate the standard emf for each of the following reactions: $$ \begin{array}{l}{\text { (a) } \mathrm{Cl}_{2}(g)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{Cl}^{-}(a q)+\mathrm{I}_{2}(s)} \\ {\text { (b) } \mathrm{Ni}(s)+2 \mathrm{Ce}^{4+}(a q) \longrightarrow \mathrm{Ni}^{2+}(a q)+2 \mathrm{Ce}^{3+}(a q)} \\ {\text { (c) } \mathrm{Fe}(s)+2 \mathrm{Fe}^{3+}(a q) \longrightarrow 3 \mathrm{Fe}^{2+}(a q)} \\ {\text { (d) } 2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+3 \mathrm{Cu}(s) \longrightarrow 2 \mathrm{NO}(g)+} \\ \quad {4 \mathrm{H}_{2} \mathrm{O}(l)+3 \mathrm{Cu}^{2+}(a q)}\end{array} $$

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 .

Indicate whether each statement is true or false: (a) The cathode is the electrode at which oxidation takes place. (b) A galvanic cell is another name for a voltaic cell. (c) Electrons flow spontaneously from anode to cathode in a voltaic cell.

A voltaic cell utilizes the following reaction: $$ 2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}^{+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf for this cell when \(\left[\mathrm{Fe}^{3+}\right]=3.50 M, P_{\mathrm{H}_{2}}=\) \(0.95 \mathrm{atm},\left[\mathrm{Fe}^{2+}\right]=0.0010 M,\) and the \(\mathrm{pH}\) in both half-cells is 4.00\(?\)

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