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

Short Answer

Expert verified
The equilibrium constant for the cell reaction is: (a) \(K \approx 3.741\times10^2\) for \(n = 1\) (b) \(K \approx 1.401\times10^5\) for \(n = 2\) (c) \(K \approx 5.242\times10^7\) for \(n = 3\)

Step by step solution

01

Write down the Nernst equation

The Nernst equation is given by: \[\Delta G^\circ = -nFE^\circ\] where, \(\Delta G^\circ\) = standard free energy change \(n\) = number of electrons transferred \(F\) = Faraday's constant (\(9.648\times10^4\ \mathrm{C\ mol^{-1}}\)) \(E^\circ\) = standard cell emf Furthermore, we know that: \[\Delta G^\circ = -RT\ln K\] where, \(R\) = gas constant (\(8.314\ \mathrm{J\ mol^{-1} K^{-1}}\)) \(T\) = temperature \(K\) = equilibrium constant Combining these two equations, we get: \[-nFE^\circ = -RT\ln K\] This equation will be used to calculate the equilibrium constant (K) for different values of n.
02

Calculate K for n = 1

Given, \(E^\circ = 0.177\ \mathrm{V}\) and \(T = 298\ \mathrm{K}\) Plug these values and \(n=1\) in the equation obtained in Step 1: \[-(1)(9.648\times10^4\ \mathrm{C\ mol^{-1}})(0.177\ \mathrm{V}) = -(8.314\ \mathrm{J\ mol^{-1} K^{-1}})(298\ \mathrm{K})\ln K\] Solve for K: \[K = \exp{\frac{(9.648\times10^4\ \mathrm{C\ mol^{-1}})(0.177\ \mathrm{V})}{(8.314\ \mathrm{J\ mol^{-1} K^{-1}})(298\ \mathrm{K})}}\] \[K \approx 3.741\times10^2\] So, for \(n = 1\), the equilibrium constant is \(K \approx 3.741\times10^2\).
03

Calculate K for n = 2

Now, plug the given values and \(n=2\) in the equation obtained in Step 1: \[-(2)(9.648\times10^4\ \mathrm{C\ mol^{-1}})(0.177\ \mathrm{V}) = -(8.314\ \mathrm{J\ mol^{-1} K^{-1}})(298\ \mathrm{K})\ln K\] Solve for K: \[K = \exp{\frac{(9.648\times10^4\ \mathrm{C\ mol^{-1}})(0.177\ \mathrm{V})}{(2)(8.314\ \mathrm{J\ mol^{-1} K^{-1}})(298\ \mathrm{K})}}\] \[K \approx 1.401\times10^5\] So, for \(n = 2\), the equilibrium constant is \(K \approx 1.401\times10^5\).
04

Calculate K for n = 3

Finally, plug the given values and \(n=3\) in the equation obtained in Step 1: \[-(3)(9.648\times10^4\ \mathrm{C\ mol^{-1}})(0.177\ \mathrm{V}) = -(8.314\ \mathrm{J\ mol^{-1} K^{-1}})(298\ \mathrm{K})\ln K\] Solve for K: \[K = \exp{\frac{(9.648\times10^4\ \mathrm{C\ mol^{-1}})(0.177\ \mathrm{V})}{(3)(8.314\ \mathrm{J\ mol^{-1} K^{-1}})(298\ \mathrm{K})}}\] \[K \approx 5.242\times10^7\] So, for \(n = 3\), the equilibrium constant is \(K \approx 5.242\times10^7\). In conclusion, the equilibrium constant for the cell reaction is: (a) \(K \approx 3.741\times10^2\) for \(n = 1\) (b) \(K \approx 1.401\times10^5\) for \(n = 2\) (c) \(K \approx 5.242\times10^7\) for \(n = 3\)

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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 a critical tool in electrochemistry. It helps in understanding how the electromotive force (EMF) of a cell varies with the concentration of ions. The basic form of the Nernst equation is \[ E = E^\circ - \frac{RT}{nF} \ln Q \] where:
  • E is the electromotive force of the cell at non-standard conditions
  • \(E^\circ\) is the standard EMF
  • \(R\) is the gas constant
  • \(T\) is the temperature in Kelvin
  • \(n\) is the number of electrons transferred per molecule in the reaction
  • \(F\) is the Faraday's constant
  • \(Q\) is the reaction quotient
This equation shows the relationship between chemical reactions at non-standard states and their corresponding EMF. The Nernst equation is derived from combining the equations for Gibbs free energy and equilibrium constants. Understanding the Nernst equation allows a deeper understanding of the relationship between equilibrium constants and the voltage in electrochemical reactions.
It emphasizes how factors like temperature, number of electrons transferred, and ion concentrations can influence the EMF of a reaction.
Electromotive Force (EMF)
Electromotive force, often referred to as EMF, represents the voltage generated by a chemical cell or a battery when no current flows through it. It's a measure of the energy that promotes the movement of electrons through a conductor within a circuit. The standard EMF \(E^\circ\) under standard conditions (1 atm, 1 M concentration, and 25°C) provides valuable information about the inherent energy changes of an electrochemical reaction.
The standard cell potential gives insight into the reaction's favorability. A positive EMF indicates a spontaneous reaction under standard conditions, suggesting that the product formation is favorable. Conversely, a negative EMF suggests non-spontaneity under such conditions, meaning the reactants are more thermodynamically advantageous.
In practical scenarios, the Nernst equation is employed to adjust the value of EMF when conditions deviate from the standard. Such real-life applications make EMF a central player in technologies like batteries, fuel cells, and electrolysis processes.
Number of Electrons Transferred (n)
In any redox reaction taking place in an electrochemical cell, the number of electrons transferred, denoted by \(n\), is a key component.
It denotes the number of moles of electrons that pass from the reducing agent to the oxidizing agent during the reaction. This value is crucial for accurately determining the Nernst equation and calculating the cell's EMF.\(n\) directly influences the equilibrium constant of the reaction, as shown by the formula:\[ -nFE^\circ = -RT\ln K \]A larger \(n\) means more electrons are available to participate in the reaction, potentially leading to a higher equilibrium constant. Therefore, understanding the role of \(n\) helps in predicting the spontaneity of electrochemical reactions and optimizing them for various applications.
It also aids in tailoring the reactions for different industrial and scientific uses by altering the conditions and hence the value of \(n\). This directly allows scientists and engineers to evaluate the efficiency and viability of electrochemical processes.
Faraday's Constant
Faraday's constant \(F\) is one of the fundamental constants in electrochemistry. It represents the charge of one mole of electrons, approximately \(9.648 \times 10^4 \, \mathrm{C \cdot mol^{-1}}\). Faraday's constant is essential for calculating the amount of substance altered in an electrochemical reaction, as it bridges the relationship between the quantity of charge and the number of moles of electrons.In the Nernst equation, Faraday's constant aids in determining the driving force behind the reaction. It is crucial while converting between quantities of substances to the electrical charge needed for reactions. Essentially, \(F\) helps quantify how much charge is required to move a certain number of electrons in redox reactions.
This constant plays an integral role in fields like electroplating, batteries, and general electrochemical cells.
It helps to quantitatively predict how much material will be deposited in electro-deposition or consumed in an electrochemical reaction. This understanding enables researchers and engineers to design more efficient and precise electrochemical systems.

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Most popular questions from this chapter

(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 \(10.0 \mathrm{~s} ?(\mathrm{c})\) 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?

A voltaic cell utilizes the following reaction: $$ 2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2}(g) \rightarrow \rightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}^{+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf for this cell when \(\left[\mathrm{Fe}^{3+}\right]=2.50 \mathrm{M}\), \(P_{\mathrm{H}_{2}}=0.85 \mathrm{~atm},\left[\mathrm{Fe}^{2+}\right]=0.0010 M\), and the \(\mathrm{pH}\) in both compartments is \(5.00 ?\)

(a) Why is it impossible to measure the standard reduction potential of a single half-reaction? (b) Describe how the standard reduction potential of a half-reaction can be determined.

A plumber's handbook states that you should not connect a brass pipe directly to a galvanized steel pipe because electrochemical reactions between the two metals will cause corrosion. The handbook recommends you use, instead, an insulating fitting to connect them. Brass is a mixture of copper and zinc. What spontaneous redox reaction(s) might cause the corrosion? Justify your answer with standard emf calculations.

(a) What is meant by the term oxidation? (b) On which side of an oxidation half-reaction do the electrons appear? (c) What is meant by the term oxidant? (d) What is meant by the term oxidizing agent?

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